are quotients by equivalence relations “better” than surjections?

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This might be a load of old nonsense.



I have always had it in my head that if $f:Xto Y$ is an injection, then $f$ has some sort of "canonical factorization" as a bijection $Xto f(X)$ followed by an inclusion $f(X)subseteq Y$. Similarly if $g:Xto Y$ is a surjection, and if we define an equivalence relation on $X$ by $asim biff g(a)=g(b)$ and let $Q$ be the set of equivalence classes, then $g$ has a "canonical factorization" as a quotient $Xto Q$ followed by a bijection $Qto B$. Furthermore I'd always suspected that these two "canonical" factorizations were in some way dual to each other.



I mentioned this in passing to a room full of smart undergraduates today and one of them called me up on it afterwards, and I realised that I could not attach any real meaning to what I've just said above. I half-wondered whether subobject classifiers might have something to do with it but having looked up the definition I am not so sure that they help at all.



Are inclusions in some way better than arbitrary injections? (in my mind they've always been the "best kind of injections" somehow). Are maps to sets of equivalence classes somehow better than arbitrary quotients? I can't help thinking that there might be something in these ideas but I am not sure I have the language to express it. Maybe I'm just wrong, or maybe there's some ncatlab page somewhere which will explain to me what I'm trying to formalise here.










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  • I guess one could go all the way and say that if $f:Xto Y$ is an arbitrary map of sets then $f$ factors as a quotient by an equivalence relation $Xto X/sim$, followed by a bijection $X/simto f(X)$, followed by an inclusion $f(X)subseteq Y$. Is this a thing?
    – Kevin Buzzard
    Dec 7 at 21:23






  • 2




    How could category theory ever tell the difference between an inclusion and an arbitrary injection? Dually: how could it tell the difference between "the" quotient map by an equivalence relation and an arbitrary surjection? If you move away from the category of sets, there is some distinction amongst good and bad epi/monomorphisms. See, e.g. "effective epimorphism" and related notions.
    – Dylan Wilson
    Dec 7 at 21:41










  • from the category theory perspective, the main fact for sets and functions is that every function factors into a surjection followed by an injection. However, from the perspective of category theory, you can't distinguish inclusion maps from injections, because you are only looking at objects and morphisms: the notion of 'element' is lost. However, for your other comment: there is the notion of factorization system. The same idea holds for groups or rings too: a group homomorphism $phi: G to H$ factors into $G to G/ker phi to phi(G) to H$.
    – Jacob White
    Dec 7 at 21:47







  • 1




    A subobject of $A$ is an isomorphism class of monic arrows into $A$ ($fcolon Bto A$ and $f'colon B' to A$ are isomorphic if there is an isomorphism $gcolon Bto B'$ such that $f = f'circ g$). Dually, a quotient object of $A$ is an isomorphism class of epic arrows out of $A$. I suppose you could view the category of sets as being equipped with an extra structure: a choice of a unique representative in each subobject class (the inclusion) and each quotient object class (the "actual quotient"). Then each monic arrow factors uniquely as an isomorphism followed by an inclusion, ...
    – Alex Kruckman
    Dec 7 at 22:20






  • 3




    You may be thinking of a context that abstracts to an M-category (ncatlab.org/nlab/show/M-category). The 'tight morphisms' are the literal subset inclusions, and so have a role distinguished from arbitrary inclusions. In your setup one also seems to be asking that the class of tight morphisms are all regular monomorphisms (ncatlab.org/nlab/show/regular+monomorphism), that is, literally subsets carved out by equations (say by constructing the pullback of 1 --> 0,1 along a characteristic function by an equaliser).
    – David Roberts
    Dec 7 at 23:24














up vote
21
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This might be a load of old nonsense.



I have always had it in my head that if $f:Xto Y$ is an injection, then $f$ has some sort of "canonical factorization" as a bijection $Xto f(X)$ followed by an inclusion $f(X)subseteq Y$. Similarly if $g:Xto Y$ is a surjection, and if we define an equivalence relation on $X$ by $asim biff g(a)=g(b)$ and let $Q$ be the set of equivalence classes, then $g$ has a "canonical factorization" as a quotient $Xto Q$ followed by a bijection $Qto B$. Furthermore I'd always suspected that these two "canonical" factorizations were in some way dual to each other.



I mentioned this in passing to a room full of smart undergraduates today and one of them called me up on it afterwards, and I realised that I could not attach any real meaning to what I've just said above. I half-wondered whether subobject classifiers might have something to do with it but having looked up the definition I am not so sure that they help at all.



Are inclusions in some way better than arbitrary injections? (in my mind they've always been the "best kind of injections" somehow). Are maps to sets of equivalence classes somehow better than arbitrary quotients? I can't help thinking that there might be something in these ideas but I am not sure I have the language to express it. Maybe I'm just wrong, or maybe there's some ncatlab page somewhere which will explain to me what I'm trying to formalise here.










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  • I guess one could go all the way and say that if $f:Xto Y$ is an arbitrary map of sets then $f$ factors as a quotient by an equivalence relation $Xto X/sim$, followed by a bijection $X/simto f(X)$, followed by an inclusion $f(X)subseteq Y$. Is this a thing?
    – Kevin Buzzard
    Dec 7 at 21:23






  • 2




    How could category theory ever tell the difference between an inclusion and an arbitrary injection? Dually: how could it tell the difference between "the" quotient map by an equivalence relation and an arbitrary surjection? If you move away from the category of sets, there is some distinction amongst good and bad epi/monomorphisms. See, e.g. "effective epimorphism" and related notions.
    – Dylan Wilson
    Dec 7 at 21:41










  • from the category theory perspective, the main fact for sets and functions is that every function factors into a surjection followed by an injection. However, from the perspective of category theory, you can't distinguish inclusion maps from injections, because you are only looking at objects and morphisms: the notion of 'element' is lost. However, for your other comment: there is the notion of factorization system. The same idea holds for groups or rings too: a group homomorphism $phi: G to H$ factors into $G to G/ker phi to phi(G) to H$.
    – Jacob White
    Dec 7 at 21:47







  • 1




    A subobject of $A$ is an isomorphism class of monic arrows into $A$ ($fcolon Bto A$ and $f'colon B' to A$ are isomorphic if there is an isomorphism $gcolon Bto B'$ such that $f = f'circ g$). Dually, a quotient object of $A$ is an isomorphism class of epic arrows out of $A$. I suppose you could view the category of sets as being equipped with an extra structure: a choice of a unique representative in each subobject class (the inclusion) and each quotient object class (the "actual quotient"). Then each monic arrow factors uniquely as an isomorphism followed by an inclusion, ...
    – Alex Kruckman
    Dec 7 at 22:20






  • 3




    You may be thinking of a context that abstracts to an M-category (ncatlab.org/nlab/show/M-category). The 'tight morphisms' are the literal subset inclusions, and so have a role distinguished from arbitrary inclusions. In your setup one also seems to be asking that the class of tight morphisms are all regular monomorphisms (ncatlab.org/nlab/show/regular+monomorphism), that is, literally subsets carved out by equations (say by constructing the pullback of 1 --> 0,1 along a characteristic function by an equaliser).
    – David Roberts
    Dec 7 at 23:24












up vote
21
down vote

favorite
10









up vote
21
down vote

favorite
10






10





This might be a load of old nonsense.



I have always had it in my head that if $f:Xto Y$ is an injection, then $f$ has some sort of "canonical factorization" as a bijection $Xto f(X)$ followed by an inclusion $f(X)subseteq Y$. Similarly if $g:Xto Y$ is a surjection, and if we define an equivalence relation on $X$ by $asim biff g(a)=g(b)$ and let $Q$ be the set of equivalence classes, then $g$ has a "canonical factorization" as a quotient $Xto Q$ followed by a bijection $Qto B$. Furthermore I'd always suspected that these two "canonical" factorizations were in some way dual to each other.



I mentioned this in passing to a room full of smart undergraduates today and one of them called me up on it afterwards, and I realised that I could not attach any real meaning to what I've just said above. I half-wondered whether subobject classifiers might have something to do with it but having looked up the definition I am not so sure that they help at all.



Are inclusions in some way better than arbitrary injections? (in my mind they've always been the "best kind of injections" somehow). Are maps to sets of equivalence classes somehow better than arbitrary quotients? I can't help thinking that there might be something in these ideas but I am not sure I have the language to express it. Maybe I'm just wrong, or maybe there's some ncatlab page somewhere which will explain to me what I'm trying to formalise here.










share|cite|improve this question













This might be a load of old nonsense.



I have always had it in my head that if $f:Xto Y$ is an injection, then $f$ has some sort of "canonical factorization" as a bijection $Xto f(X)$ followed by an inclusion $f(X)subseteq Y$. Similarly if $g:Xto Y$ is a surjection, and if we define an equivalence relation on $X$ by $asim biff g(a)=g(b)$ and let $Q$ be the set of equivalence classes, then $g$ has a "canonical factorization" as a quotient $Xto Q$ followed by a bijection $Qto B$. Furthermore I'd always suspected that these two "canonical" factorizations were in some way dual to each other.



I mentioned this in passing to a room full of smart undergraduates today and one of them called me up on it afterwards, and I realised that I could not attach any real meaning to what I've just said above. I half-wondered whether subobject classifiers might have something to do with it but having looked up the definition I am not so sure that they help at all.



Are inclusions in some way better than arbitrary injections? (in my mind they've always been the "best kind of injections" somehow). Are maps to sets of equivalence classes somehow better than arbitrary quotients? I can't help thinking that there might be something in these ideas but I am not sure I have the language to express it. Maybe I'm just wrong, or maybe there's some ncatlab page somewhere which will explain to me what I'm trying to formalise here.







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asked Dec 7 at 21:09









Kevin Buzzard

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  • I guess one could go all the way and say that if $f:Xto Y$ is an arbitrary map of sets then $f$ factors as a quotient by an equivalence relation $Xto X/sim$, followed by a bijection $X/simto f(X)$, followed by an inclusion $f(X)subseteq Y$. Is this a thing?
    – Kevin Buzzard
    Dec 7 at 21:23






  • 2




    How could category theory ever tell the difference between an inclusion and an arbitrary injection? Dually: how could it tell the difference between "the" quotient map by an equivalence relation and an arbitrary surjection? If you move away from the category of sets, there is some distinction amongst good and bad epi/monomorphisms. See, e.g. "effective epimorphism" and related notions.
    – Dylan Wilson
    Dec 7 at 21:41










  • from the category theory perspective, the main fact for sets and functions is that every function factors into a surjection followed by an injection. However, from the perspective of category theory, you can't distinguish inclusion maps from injections, because you are only looking at objects and morphisms: the notion of 'element' is lost. However, for your other comment: there is the notion of factorization system. The same idea holds for groups or rings too: a group homomorphism $phi: G to H$ factors into $G to G/ker phi to phi(G) to H$.
    – Jacob White
    Dec 7 at 21:47







  • 1




    A subobject of $A$ is an isomorphism class of monic arrows into $A$ ($fcolon Bto A$ and $f'colon B' to A$ are isomorphic if there is an isomorphism $gcolon Bto B'$ such that $f = f'circ g$). Dually, a quotient object of $A$ is an isomorphism class of epic arrows out of $A$. I suppose you could view the category of sets as being equipped with an extra structure: a choice of a unique representative in each subobject class (the inclusion) and each quotient object class (the "actual quotient"). Then each monic arrow factors uniquely as an isomorphism followed by an inclusion, ...
    – Alex Kruckman
    Dec 7 at 22:20






  • 3




    You may be thinking of a context that abstracts to an M-category (ncatlab.org/nlab/show/M-category). The 'tight morphisms' are the literal subset inclusions, and so have a role distinguished from arbitrary inclusions. In your setup one also seems to be asking that the class of tight morphisms are all regular monomorphisms (ncatlab.org/nlab/show/regular+monomorphism), that is, literally subsets carved out by equations (say by constructing the pullback of 1 --> 0,1 along a characteristic function by an equaliser).
    – David Roberts
    Dec 7 at 23:24
















  • I guess one could go all the way and say that if $f:Xto Y$ is an arbitrary map of sets then $f$ factors as a quotient by an equivalence relation $Xto X/sim$, followed by a bijection $X/simto f(X)$, followed by an inclusion $f(X)subseteq Y$. Is this a thing?
    – Kevin Buzzard
    Dec 7 at 21:23






  • 2




    How could category theory ever tell the difference between an inclusion and an arbitrary injection? Dually: how could it tell the difference between "the" quotient map by an equivalence relation and an arbitrary surjection? If you move away from the category of sets, there is some distinction amongst good and bad epi/monomorphisms. See, e.g. "effective epimorphism" and related notions.
    – Dylan Wilson
    Dec 7 at 21:41










  • from the category theory perspective, the main fact for sets and functions is that every function factors into a surjection followed by an injection. However, from the perspective of category theory, you can't distinguish inclusion maps from injections, because you are only looking at objects and morphisms: the notion of 'element' is lost. However, for your other comment: there is the notion of factorization system. The same idea holds for groups or rings too: a group homomorphism $phi: G to H$ factors into $G to G/ker phi to phi(G) to H$.
    – Jacob White
    Dec 7 at 21:47







  • 1




    A subobject of $A$ is an isomorphism class of monic arrows into $A$ ($fcolon Bto A$ and $f'colon B' to A$ are isomorphic if there is an isomorphism $gcolon Bto B'$ such that $f = f'circ g$). Dually, a quotient object of $A$ is an isomorphism class of epic arrows out of $A$. I suppose you could view the category of sets as being equipped with an extra structure: a choice of a unique representative in each subobject class (the inclusion) and each quotient object class (the "actual quotient"). Then each monic arrow factors uniquely as an isomorphism followed by an inclusion, ...
    – Alex Kruckman
    Dec 7 at 22:20






  • 3




    You may be thinking of a context that abstracts to an M-category (ncatlab.org/nlab/show/M-category). The 'tight morphisms' are the literal subset inclusions, and so have a role distinguished from arbitrary inclusions. In your setup one also seems to be asking that the class of tight morphisms are all regular monomorphisms (ncatlab.org/nlab/show/regular+monomorphism), that is, literally subsets carved out by equations (say by constructing the pullback of 1 --> 0,1 along a characteristic function by an equaliser).
    – David Roberts
    Dec 7 at 23:24















I guess one could go all the way and say that if $f:Xto Y$ is an arbitrary map of sets then $f$ factors as a quotient by an equivalence relation $Xto X/sim$, followed by a bijection $X/simto f(X)$, followed by an inclusion $f(X)subseteq Y$. Is this a thing?
– Kevin Buzzard
Dec 7 at 21:23




I guess one could go all the way and say that if $f:Xto Y$ is an arbitrary map of sets then $f$ factors as a quotient by an equivalence relation $Xto X/sim$, followed by a bijection $X/simto f(X)$, followed by an inclusion $f(X)subseteq Y$. Is this a thing?
– Kevin Buzzard
Dec 7 at 21:23




2




2




How could category theory ever tell the difference between an inclusion and an arbitrary injection? Dually: how could it tell the difference between "the" quotient map by an equivalence relation and an arbitrary surjection? If you move away from the category of sets, there is some distinction amongst good and bad epi/monomorphisms. See, e.g. "effective epimorphism" and related notions.
– Dylan Wilson
Dec 7 at 21:41




How could category theory ever tell the difference between an inclusion and an arbitrary injection? Dually: how could it tell the difference between "the" quotient map by an equivalence relation and an arbitrary surjection? If you move away from the category of sets, there is some distinction amongst good and bad epi/monomorphisms. See, e.g. "effective epimorphism" and related notions.
– Dylan Wilson
Dec 7 at 21:41












from the category theory perspective, the main fact for sets and functions is that every function factors into a surjection followed by an injection. However, from the perspective of category theory, you can't distinguish inclusion maps from injections, because you are only looking at objects and morphisms: the notion of 'element' is lost. However, for your other comment: there is the notion of factorization system. The same idea holds for groups or rings too: a group homomorphism $phi: G to H$ factors into $G to G/ker phi to phi(G) to H$.
– Jacob White
Dec 7 at 21:47





from the category theory perspective, the main fact for sets and functions is that every function factors into a surjection followed by an injection. However, from the perspective of category theory, you can't distinguish inclusion maps from injections, because you are only looking at objects and morphisms: the notion of 'element' is lost. However, for your other comment: there is the notion of factorization system. The same idea holds for groups or rings too: a group homomorphism $phi: G to H$ factors into $G to G/ker phi to phi(G) to H$.
– Jacob White
Dec 7 at 21:47





1




1




A subobject of $A$ is an isomorphism class of monic arrows into $A$ ($fcolon Bto A$ and $f'colon B' to A$ are isomorphic if there is an isomorphism $gcolon Bto B'$ such that $f = f'circ g$). Dually, a quotient object of $A$ is an isomorphism class of epic arrows out of $A$. I suppose you could view the category of sets as being equipped with an extra structure: a choice of a unique representative in each subobject class (the inclusion) and each quotient object class (the "actual quotient"). Then each monic arrow factors uniquely as an isomorphism followed by an inclusion, ...
– Alex Kruckman
Dec 7 at 22:20




A subobject of $A$ is an isomorphism class of monic arrows into $A$ ($fcolon Bto A$ and $f'colon B' to A$ are isomorphic if there is an isomorphism $gcolon Bto B'$ such that $f = f'circ g$). Dually, a quotient object of $A$ is an isomorphism class of epic arrows out of $A$. I suppose you could view the category of sets as being equipped with an extra structure: a choice of a unique representative in each subobject class (the inclusion) and each quotient object class (the "actual quotient"). Then each monic arrow factors uniquely as an isomorphism followed by an inclusion, ...
– Alex Kruckman
Dec 7 at 22:20




3




3




You may be thinking of a context that abstracts to an M-category (ncatlab.org/nlab/show/M-category). The 'tight morphisms' are the literal subset inclusions, and so have a role distinguished from arbitrary inclusions. In your setup one also seems to be asking that the class of tight morphisms are all regular monomorphisms (ncatlab.org/nlab/show/regular+monomorphism), that is, literally subsets carved out by equations (say by constructing the pullback of 1 --> 0,1 along a characteristic function by an equaliser).
– David Roberts
Dec 7 at 23:24




You may be thinking of a context that abstracts to an M-category (ncatlab.org/nlab/show/M-category). The 'tight morphisms' are the literal subset inclusions, and so have a role distinguished from arbitrary inclusions. In your setup one also seems to be asking that the class of tight morphisms are all regular monomorphisms (ncatlab.org/nlab/show/regular+monomorphism), that is, literally subsets carved out by equations (say by constructing the pullback of 1 --> 0,1 along a characteristic function by an equaliser).
– David Roberts
Dec 7 at 23:24










5 Answers
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Are inclusions in some way better than arbitrary injections? Are maps to sets of equivalence classes somehow better than arbitrary quotients?




This is a simple-minded answer, but I would say yes because there is a proper class of distinct injections into $Y$ --- the injecting set $X$ can live anywhere in the set-theoretic universe --- whereas inclusions into $Y$ are effectively just subsets of $Y$. Under the natural notion of equivalence for injections, distinct subsets give inequivalent injections, so you could say that going from injections to inclusions is a matter of selecting one distinguished, canonical element from each equivalence class of injections.



I would say that the injections into $Y$ are classified by the subsets of $Y$, and similarly the quotients of $X$ are classified by the equivalence relations on $X$.






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    I completely understand that I am choosing one element from each equivalence class of injections. What I am unsure about is whether it is "canonical" and indeed what "canonical" even means in this context.
    – Kevin Buzzard
    Dec 9 at 1:33










  • Ah, I see. I suppose "canonical" has a fairly straightforward informal meaning, but if you want a rigorous definition ... I don't know, it doesn't seem likely.
    – Nik Weaver
    Dec 9 at 2:03






  • 1




    What does your answer mean if you cannot give a rigorous definition of "canonical"? Taking this further -- what does my question mean? Is this mathematics at all? This is what I do not understand. I know what mathematics is supposed to be, and there are not supposed to be things which are open to discussion.
    – Kevin Buzzard
    Dec 9 at 11:01






  • 1




    @KevinBuzzard Since the category of sets is equivalent to its skeleton, this question is not 'mathematicial' (definable in any invariant way) without additional structure (as in Mike's answer and David's comment). If you are working with ZFC, you can promote the category of ZFC sets to an M-Category in a canonical way, and that is how you make it mathematical.
    – Harry Gindi
    Dec 9 at 17:46







  • 1




    And in Harry's comment, one can replace ZFC by any set theory in which is meaningful to define subsets $xsubseteq y$ as $forall z, zin x Rightarrow zin y$.
    – David Roberts
    Dec 9 at 19:49

















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It seems that you've got factorization of maps covered, so let me address the question of why canonical quotient maps and canonical inclusions are "better".



Given a set $X$, in general there is a proper class of injections $Y to X$. However, many of these are isomorphic, where injections $i : Y to X$ and $j : Z to X$ are isomorphic if there is an isomorphism $k : Y to Z$ such that $i = j circ k$. The isomorphism classes of injections into $X$ are the subobjects of $X$. In fact, there are only set-many subobjects of $X$ (in category-theoretic language, sets form a well-powered category). It is pesky to work with set-many proper equivalence classes, so we instead look for a set $P(X)$ of injections into $X$, one from each isomorphism class. We may additionally require some nice properties, for instance, if $i : X to Y$ is in $P(Y)$ and $j : Y to Z$ is in $P(Y)$, we would expect $j circ i : X to Z$ to be in $P(Z)$. One can come up with a wish list of such nice closure conditions, here's another one: if $i : Y to X$ and $j : Z to X$ are in $P(X)$, and there is (a unique) $k : X to Z$ such that $i = j circ k$, then $k$ is in $P(Z)$.



We know the answer, of course, just take $P(X)$ to be the canonical subset inclusions into $X$. This is not the only choice of such representative inclusions, but it's a pretty good one.



We may therefore say that the canonical inclusions of subsets are "better" because they are the canonical representatives of subobjects (equivalence classes of injections).



The answer for quotient maps and surjections is dual. Consider equivalence classes of surjections, quotiented by isomorphism. There are only set-many such classes, therefore sets form a well-copowered category. (Some people say "cowell-powered" but then why not call it "ill-powered"?) This time we look for a set $Q(X)$ of surjections from $X$, each representing one equivalence class of surjections from $X$. We may take $Q(X)$ to be the set of all canonical quotient maps $X to X/sim$, or just the set of all equivalence classes on $X$. Once again, canonical quotient maps are "better" because they are the distinguished representatives of isomorphism classes of surjections.






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  • "This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely.
    – Kevin Buzzard
    Dec 8 at 0:11










  • For any set $S$ let $barS = x mid x in S$ and let $b_S : barS to S$ be given by $b_S(x) = x$. Given $S subseteq X$ let $i_S : S to X$ be the canonical inclusion. Define $P(X) = i_S circ b_S : barS to X mid S subseteq X$. Then I think we get representatives of injections that are closed under composition (because they're not composable).
    – Andrej Bauer
    Dec 8 at 1:04










  • @AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$?
    – Santana Afton
    Dec 8 at 4:03










  • Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Yto Z$ in $P(Z)$ with $Y=overlineY_0$, I choose a subset $X_0$ of $Y$, I get a map $Xto Y$ in $P(Y)$, but I don't think that $Xto Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right?
    – Kevin Buzzard
    Dec 8 at 8:05










  • It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow".
    – Kevin Buzzard
    Dec 8 at 8:09

















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The nLab page you're looking for is called factorization systems. Here is my favorite one, which I think answers your question. In any category with finite limits and colimits, every morphism $f : X to Y$ has a canonical factorization



$$X to textcoim(f) to textim(f) to Y$$



where $textim(f)$, the regular image, is the equalizer of the cokernel pair of $f$ (this is the "nonabelian" version of "kernel of the cokernel") and $textcoim(f)$, the regular coimage, is the coequalizer of the kernel pair of $f$ (again, the "nonabelian" version of "cokernel of the kernel"). These two constructions are categorically dual and so, among other things, the coimage-image factorization of $f$ in the opposite category is the same sequence of maps but in the opposite order.



In $textSet$, the coimage and image are both the image of a function in the usual sense, but computed in different ways, which I think match the distinction you're getting at. $textcoim(f)$ is computed, more or less, by constructing the equivalence relation on $X$ defined by $x_1 sim x_2 Leftrightarrow f(x_1) = f(x_2)$, then quotienting $X$ by it. $textim(f)$ is computed in a categorically dual way, although it looks a little strange at first: by first constructing the pushout $Y sqcup_X Y$, then isolating the subset of $Y$ of elements which are sent to the same element by both of the canonical maps $Y to Y sqcup_X Y$.



In particular, the factorization you want for an injection is the regular image factorization, and the factorization you want for a surjection is the regular coimage factorization, so they are in fact categorically dual. The full coimage-image factorization combines these.



It's a nontrivial theorem that the map $textcoim(f) to textim(f)$ is an isomorphism in $textSet$. It's also an isomorphism in any abelian category and in $textGrp$ (this is an abstract form of the first isomorphism theorem), but in general it's just both a monomorphism and an epimorphism. A very instructive example is $textTop$, where $textcoim(f)$ is the set-theoretic image topologized as a quotient of $X$, and $textim(f)$ is the set-theoretic image topologized as a subspace of $Y$. (Note that these coincide for compact Hausdorff spaces!)






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  • 1




    But these things like im and coim are I believe defined by category-theoretic properties and are hence only defined up to unique isomorphism. Aren't I trying to do something better than isolating the object up to unique isomorphism? I'm wondering if in Set there is some sort of canonical representative (and am still not sure about if there is one).
    – Kevin Buzzard
    Dec 8 at 0:06










  • @Kevin: yes, I'm eliding some philosophical points here that I think Andrej's answer tackles better, at least in $textSet$. In particular I am really referring to a particular construction of the image and coimage rather than to anything which satisfies their universal properties, which is a sin I don't normally indulge in. My excuse is that I never just isolate myself to $textSet$ if I can help it; morally the coimage is "the image as seen from $X$" and the image is "the image as seen from $Y$," as I think the example of $textTop$ makes vividly clear, and so I think of them as...
    – Qiaochu Yuan
    Dec 8 at 0:10










  • ...conceptually distinct even though they happen to agree as universal constructions in $textSet$. Said another way, I'm either talking about your state of knowledge before you learn that $textcoim(f) to textim(f)$ is an isomorphism in $textSet$, or I'm reserving the right to replace $textSet$ with another category in which that's false. If you really want to stick to $textSet$ then Andrej's answer is probably more relevant than mine.
    – Qiaochu Yuan
    Dec 8 at 0:12

















up vote
9
down vote













Andrej's answer is a good one for explaining "what is special" about injections and quotients. In terms of formalizing the notion of "subset inclusion", David Roberts has already mentioned in a comment the notion of M-category, which could be used to represent "the category $mathrmSet$ together with the information about which injections are inclusions" (or which surjections are quotients).



Another formalization which is specifically adapted to "inclusion-like" subcategories is called (appropriately enough) a system of inclusions (with enhancements that add adjectives such as "directed" and "structural"); it was introduced in this paper by Awodey, Butz, Simpson, and Streicher. One could presumably dualize this somehow to obtain a notion of "system of quotients". Note that a system of inclusions is a particular kind of M-category.



As has been noted, when the category $rm Set$ is treated "purely category-theoretically" it does not "notice" which injections are inclusions, i.e. this information does not transfer usefully across an equivalence of categories. An M-category is a way of defining a "category-theoretic" structure (namely, a certain kind of enriched category) that nevertheless can carry this sort of information.






share|cite|improve this answer




















  • Thanks for the reference, I was aware of but failed to mention it.
    – Andrej Bauer
    Dec 9 at 9:08






  • 1




    However, the standard construction of quotients doesn't naively form an M-category, as the 'set of equivalence classes with a map to it' isn't transitive: one gets sets of sets of equivalence classes. Perhaps there is a way to fix this, but I didn't see it with a (small amount of) thinking.
    – David Roberts
    Dec 9 at 19:52










  • So "M-category" does not answer the question for surjections?
    – Kevin Buzzard
    Dec 9 at 22:18






  • 1




    @DavidRoberts Good point, I was too glib about that. I guess the dual notion of "system of quotients" would have to omit the composability if you want to include the usual notion of quotient set in $rm Set$. However, it occurs to me that "quotients" of setOIDS do compose, so at least $rm Set$ is equivalent to some category that has a system of quotients with composition.
    – Mike Shulman
    Dec 9 at 22:47

















up vote
4
down vote














"Are inclusions in some way better than arbitrary injections?"




No they are not.

Anything that you can say about the one, you can say about the other.



In particular:
Given $X$, there is a set of inclusions into $X$. And there is a (large groupoid equivalent to a) set of injections into $X$. For all practical purposes these are the same thing. So they should be treated as the same thing.




Are quotient maps somehow better than arbitrary surjections?




Again no.

Same story.



    The fact that we can talk about inclusions versus injections is an artefact of our set theoretic foundations of mathematics. It would be better to use a language that disallows us from asking whether two sets are equal, and only allows us to talk about isomorphisms between sets.
    We largely already do this in practice. When we write $mathbb Z subset mathbb Q$, we don't stop to explain that elements of $mathbb Q$ are defined as pairs of elements of $mathbb Z$, and we don't distinguish between $mathbb Z$ and its image in $mathbb Q$. We just write $mathbb Z subset mathbb Q$.



    Unfortunately, I'm not knowledgeable enough in foundational issues in order to make sense of the notion of "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms". But I'm sure that such a thing exists (and that it can be tweaked so as to be exactly as powerful as ZFC).






share|cite|improve this answer




















  • "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms" One approach is FOLDS, by Makkai (cf ncatlab.org/nlab/show/FOLDS), and another approach was discussed by Bénabou projecteuclid.org/euclid.jsl/1183741771
    – David Roberts
    Dec 12 at 1:57










  • Sounds like univalent foundations to me.
    – Nik Weaver
    Dec 12 at 5:29










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5 Answers
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up vote
18
down vote














Are inclusions in some way better than arbitrary injections? Are maps to sets of equivalence classes somehow better than arbitrary quotients?




This is a simple-minded answer, but I would say yes because there is a proper class of distinct injections into $Y$ --- the injecting set $X$ can live anywhere in the set-theoretic universe --- whereas inclusions into $Y$ are effectively just subsets of $Y$. Under the natural notion of equivalence for injections, distinct subsets give inequivalent injections, so you could say that going from injections to inclusions is a matter of selecting one distinguished, canonical element from each equivalence class of injections.



I would say that the injections into $Y$ are classified by the subsets of $Y$, and similarly the quotients of $X$ are classified by the equivalence relations on $X$.






share|cite|improve this answer


















  • 1




    I completely understand that I am choosing one element from each equivalence class of injections. What I am unsure about is whether it is "canonical" and indeed what "canonical" even means in this context.
    – Kevin Buzzard
    Dec 9 at 1:33










  • Ah, I see. I suppose "canonical" has a fairly straightforward informal meaning, but if you want a rigorous definition ... I don't know, it doesn't seem likely.
    – Nik Weaver
    Dec 9 at 2:03






  • 1




    What does your answer mean if you cannot give a rigorous definition of "canonical"? Taking this further -- what does my question mean? Is this mathematics at all? This is what I do not understand. I know what mathematics is supposed to be, and there are not supposed to be things which are open to discussion.
    – Kevin Buzzard
    Dec 9 at 11:01






  • 1




    @KevinBuzzard Since the category of sets is equivalent to its skeleton, this question is not 'mathematicial' (definable in any invariant way) without additional structure (as in Mike's answer and David's comment). If you are working with ZFC, you can promote the category of ZFC sets to an M-Category in a canonical way, and that is how you make it mathematical.
    – Harry Gindi
    Dec 9 at 17:46







  • 1




    And in Harry's comment, one can replace ZFC by any set theory in which is meaningful to define subsets $xsubseteq y$ as $forall z, zin x Rightarrow zin y$.
    – David Roberts
    Dec 9 at 19:49














up vote
18
down vote














Are inclusions in some way better than arbitrary injections? Are maps to sets of equivalence classes somehow better than arbitrary quotients?




This is a simple-minded answer, but I would say yes because there is a proper class of distinct injections into $Y$ --- the injecting set $X$ can live anywhere in the set-theoretic universe --- whereas inclusions into $Y$ are effectively just subsets of $Y$. Under the natural notion of equivalence for injections, distinct subsets give inequivalent injections, so you could say that going from injections to inclusions is a matter of selecting one distinguished, canonical element from each equivalence class of injections.



I would say that the injections into $Y$ are classified by the subsets of $Y$, and similarly the quotients of $X$ are classified by the equivalence relations on $X$.






share|cite|improve this answer


















  • 1




    I completely understand that I am choosing one element from each equivalence class of injections. What I am unsure about is whether it is "canonical" and indeed what "canonical" even means in this context.
    – Kevin Buzzard
    Dec 9 at 1:33










  • Ah, I see. I suppose "canonical" has a fairly straightforward informal meaning, but if you want a rigorous definition ... I don't know, it doesn't seem likely.
    – Nik Weaver
    Dec 9 at 2:03






  • 1




    What does your answer mean if you cannot give a rigorous definition of "canonical"? Taking this further -- what does my question mean? Is this mathematics at all? This is what I do not understand. I know what mathematics is supposed to be, and there are not supposed to be things which are open to discussion.
    – Kevin Buzzard
    Dec 9 at 11:01






  • 1




    @KevinBuzzard Since the category of sets is equivalent to its skeleton, this question is not 'mathematicial' (definable in any invariant way) without additional structure (as in Mike's answer and David's comment). If you are working with ZFC, you can promote the category of ZFC sets to an M-Category in a canonical way, and that is how you make it mathematical.
    – Harry Gindi
    Dec 9 at 17:46







  • 1




    And in Harry's comment, one can replace ZFC by any set theory in which is meaningful to define subsets $xsubseteq y$ as $forall z, zin x Rightarrow zin y$.
    – David Roberts
    Dec 9 at 19:49












up vote
18
down vote










up vote
18
down vote










Are inclusions in some way better than arbitrary injections? Are maps to sets of equivalence classes somehow better than arbitrary quotients?




This is a simple-minded answer, but I would say yes because there is a proper class of distinct injections into $Y$ --- the injecting set $X$ can live anywhere in the set-theoretic universe --- whereas inclusions into $Y$ are effectively just subsets of $Y$. Under the natural notion of equivalence for injections, distinct subsets give inequivalent injections, so you could say that going from injections to inclusions is a matter of selecting one distinguished, canonical element from each equivalence class of injections.



I would say that the injections into $Y$ are classified by the subsets of $Y$, and similarly the quotients of $X$ are classified by the equivalence relations on $X$.






share|cite|improve this answer















Are inclusions in some way better than arbitrary injections? Are maps to sets of equivalence classes somehow better than arbitrary quotients?




This is a simple-minded answer, but I would say yes because there is a proper class of distinct injections into $Y$ --- the injecting set $X$ can live anywhere in the set-theoretic universe --- whereas inclusions into $Y$ are effectively just subsets of $Y$. Under the natural notion of equivalence for injections, distinct subsets give inequivalent injections, so you could say that going from injections to inclusions is a matter of selecting one distinguished, canonical element from each equivalence class of injections.



I would say that the injections into $Y$ are classified by the subsets of $Y$, and similarly the quotients of $X$ are classified by the equivalence relations on $X$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 at 4:25

























answered Dec 7 at 22:32









Nik Weaver

19.8k145123




19.8k145123







  • 1




    I completely understand that I am choosing one element from each equivalence class of injections. What I am unsure about is whether it is "canonical" and indeed what "canonical" even means in this context.
    – Kevin Buzzard
    Dec 9 at 1:33










  • Ah, I see. I suppose "canonical" has a fairly straightforward informal meaning, but if you want a rigorous definition ... I don't know, it doesn't seem likely.
    – Nik Weaver
    Dec 9 at 2:03






  • 1




    What does your answer mean if you cannot give a rigorous definition of "canonical"? Taking this further -- what does my question mean? Is this mathematics at all? This is what I do not understand. I know what mathematics is supposed to be, and there are not supposed to be things which are open to discussion.
    – Kevin Buzzard
    Dec 9 at 11:01






  • 1




    @KevinBuzzard Since the category of sets is equivalent to its skeleton, this question is not 'mathematicial' (definable in any invariant way) without additional structure (as in Mike's answer and David's comment). If you are working with ZFC, you can promote the category of ZFC sets to an M-Category in a canonical way, and that is how you make it mathematical.
    – Harry Gindi
    Dec 9 at 17:46







  • 1




    And in Harry's comment, one can replace ZFC by any set theory in which is meaningful to define subsets $xsubseteq y$ as $forall z, zin x Rightarrow zin y$.
    – David Roberts
    Dec 9 at 19:49












  • 1




    I completely understand that I am choosing one element from each equivalence class of injections. What I am unsure about is whether it is "canonical" and indeed what "canonical" even means in this context.
    – Kevin Buzzard
    Dec 9 at 1:33










  • Ah, I see. I suppose "canonical" has a fairly straightforward informal meaning, but if you want a rigorous definition ... I don't know, it doesn't seem likely.
    – Nik Weaver
    Dec 9 at 2:03






  • 1




    What does your answer mean if you cannot give a rigorous definition of "canonical"? Taking this further -- what does my question mean? Is this mathematics at all? This is what I do not understand. I know what mathematics is supposed to be, and there are not supposed to be things which are open to discussion.
    – Kevin Buzzard
    Dec 9 at 11:01






  • 1




    @KevinBuzzard Since the category of sets is equivalent to its skeleton, this question is not 'mathematicial' (definable in any invariant way) without additional structure (as in Mike's answer and David's comment). If you are working with ZFC, you can promote the category of ZFC sets to an M-Category in a canonical way, and that is how you make it mathematical.
    – Harry Gindi
    Dec 9 at 17:46







  • 1




    And in Harry's comment, one can replace ZFC by any set theory in which is meaningful to define subsets $xsubseteq y$ as $forall z, zin x Rightarrow zin y$.
    – David Roberts
    Dec 9 at 19:49







1




1




I completely understand that I am choosing one element from each equivalence class of injections. What I am unsure about is whether it is "canonical" and indeed what "canonical" even means in this context.
– Kevin Buzzard
Dec 9 at 1:33




I completely understand that I am choosing one element from each equivalence class of injections. What I am unsure about is whether it is "canonical" and indeed what "canonical" even means in this context.
– Kevin Buzzard
Dec 9 at 1:33












Ah, I see. I suppose "canonical" has a fairly straightforward informal meaning, but if you want a rigorous definition ... I don't know, it doesn't seem likely.
– Nik Weaver
Dec 9 at 2:03




Ah, I see. I suppose "canonical" has a fairly straightforward informal meaning, but if you want a rigorous definition ... I don't know, it doesn't seem likely.
– Nik Weaver
Dec 9 at 2:03




1




1




What does your answer mean if you cannot give a rigorous definition of "canonical"? Taking this further -- what does my question mean? Is this mathematics at all? This is what I do not understand. I know what mathematics is supposed to be, and there are not supposed to be things which are open to discussion.
– Kevin Buzzard
Dec 9 at 11:01




What does your answer mean if you cannot give a rigorous definition of "canonical"? Taking this further -- what does my question mean? Is this mathematics at all? This is what I do not understand. I know what mathematics is supposed to be, and there are not supposed to be things which are open to discussion.
– Kevin Buzzard
Dec 9 at 11:01




1




1




@KevinBuzzard Since the category of sets is equivalent to its skeleton, this question is not 'mathematicial' (definable in any invariant way) without additional structure (as in Mike's answer and David's comment). If you are working with ZFC, you can promote the category of ZFC sets to an M-Category in a canonical way, and that is how you make it mathematical.
– Harry Gindi
Dec 9 at 17:46





@KevinBuzzard Since the category of sets is equivalent to its skeleton, this question is not 'mathematicial' (definable in any invariant way) without additional structure (as in Mike's answer and David's comment). If you are working with ZFC, you can promote the category of ZFC sets to an M-Category in a canonical way, and that is how you make it mathematical.
– Harry Gindi
Dec 9 at 17:46





1




1




And in Harry's comment, one can replace ZFC by any set theory in which is meaningful to define subsets $xsubseteq y$ as $forall z, zin x Rightarrow zin y$.
– David Roberts
Dec 9 at 19:49




And in Harry's comment, one can replace ZFC by any set theory in which is meaningful to define subsets $xsubseteq y$ as $forall z, zin x Rightarrow zin y$.
– David Roberts
Dec 9 at 19:49










up vote
14
down vote













It seems that you've got factorization of maps covered, so let me address the question of why canonical quotient maps and canonical inclusions are "better".



Given a set $X$, in general there is a proper class of injections $Y to X$. However, many of these are isomorphic, where injections $i : Y to X$ and $j : Z to X$ are isomorphic if there is an isomorphism $k : Y to Z$ such that $i = j circ k$. The isomorphism classes of injections into $X$ are the subobjects of $X$. In fact, there are only set-many subobjects of $X$ (in category-theoretic language, sets form a well-powered category). It is pesky to work with set-many proper equivalence classes, so we instead look for a set $P(X)$ of injections into $X$, one from each isomorphism class. We may additionally require some nice properties, for instance, if $i : X to Y$ is in $P(Y)$ and $j : Y to Z$ is in $P(Y)$, we would expect $j circ i : X to Z$ to be in $P(Z)$. One can come up with a wish list of such nice closure conditions, here's another one: if $i : Y to X$ and $j : Z to X$ are in $P(X)$, and there is (a unique) $k : X to Z$ such that $i = j circ k$, then $k$ is in $P(Z)$.



We know the answer, of course, just take $P(X)$ to be the canonical subset inclusions into $X$. This is not the only choice of such representative inclusions, but it's a pretty good one.



We may therefore say that the canonical inclusions of subsets are "better" because they are the canonical representatives of subobjects (equivalence classes of injections).



The answer for quotient maps and surjections is dual. Consider equivalence classes of surjections, quotiented by isomorphism. There are only set-many such classes, therefore sets form a well-copowered category. (Some people say "cowell-powered" but then why not call it "ill-powered"?) This time we look for a set $Q(X)$ of surjections from $X$, each representing one equivalence class of surjections from $X$. We may take $Q(X)$ to be the set of all canonical quotient maps $X to X/sim$, or just the set of all equivalence classes on $X$. Once again, canonical quotient maps are "better" because they are the distinguished representatives of isomorphism classes of surjections.






share|cite|improve this answer






















  • "This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely.
    – Kevin Buzzard
    Dec 8 at 0:11










  • For any set $S$ let $barS = x mid x in S$ and let $b_S : barS to S$ be given by $b_S(x) = x$. Given $S subseteq X$ let $i_S : S to X$ be the canonical inclusion. Define $P(X) = i_S circ b_S : barS to X mid S subseteq X$. Then I think we get representatives of injections that are closed under composition (because they're not composable).
    – Andrej Bauer
    Dec 8 at 1:04










  • @AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$?
    – Santana Afton
    Dec 8 at 4:03










  • Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Yto Z$ in $P(Z)$ with $Y=overlineY_0$, I choose a subset $X_0$ of $Y$, I get a map $Xto Y$ in $P(Y)$, but I don't think that $Xto Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right?
    – Kevin Buzzard
    Dec 8 at 8:05










  • It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow".
    – Kevin Buzzard
    Dec 8 at 8:09














up vote
14
down vote













It seems that you've got factorization of maps covered, so let me address the question of why canonical quotient maps and canonical inclusions are "better".



Given a set $X$, in general there is a proper class of injections $Y to X$. However, many of these are isomorphic, where injections $i : Y to X$ and $j : Z to X$ are isomorphic if there is an isomorphism $k : Y to Z$ such that $i = j circ k$. The isomorphism classes of injections into $X$ are the subobjects of $X$. In fact, there are only set-many subobjects of $X$ (in category-theoretic language, sets form a well-powered category). It is pesky to work with set-many proper equivalence classes, so we instead look for a set $P(X)$ of injections into $X$, one from each isomorphism class. We may additionally require some nice properties, for instance, if $i : X to Y$ is in $P(Y)$ and $j : Y to Z$ is in $P(Y)$, we would expect $j circ i : X to Z$ to be in $P(Z)$. One can come up with a wish list of such nice closure conditions, here's another one: if $i : Y to X$ and $j : Z to X$ are in $P(X)$, and there is (a unique) $k : X to Z$ such that $i = j circ k$, then $k$ is in $P(Z)$.



We know the answer, of course, just take $P(X)$ to be the canonical subset inclusions into $X$. This is not the only choice of such representative inclusions, but it's a pretty good one.



We may therefore say that the canonical inclusions of subsets are "better" because they are the canonical representatives of subobjects (equivalence classes of injections).



The answer for quotient maps and surjections is dual. Consider equivalence classes of surjections, quotiented by isomorphism. There are only set-many such classes, therefore sets form a well-copowered category. (Some people say "cowell-powered" but then why not call it "ill-powered"?) This time we look for a set $Q(X)$ of surjections from $X$, each representing one equivalence class of surjections from $X$. We may take $Q(X)$ to be the set of all canonical quotient maps $X to X/sim$, or just the set of all equivalence classes on $X$. Once again, canonical quotient maps are "better" because they are the distinguished representatives of isomorphism classes of surjections.






share|cite|improve this answer






















  • "This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely.
    – Kevin Buzzard
    Dec 8 at 0:11










  • For any set $S$ let $barS = x mid x in S$ and let $b_S : barS to S$ be given by $b_S(x) = x$. Given $S subseteq X$ let $i_S : S to X$ be the canonical inclusion. Define $P(X) = i_S circ b_S : barS to X mid S subseteq X$. Then I think we get representatives of injections that are closed under composition (because they're not composable).
    – Andrej Bauer
    Dec 8 at 1:04










  • @AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$?
    – Santana Afton
    Dec 8 at 4:03










  • Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Yto Z$ in $P(Z)$ with $Y=overlineY_0$, I choose a subset $X_0$ of $Y$, I get a map $Xto Y$ in $P(Y)$, but I don't think that $Xto Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right?
    – Kevin Buzzard
    Dec 8 at 8:05










  • It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow".
    – Kevin Buzzard
    Dec 8 at 8:09












up vote
14
down vote










up vote
14
down vote









It seems that you've got factorization of maps covered, so let me address the question of why canonical quotient maps and canonical inclusions are "better".



Given a set $X$, in general there is a proper class of injections $Y to X$. However, many of these are isomorphic, where injections $i : Y to X$ and $j : Z to X$ are isomorphic if there is an isomorphism $k : Y to Z$ such that $i = j circ k$. The isomorphism classes of injections into $X$ are the subobjects of $X$. In fact, there are only set-many subobjects of $X$ (in category-theoretic language, sets form a well-powered category). It is pesky to work with set-many proper equivalence classes, so we instead look for a set $P(X)$ of injections into $X$, one from each isomorphism class. We may additionally require some nice properties, for instance, if $i : X to Y$ is in $P(Y)$ and $j : Y to Z$ is in $P(Y)$, we would expect $j circ i : X to Z$ to be in $P(Z)$. One can come up with a wish list of such nice closure conditions, here's another one: if $i : Y to X$ and $j : Z to X$ are in $P(X)$, and there is (a unique) $k : X to Z$ such that $i = j circ k$, then $k$ is in $P(Z)$.



We know the answer, of course, just take $P(X)$ to be the canonical subset inclusions into $X$. This is not the only choice of such representative inclusions, but it's a pretty good one.



We may therefore say that the canonical inclusions of subsets are "better" because they are the canonical representatives of subobjects (equivalence classes of injections).



The answer for quotient maps and surjections is dual. Consider equivalence classes of surjections, quotiented by isomorphism. There are only set-many such classes, therefore sets form a well-copowered category. (Some people say "cowell-powered" but then why not call it "ill-powered"?) This time we look for a set $Q(X)$ of surjections from $X$, each representing one equivalence class of surjections from $X$. We may take $Q(X)$ to be the set of all canonical quotient maps $X to X/sim$, or just the set of all equivalence classes on $X$. Once again, canonical quotient maps are "better" because they are the distinguished representatives of isomorphism classes of surjections.






share|cite|improve this answer














It seems that you've got factorization of maps covered, so let me address the question of why canonical quotient maps and canonical inclusions are "better".



Given a set $X$, in general there is a proper class of injections $Y to X$. However, many of these are isomorphic, where injections $i : Y to X$ and $j : Z to X$ are isomorphic if there is an isomorphism $k : Y to Z$ such that $i = j circ k$. The isomorphism classes of injections into $X$ are the subobjects of $X$. In fact, there are only set-many subobjects of $X$ (in category-theoretic language, sets form a well-powered category). It is pesky to work with set-many proper equivalence classes, so we instead look for a set $P(X)$ of injections into $X$, one from each isomorphism class. We may additionally require some nice properties, for instance, if $i : X to Y$ is in $P(Y)$ and $j : Y to Z$ is in $P(Y)$, we would expect $j circ i : X to Z$ to be in $P(Z)$. One can come up with a wish list of such nice closure conditions, here's another one: if $i : Y to X$ and $j : Z to X$ are in $P(X)$, and there is (a unique) $k : X to Z$ such that $i = j circ k$, then $k$ is in $P(Z)$.



We know the answer, of course, just take $P(X)$ to be the canonical subset inclusions into $X$. This is not the only choice of such representative inclusions, but it's a pretty good one.



We may therefore say that the canonical inclusions of subsets are "better" because they are the canonical representatives of subobjects (equivalence classes of injections).



The answer for quotient maps and surjections is dual. Consider equivalence classes of surjections, quotiented by isomorphism. There are only set-many such classes, therefore sets form a well-copowered category. (Some people say "cowell-powered" but then why not call it "ill-powered"?) This time we look for a set $Q(X)$ of surjections from $X$, each representing one equivalence class of surjections from $X$. We may take $Q(X)$ to be the set of all canonical quotient maps $X to X/sim$, or just the set of all equivalence classes on $X$. Once again, canonical quotient maps are "better" because they are the distinguished representatives of isomorphism classes of surjections.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 at 1:05

























answered Dec 7 at 23:15









Andrej Bauer

29.9k477164




29.9k477164











  • "This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely.
    – Kevin Buzzard
    Dec 8 at 0:11










  • For any set $S$ let $barS = x mid x in S$ and let $b_S : barS to S$ be given by $b_S(x) = x$. Given $S subseteq X$ let $i_S : S to X$ be the canonical inclusion. Define $P(X) = i_S circ b_S : barS to X mid S subseteq X$. Then I think we get representatives of injections that are closed under composition (because they're not composable).
    – Andrej Bauer
    Dec 8 at 1:04










  • @AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$?
    – Santana Afton
    Dec 8 at 4:03










  • Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Yto Z$ in $P(Z)$ with $Y=overlineY_0$, I choose a subset $X_0$ of $Y$, I get a map $Xto Y$ in $P(Y)$, but I don't think that $Xto Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right?
    – Kevin Buzzard
    Dec 8 at 8:05










  • It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow".
    – Kevin Buzzard
    Dec 8 at 8:09
















  • "This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely.
    – Kevin Buzzard
    Dec 8 at 0:11










  • For any set $S$ let $barS = x mid x in S$ and let $b_S : barS to S$ be given by $b_S(x) = x$. Given $S subseteq X$ let $i_S : S to X$ be the canonical inclusion. Define $P(X) = i_S circ b_S : barS to X mid S subseteq X$. Then I think we get representatives of injections that are closed under composition (because they're not composable).
    – Andrej Bauer
    Dec 8 at 1:04










  • @AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$?
    – Santana Afton
    Dec 8 at 4:03










  • Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Yto Z$ in $P(Z)$ with $Y=overlineY_0$, I choose a subset $X_0$ of $Y$, I get a map $Xto Y$ in $P(Y)$, but I don't think that $Xto Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right?
    – Kevin Buzzard
    Dec 8 at 8:05










  • It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow".
    – Kevin Buzzard
    Dec 8 at 8:09















"This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely.
– Kevin Buzzard
Dec 8 at 0:11




"This is not the only choice of such representative inclusions, but it's a pretty good one.". Can you give an example of another one? In particular do you know one satisfying all the properties you mention, as well as all the further properties that we might think of once you've come up with one? It would be really cool if we could either classify the subset inclusions uniquely or prove that there is no chance of classifying them uniquely.
– Kevin Buzzard
Dec 8 at 0:11












For any set $S$ let $barS = x mid x in S$ and let $b_S : barS to S$ be given by $b_S(x) = x$. Given $S subseteq X$ let $i_S : S to X$ be the canonical inclusion. Define $P(X) = i_S circ b_S : barS to X mid S subseteq X$. Then I think we get representatives of injections that are closed under composition (because they're not composable).
– Andrej Bauer
Dec 8 at 1:04




For any set $S$ let $barS = x mid x in S$ and let $b_S : barS to S$ be given by $b_S(x) = x$. Given $S subseteq X$ let $i_S : S to X$ be the canonical inclusion. Define $P(X) = i_S circ b_S : barS to X mid S subseteq X$. Then I think we get representatives of injections that are closed under composition (because they're not composable).
– Andrej Bauer
Dec 8 at 1:04












@AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$?
– Santana Afton
Dec 8 at 4:03




@AndrejBauer At the end of your second paragraph, do you want the source of $k$ to be $Y$ and not $X$?
– Santana Afton
Dec 8 at 4:03












Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Yto Z$ in $P(Z)$ with $Y=overlineY_0$, I choose a subset $X_0$ of $Y$, I get a map $Xto Y$ in $P(Y)$, but I don't think that $Xto Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right?
– Kevin Buzzard
Dec 8 at 8:05




Is this bar example really closed under composition? I start with $Z$, I choose an actual subset $Y_0$, I get a corresponding map $Yto Z$ in $P(Z)$ with $Y=overlineY_0$, I choose a subset $X_0$ of $Y$, I get a map $Xto Y$ in $P(Y)$, but I don't think that $Xto Z$ will be in $P(Z)$ because I'll have to strip squiggly brackets twice, right?
– Kevin Buzzard
Dec 8 at 8:05












It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow".
– Kevin Buzzard
Dec 8 at 8:09




It would be a bit depressing (but perhaps not surprising) if there was some non-constructive "of course other choices for $P$ exist, just start off with a map from a set with one element to a set with two elements which isn't an inclusion and then extend by Zorn's Lemma somehow".
– Kevin Buzzard
Dec 8 at 8:09










up vote
11
down vote













The nLab page you're looking for is called factorization systems. Here is my favorite one, which I think answers your question. In any category with finite limits and colimits, every morphism $f : X to Y$ has a canonical factorization



$$X to textcoim(f) to textim(f) to Y$$



where $textim(f)$, the regular image, is the equalizer of the cokernel pair of $f$ (this is the "nonabelian" version of "kernel of the cokernel") and $textcoim(f)$, the regular coimage, is the coequalizer of the kernel pair of $f$ (again, the "nonabelian" version of "cokernel of the kernel"). These two constructions are categorically dual and so, among other things, the coimage-image factorization of $f$ in the opposite category is the same sequence of maps but in the opposite order.



In $textSet$, the coimage and image are both the image of a function in the usual sense, but computed in different ways, which I think match the distinction you're getting at. $textcoim(f)$ is computed, more or less, by constructing the equivalence relation on $X$ defined by $x_1 sim x_2 Leftrightarrow f(x_1) = f(x_2)$, then quotienting $X$ by it. $textim(f)$ is computed in a categorically dual way, although it looks a little strange at first: by first constructing the pushout $Y sqcup_X Y$, then isolating the subset of $Y$ of elements which are sent to the same element by both of the canonical maps $Y to Y sqcup_X Y$.



In particular, the factorization you want for an injection is the regular image factorization, and the factorization you want for a surjection is the regular coimage factorization, so they are in fact categorically dual. The full coimage-image factorization combines these.



It's a nontrivial theorem that the map $textcoim(f) to textim(f)$ is an isomorphism in $textSet$. It's also an isomorphism in any abelian category and in $textGrp$ (this is an abstract form of the first isomorphism theorem), but in general it's just both a monomorphism and an epimorphism. A very instructive example is $textTop$, where $textcoim(f)$ is the set-theoretic image topologized as a quotient of $X$, and $textim(f)$ is the set-theoretic image topologized as a subspace of $Y$. (Note that these coincide for compact Hausdorff spaces!)






share|cite|improve this answer


















  • 1




    But these things like im and coim are I believe defined by category-theoretic properties and are hence only defined up to unique isomorphism. Aren't I trying to do something better than isolating the object up to unique isomorphism? I'm wondering if in Set there is some sort of canonical representative (and am still not sure about if there is one).
    – Kevin Buzzard
    Dec 8 at 0:06










  • @Kevin: yes, I'm eliding some philosophical points here that I think Andrej's answer tackles better, at least in $textSet$. In particular I am really referring to a particular construction of the image and coimage rather than to anything which satisfies their universal properties, which is a sin I don't normally indulge in. My excuse is that I never just isolate myself to $textSet$ if I can help it; morally the coimage is "the image as seen from $X$" and the image is "the image as seen from $Y$," as I think the example of $textTop$ makes vividly clear, and so I think of them as...
    – Qiaochu Yuan
    Dec 8 at 0:10










  • ...conceptually distinct even though they happen to agree as universal constructions in $textSet$. Said another way, I'm either talking about your state of knowledge before you learn that $textcoim(f) to textim(f)$ is an isomorphism in $textSet$, or I'm reserving the right to replace $textSet$ with another category in which that's false. If you really want to stick to $textSet$ then Andrej's answer is probably more relevant than mine.
    – Qiaochu Yuan
    Dec 8 at 0:12














up vote
11
down vote













The nLab page you're looking for is called factorization systems. Here is my favorite one, which I think answers your question. In any category with finite limits and colimits, every morphism $f : X to Y$ has a canonical factorization



$$X to textcoim(f) to textim(f) to Y$$



where $textim(f)$, the regular image, is the equalizer of the cokernel pair of $f$ (this is the "nonabelian" version of "kernel of the cokernel") and $textcoim(f)$, the regular coimage, is the coequalizer of the kernel pair of $f$ (again, the "nonabelian" version of "cokernel of the kernel"). These two constructions are categorically dual and so, among other things, the coimage-image factorization of $f$ in the opposite category is the same sequence of maps but in the opposite order.



In $textSet$, the coimage and image are both the image of a function in the usual sense, but computed in different ways, which I think match the distinction you're getting at. $textcoim(f)$ is computed, more or less, by constructing the equivalence relation on $X$ defined by $x_1 sim x_2 Leftrightarrow f(x_1) = f(x_2)$, then quotienting $X$ by it. $textim(f)$ is computed in a categorically dual way, although it looks a little strange at first: by first constructing the pushout $Y sqcup_X Y$, then isolating the subset of $Y$ of elements which are sent to the same element by both of the canonical maps $Y to Y sqcup_X Y$.



In particular, the factorization you want for an injection is the regular image factorization, and the factorization you want for a surjection is the regular coimage factorization, so they are in fact categorically dual. The full coimage-image factorization combines these.



It's a nontrivial theorem that the map $textcoim(f) to textim(f)$ is an isomorphism in $textSet$. It's also an isomorphism in any abelian category and in $textGrp$ (this is an abstract form of the first isomorphism theorem), but in general it's just both a monomorphism and an epimorphism. A very instructive example is $textTop$, where $textcoim(f)$ is the set-theoretic image topologized as a quotient of $X$, and $textim(f)$ is the set-theoretic image topologized as a subspace of $Y$. (Note that these coincide for compact Hausdorff spaces!)






share|cite|improve this answer


















  • 1




    But these things like im and coim are I believe defined by category-theoretic properties and are hence only defined up to unique isomorphism. Aren't I trying to do something better than isolating the object up to unique isomorphism? I'm wondering if in Set there is some sort of canonical representative (and am still not sure about if there is one).
    – Kevin Buzzard
    Dec 8 at 0:06










  • @Kevin: yes, I'm eliding some philosophical points here that I think Andrej's answer tackles better, at least in $textSet$. In particular I am really referring to a particular construction of the image and coimage rather than to anything which satisfies their universal properties, which is a sin I don't normally indulge in. My excuse is that I never just isolate myself to $textSet$ if I can help it; morally the coimage is "the image as seen from $X$" and the image is "the image as seen from $Y$," as I think the example of $textTop$ makes vividly clear, and so I think of them as...
    – Qiaochu Yuan
    Dec 8 at 0:10










  • ...conceptually distinct even though they happen to agree as universal constructions in $textSet$. Said another way, I'm either talking about your state of knowledge before you learn that $textcoim(f) to textim(f)$ is an isomorphism in $textSet$, or I'm reserving the right to replace $textSet$ with another category in which that's false. If you really want to stick to $textSet$ then Andrej's answer is probably more relevant than mine.
    – Qiaochu Yuan
    Dec 8 at 0:12












up vote
11
down vote










up vote
11
down vote









The nLab page you're looking for is called factorization systems. Here is my favorite one, which I think answers your question. In any category with finite limits and colimits, every morphism $f : X to Y$ has a canonical factorization



$$X to textcoim(f) to textim(f) to Y$$



where $textim(f)$, the regular image, is the equalizer of the cokernel pair of $f$ (this is the "nonabelian" version of "kernel of the cokernel") and $textcoim(f)$, the regular coimage, is the coequalizer of the kernel pair of $f$ (again, the "nonabelian" version of "cokernel of the kernel"). These two constructions are categorically dual and so, among other things, the coimage-image factorization of $f$ in the opposite category is the same sequence of maps but in the opposite order.



In $textSet$, the coimage and image are both the image of a function in the usual sense, but computed in different ways, which I think match the distinction you're getting at. $textcoim(f)$ is computed, more or less, by constructing the equivalence relation on $X$ defined by $x_1 sim x_2 Leftrightarrow f(x_1) = f(x_2)$, then quotienting $X$ by it. $textim(f)$ is computed in a categorically dual way, although it looks a little strange at first: by first constructing the pushout $Y sqcup_X Y$, then isolating the subset of $Y$ of elements which are sent to the same element by both of the canonical maps $Y to Y sqcup_X Y$.



In particular, the factorization you want for an injection is the regular image factorization, and the factorization you want for a surjection is the regular coimage factorization, so they are in fact categorically dual. The full coimage-image factorization combines these.



It's a nontrivial theorem that the map $textcoim(f) to textim(f)$ is an isomorphism in $textSet$. It's also an isomorphism in any abelian category and in $textGrp$ (this is an abstract form of the first isomorphism theorem), but in general it's just both a monomorphism and an epimorphism. A very instructive example is $textTop$, where $textcoim(f)$ is the set-theoretic image topologized as a quotient of $X$, and $textim(f)$ is the set-theoretic image topologized as a subspace of $Y$. (Note that these coincide for compact Hausdorff spaces!)






share|cite|improve this answer














The nLab page you're looking for is called factorization systems. Here is my favorite one, which I think answers your question. In any category with finite limits and colimits, every morphism $f : X to Y$ has a canonical factorization



$$X to textcoim(f) to textim(f) to Y$$



where $textim(f)$, the regular image, is the equalizer of the cokernel pair of $f$ (this is the "nonabelian" version of "kernel of the cokernel") and $textcoim(f)$, the regular coimage, is the coequalizer of the kernel pair of $f$ (again, the "nonabelian" version of "cokernel of the kernel"). These two constructions are categorically dual and so, among other things, the coimage-image factorization of $f$ in the opposite category is the same sequence of maps but in the opposite order.



In $textSet$, the coimage and image are both the image of a function in the usual sense, but computed in different ways, which I think match the distinction you're getting at. $textcoim(f)$ is computed, more or less, by constructing the equivalence relation on $X$ defined by $x_1 sim x_2 Leftrightarrow f(x_1) = f(x_2)$, then quotienting $X$ by it. $textim(f)$ is computed in a categorically dual way, although it looks a little strange at first: by first constructing the pushout $Y sqcup_X Y$, then isolating the subset of $Y$ of elements which are sent to the same element by both of the canonical maps $Y to Y sqcup_X Y$.



In particular, the factorization you want for an injection is the regular image factorization, and the factorization you want for a surjection is the regular coimage factorization, so they are in fact categorically dual. The full coimage-image factorization combines these.



It's a nontrivial theorem that the map $textcoim(f) to textim(f)$ is an isomorphism in $textSet$. It's also an isomorphism in any abelian category and in $textGrp$ (this is an abstract form of the first isomorphism theorem), but in general it's just both a monomorphism and an epimorphism. A very instructive example is $textTop$, where $textcoim(f)$ is the set-theoretic image topologized as a quotient of $X$, and $textim(f)$ is the set-theoretic image topologized as a subspace of $Y$. (Note that these coincide for compact Hausdorff spaces!)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 at 22:42

























answered Dec 7 at 22:29









Qiaochu Yuan

76.7k25317598




76.7k25317598







  • 1




    But these things like im and coim are I believe defined by category-theoretic properties and are hence only defined up to unique isomorphism. Aren't I trying to do something better than isolating the object up to unique isomorphism? I'm wondering if in Set there is some sort of canonical representative (and am still not sure about if there is one).
    – Kevin Buzzard
    Dec 8 at 0:06










  • @Kevin: yes, I'm eliding some philosophical points here that I think Andrej's answer tackles better, at least in $textSet$. In particular I am really referring to a particular construction of the image and coimage rather than to anything which satisfies their universal properties, which is a sin I don't normally indulge in. My excuse is that I never just isolate myself to $textSet$ if I can help it; morally the coimage is "the image as seen from $X$" and the image is "the image as seen from $Y$," as I think the example of $textTop$ makes vividly clear, and so I think of them as...
    – Qiaochu Yuan
    Dec 8 at 0:10










  • ...conceptually distinct even though they happen to agree as universal constructions in $textSet$. Said another way, I'm either talking about your state of knowledge before you learn that $textcoim(f) to textim(f)$ is an isomorphism in $textSet$, or I'm reserving the right to replace $textSet$ with another category in which that's false. If you really want to stick to $textSet$ then Andrej's answer is probably more relevant than mine.
    – Qiaochu Yuan
    Dec 8 at 0:12












  • 1




    But these things like im and coim are I believe defined by category-theoretic properties and are hence only defined up to unique isomorphism. Aren't I trying to do something better than isolating the object up to unique isomorphism? I'm wondering if in Set there is some sort of canonical representative (and am still not sure about if there is one).
    – Kevin Buzzard
    Dec 8 at 0:06










  • @Kevin: yes, I'm eliding some philosophical points here that I think Andrej's answer tackles better, at least in $textSet$. In particular I am really referring to a particular construction of the image and coimage rather than to anything which satisfies their universal properties, which is a sin I don't normally indulge in. My excuse is that I never just isolate myself to $textSet$ if I can help it; morally the coimage is "the image as seen from $X$" and the image is "the image as seen from $Y$," as I think the example of $textTop$ makes vividly clear, and so I think of them as...
    – Qiaochu Yuan
    Dec 8 at 0:10










  • ...conceptually distinct even though they happen to agree as universal constructions in $textSet$. Said another way, I'm either talking about your state of knowledge before you learn that $textcoim(f) to textim(f)$ is an isomorphism in $textSet$, or I'm reserving the right to replace $textSet$ with another category in which that's false. If you really want to stick to $textSet$ then Andrej's answer is probably more relevant than mine.
    – Qiaochu Yuan
    Dec 8 at 0:12







1




1




But these things like im and coim are I believe defined by category-theoretic properties and are hence only defined up to unique isomorphism. Aren't I trying to do something better than isolating the object up to unique isomorphism? I'm wondering if in Set there is some sort of canonical representative (and am still not sure about if there is one).
– Kevin Buzzard
Dec 8 at 0:06




But these things like im and coim are I believe defined by category-theoretic properties and are hence only defined up to unique isomorphism. Aren't I trying to do something better than isolating the object up to unique isomorphism? I'm wondering if in Set there is some sort of canonical representative (and am still not sure about if there is one).
– Kevin Buzzard
Dec 8 at 0:06












@Kevin: yes, I'm eliding some philosophical points here that I think Andrej's answer tackles better, at least in $textSet$. In particular I am really referring to a particular construction of the image and coimage rather than to anything which satisfies their universal properties, which is a sin I don't normally indulge in. My excuse is that I never just isolate myself to $textSet$ if I can help it; morally the coimage is "the image as seen from $X$" and the image is "the image as seen from $Y$," as I think the example of $textTop$ makes vividly clear, and so I think of them as...
– Qiaochu Yuan
Dec 8 at 0:10




@Kevin: yes, I'm eliding some philosophical points here that I think Andrej's answer tackles better, at least in $textSet$. In particular I am really referring to a particular construction of the image and coimage rather than to anything which satisfies their universal properties, which is a sin I don't normally indulge in. My excuse is that I never just isolate myself to $textSet$ if I can help it; morally the coimage is "the image as seen from $X$" and the image is "the image as seen from $Y$," as I think the example of $textTop$ makes vividly clear, and so I think of them as...
– Qiaochu Yuan
Dec 8 at 0:10












...conceptually distinct even though they happen to agree as universal constructions in $textSet$. Said another way, I'm either talking about your state of knowledge before you learn that $textcoim(f) to textim(f)$ is an isomorphism in $textSet$, or I'm reserving the right to replace $textSet$ with another category in which that's false. If you really want to stick to $textSet$ then Andrej's answer is probably more relevant than mine.
– Qiaochu Yuan
Dec 8 at 0:12




...conceptually distinct even though they happen to agree as universal constructions in $textSet$. Said another way, I'm either talking about your state of knowledge before you learn that $textcoim(f) to textim(f)$ is an isomorphism in $textSet$, or I'm reserving the right to replace $textSet$ with another category in which that's false. If you really want to stick to $textSet$ then Andrej's answer is probably more relevant than mine.
– Qiaochu Yuan
Dec 8 at 0:12










up vote
9
down vote













Andrej's answer is a good one for explaining "what is special" about injections and quotients. In terms of formalizing the notion of "subset inclusion", David Roberts has already mentioned in a comment the notion of M-category, which could be used to represent "the category $mathrmSet$ together with the information about which injections are inclusions" (or which surjections are quotients).



Another formalization which is specifically adapted to "inclusion-like" subcategories is called (appropriately enough) a system of inclusions (with enhancements that add adjectives such as "directed" and "structural"); it was introduced in this paper by Awodey, Butz, Simpson, and Streicher. One could presumably dualize this somehow to obtain a notion of "system of quotients". Note that a system of inclusions is a particular kind of M-category.



As has been noted, when the category $rm Set$ is treated "purely category-theoretically" it does not "notice" which injections are inclusions, i.e. this information does not transfer usefully across an equivalence of categories. An M-category is a way of defining a "category-theoretic" structure (namely, a certain kind of enriched category) that nevertheless can carry this sort of information.






share|cite|improve this answer




















  • Thanks for the reference, I was aware of but failed to mention it.
    – Andrej Bauer
    Dec 9 at 9:08






  • 1




    However, the standard construction of quotients doesn't naively form an M-category, as the 'set of equivalence classes with a map to it' isn't transitive: one gets sets of sets of equivalence classes. Perhaps there is a way to fix this, but I didn't see it with a (small amount of) thinking.
    – David Roberts
    Dec 9 at 19:52










  • So "M-category" does not answer the question for surjections?
    – Kevin Buzzard
    Dec 9 at 22:18






  • 1




    @DavidRoberts Good point, I was too glib about that. I guess the dual notion of "system of quotients" would have to omit the composability if you want to include the usual notion of quotient set in $rm Set$. However, it occurs to me that "quotients" of setOIDS do compose, so at least $rm Set$ is equivalent to some category that has a system of quotients with composition.
    – Mike Shulman
    Dec 9 at 22:47














up vote
9
down vote













Andrej's answer is a good one for explaining "what is special" about injections and quotients. In terms of formalizing the notion of "subset inclusion", David Roberts has already mentioned in a comment the notion of M-category, which could be used to represent "the category $mathrmSet$ together with the information about which injections are inclusions" (or which surjections are quotients).



Another formalization which is specifically adapted to "inclusion-like" subcategories is called (appropriately enough) a system of inclusions (with enhancements that add adjectives such as "directed" and "structural"); it was introduced in this paper by Awodey, Butz, Simpson, and Streicher. One could presumably dualize this somehow to obtain a notion of "system of quotients". Note that a system of inclusions is a particular kind of M-category.



As has been noted, when the category $rm Set$ is treated "purely category-theoretically" it does not "notice" which injections are inclusions, i.e. this information does not transfer usefully across an equivalence of categories. An M-category is a way of defining a "category-theoretic" structure (namely, a certain kind of enriched category) that nevertheless can carry this sort of information.






share|cite|improve this answer




















  • Thanks for the reference, I was aware of but failed to mention it.
    – Andrej Bauer
    Dec 9 at 9:08






  • 1




    However, the standard construction of quotients doesn't naively form an M-category, as the 'set of equivalence classes with a map to it' isn't transitive: one gets sets of sets of equivalence classes. Perhaps there is a way to fix this, but I didn't see it with a (small amount of) thinking.
    – David Roberts
    Dec 9 at 19:52










  • So "M-category" does not answer the question for surjections?
    – Kevin Buzzard
    Dec 9 at 22:18






  • 1




    @DavidRoberts Good point, I was too glib about that. I guess the dual notion of "system of quotients" would have to omit the composability if you want to include the usual notion of quotient set in $rm Set$. However, it occurs to me that "quotients" of setOIDS do compose, so at least $rm Set$ is equivalent to some category that has a system of quotients with composition.
    – Mike Shulman
    Dec 9 at 22:47












up vote
9
down vote










up vote
9
down vote









Andrej's answer is a good one for explaining "what is special" about injections and quotients. In terms of formalizing the notion of "subset inclusion", David Roberts has already mentioned in a comment the notion of M-category, which could be used to represent "the category $mathrmSet$ together with the information about which injections are inclusions" (or which surjections are quotients).



Another formalization which is specifically adapted to "inclusion-like" subcategories is called (appropriately enough) a system of inclusions (with enhancements that add adjectives such as "directed" and "structural"); it was introduced in this paper by Awodey, Butz, Simpson, and Streicher. One could presumably dualize this somehow to obtain a notion of "system of quotients". Note that a system of inclusions is a particular kind of M-category.



As has been noted, when the category $rm Set$ is treated "purely category-theoretically" it does not "notice" which injections are inclusions, i.e. this information does not transfer usefully across an equivalence of categories. An M-category is a way of defining a "category-theoretic" structure (namely, a certain kind of enriched category) that nevertheless can carry this sort of information.






share|cite|improve this answer












Andrej's answer is a good one for explaining "what is special" about injections and quotients. In terms of formalizing the notion of "subset inclusion", David Roberts has already mentioned in a comment the notion of M-category, which could be used to represent "the category $mathrmSet$ together with the information about which injections are inclusions" (or which surjections are quotients).



Another formalization which is specifically adapted to "inclusion-like" subcategories is called (appropriately enough) a system of inclusions (with enhancements that add adjectives such as "directed" and "structural"); it was introduced in this paper by Awodey, Butz, Simpson, and Streicher. One could presumably dualize this somehow to obtain a notion of "system of quotients". Note that a system of inclusions is a particular kind of M-category.



As has been noted, when the category $rm Set$ is treated "purely category-theoretically" it does not "notice" which injections are inclusions, i.e. this information does not transfer usefully across an equivalence of categories. An M-category is a way of defining a "category-theoretic" structure (namely, a certain kind of enriched category) that nevertheless can carry this sort of information.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 at 19:19









Mike Shulman

35.7k480217




35.7k480217











  • Thanks for the reference, I was aware of but failed to mention it.
    – Andrej Bauer
    Dec 9 at 9:08






  • 1




    However, the standard construction of quotients doesn't naively form an M-category, as the 'set of equivalence classes with a map to it' isn't transitive: one gets sets of sets of equivalence classes. Perhaps there is a way to fix this, but I didn't see it with a (small amount of) thinking.
    – David Roberts
    Dec 9 at 19:52










  • So "M-category" does not answer the question for surjections?
    – Kevin Buzzard
    Dec 9 at 22:18






  • 1




    @DavidRoberts Good point, I was too glib about that. I guess the dual notion of "system of quotients" would have to omit the composability if you want to include the usual notion of quotient set in $rm Set$. However, it occurs to me that "quotients" of setOIDS do compose, so at least $rm Set$ is equivalent to some category that has a system of quotients with composition.
    – Mike Shulman
    Dec 9 at 22:47
















  • Thanks for the reference, I was aware of but failed to mention it.
    – Andrej Bauer
    Dec 9 at 9:08






  • 1




    However, the standard construction of quotients doesn't naively form an M-category, as the 'set of equivalence classes with a map to it' isn't transitive: one gets sets of sets of equivalence classes. Perhaps there is a way to fix this, but I didn't see it with a (small amount of) thinking.
    – David Roberts
    Dec 9 at 19:52










  • So "M-category" does not answer the question for surjections?
    – Kevin Buzzard
    Dec 9 at 22:18






  • 1




    @DavidRoberts Good point, I was too glib about that. I guess the dual notion of "system of quotients" would have to omit the composability if you want to include the usual notion of quotient set in $rm Set$. However, it occurs to me that "quotients" of setOIDS do compose, so at least $rm Set$ is equivalent to some category that has a system of quotients with composition.
    – Mike Shulman
    Dec 9 at 22:47















Thanks for the reference, I was aware of but failed to mention it.
– Andrej Bauer
Dec 9 at 9:08




Thanks for the reference, I was aware of but failed to mention it.
– Andrej Bauer
Dec 9 at 9:08




1




1




However, the standard construction of quotients doesn't naively form an M-category, as the 'set of equivalence classes with a map to it' isn't transitive: one gets sets of sets of equivalence classes. Perhaps there is a way to fix this, but I didn't see it with a (small amount of) thinking.
– David Roberts
Dec 9 at 19:52




However, the standard construction of quotients doesn't naively form an M-category, as the 'set of equivalence classes with a map to it' isn't transitive: one gets sets of sets of equivalence classes. Perhaps there is a way to fix this, but I didn't see it with a (small amount of) thinking.
– David Roberts
Dec 9 at 19:52












So "M-category" does not answer the question for surjections?
– Kevin Buzzard
Dec 9 at 22:18




So "M-category" does not answer the question for surjections?
– Kevin Buzzard
Dec 9 at 22:18




1




1




@DavidRoberts Good point, I was too glib about that. I guess the dual notion of "system of quotients" would have to omit the composability if you want to include the usual notion of quotient set in $rm Set$. However, it occurs to me that "quotients" of setOIDS do compose, so at least $rm Set$ is equivalent to some category that has a system of quotients with composition.
– Mike Shulman
Dec 9 at 22:47




@DavidRoberts Good point, I was too glib about that. I guess the dual notion of "system of quotients" would have to omit the composability if you want to include the usual notion of quotient set in $rm Set$. However, it occurs to me that "quotients" of setOIDS do compose, so at least $rm Set$ is equivalent to some category that has a system of quotients with composition.
– Mike Shulman
Dec 9 at 22:47










up vote
4
down vote














"Are inclusions in some way better than arbitrary injections?"




No they are not.

Anything that you can say about the one, you can say about the other.



In particular:
Given $X$, there is a set of inclusions into $X$. And there is a (large groupoid equivalent to a) set of injections into $X$. For all practical purposes these are the same thing. So they should be treated as the same thing.




Are quotient maps somehow better than arbitrary surjections?




Again no.

Same story.



    The fact that we can talk about inclusions versus injections is an artefact of our set theoretic foundations of mathematics. It would be better to use a language that disallows us from asking whether two sets are equal, and only allows us to talk about isomorphisms between sets.
    We largely already do this in practice. When we write $mathbb Z subset mathbb Q$, we don't stop to explain that elements of $mathbb Q$ are defined as pairs of elements of $mathbb Z$, and we don't distinguish between $mathbb Z$ and its image in $mathbb Q$. We just write $mathbb Z subset mathbb Q$.



    Unfortunately, I'm not knowledgeable enough in foundational issues in order to make sense of the notion of "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms". But I'm sure that such a thing exists (and that it can be tweaked so as to be exactly as powerful as ZFC).






share|cite|improve this answer




















  • "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms" One approach is FOLDS, by Makkai (cf ncatlab.org/nlab/show/FOLDS), and another approach was discussed by Bénabou projecteuclid.org/euclid.jsl/1183741771
    – David Roberts
    Dec 12 at 1:57










  • Sounds like univalent foundations to me.
    – Nik Weaver
    Dec 12 at 5:29














up vote
4
down vote














"Are inclusions in some way better than arbitrary injections?"




No they are not.

Anything that you can say about the one, you can say about the other.



In particular:
Given $X$, there is a set of inclusions into $X$. And there is a (large groupoid equivalent to a) set of injections into $X$. For all practical purposes these are the same thing. So they should be treated as the same thing.




Are quotient maps somehow better than arbitrary surjections?




Again no.

Same story.



    The fact that we can talk about inclusions versus injections is an artefact of our set theoretic foundations of mathematics. It would be better to use a language that disallows us from asking whether two sets are equal, and only allows us to talk about isomorphisms between sets.
    We largely already do this in practice. When we write $mathbb Z subset mathbb Q$, we don't stop to explain that elements of $mathbb Q$ are defined as pairs of elements of $mathbb Z$, and we don't distinguish between $mathbb Z$ and its image in $mathbb Q$. We just write $mathbb Z subset mathbb Q$.



    Unfortunately, I'm not knowledgeable enough in foundational issues in order to make sense of the notion of "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms". But I'm sure that such a thing exists (and that it can be tweaked so as to be exactly as powerful as ZFC).






share|cite|improve this answer




















  • "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms" One approach is FOLDS, by Makkai (cf ncatlab.org/nlab/show/FOLDS), and another approach was discussed by Bénabou projecteuclid.org/euclid.jsl/1183741771
    – David Roberts
    Dec 12 at 1:57










  • Sounds like univalent foundations to me.
    – Nik Weaver
    Dec 12 at 5:29












up vote
4
down vote










up vote
4
down vote










"Are inclusions in some way better than arbitrary injections?"




No they are not.

Anything that you can say about the one, you can say about the other.



In particular:
Given $X$, there is a set of inclusions into $X$. And there is a (large groupoid equivalent to a) set of injections into $X$. For all practical purposes these are the same thing. So they should be treated as the same thing.




Are quotient maps somehow better than arbitrary surjections?




Again no.

Same story.



    The fact that we can talk about inclusions versus injections is an artefact of our set theoretic foundations of mathematics. It would be better to use a language that disallows us from asking whether two sets are equal, and only allows us to talk about isomorphisms between sets.
    We largely already do this in practice. When we write $mathbb Z subset mathbb Q$, we don't stop to explain that elements of $mathbb Q$ are defined as pairs of elements of $mathbb Z$, and we don't distinguish between $mathbb Z$ and its image in $mathbb Q$. We just write $mathbb Z subset mathbb Q$.



    Unfortunately, I'm not knowledgeable enough in foundational issues in order to make sense of the notion of "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms". But I'm sure that such a thing exists (and that it can be tweaked so as to be exactly as powerful as ZFC).






share|cite|improve this answer













"Are inclusions in some way better than arbitrary injections?"




No they are not.

Anything that you can say about the one, you can say about the other.



In particular:
Given $X$, there is a set of inclusions into $X$. And there is a (large groupoid equivalent to a) set of injections into $X$. For all practical purposes these are the same thing. So they should be treated as the same thing.




Are quotient maps somehow better than arbitrary surjections?




Again no.

Same story.



    The fact that we can talk about inclusions versus injections is an artefact of our set theoretic foundations of mathematics. It would be better to use a language that disallows us from asking whether two sets are equal, and only allows us to talk about isomorphisms between sets.
    We largely already do this in practice. When we write $mathbb Z subset mathbb Q$, we don't stop to explain that elements of $mathbb Q$ are defined as pairs of elements of $mathbb Z$, and we don't distinguish between $mathbb Z$ and its image in $mathbb Q$. We just write $mathbb Z subset mathbb Q$.



    Unfortunately, I'm not knowledgeable enough in foundational issues in order to make sense of the notion of "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms". But I'm sure that such a thing exists (and that it can be tweaked so as to be exactly as powerful as ZFC).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 at 1:11









André Henriques

27.5k484210




27.5k484210











  • "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms" One approach is FOLDS, by Makkai (cf ncatlab.org/nlab/show/FOLDS), and another approach was discussed by Bénabou projecteuclid.org/euclid.jsl/1183741771
    – David Roberts
    Dec 12 at 1:57










  • Sounds like univalent foundations to me.
    – Nik Weaver
    Dec 12 at 5:29
















  • "a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms" One approach is FOLDS, by Makkai (cf ncatlab.org/nlab/show/FOLDS), and another approach was discussed by Bénabou projecteuclid.org/euclid.jsl/1183741771
    – David Roberts
    Dec 12 at 1:57










  • Sounds like univalent foundations to me.
    – Nik Weaver
    Dec 12 at 5:29















"a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms" One approach is FOLDS, by Makkai (cf ncatlab.org/nlab/show/FOLDS), and another approach was discussed by Bénabou projecteuclid.org/euclid.jsl/1183741771
– David Roberts
Dec 12 at 1:57




"a language that disallows asking whether two sets are equal, and only allows to talk about isomorphisms" One approach is FOLDS, by Makkai (cf ncatlab.org/nlab/show/FOLDS), and another approach was discussed by Bénabou projecteuclid.org/euclid.jsl/1183741771
– David Roberts
Dec 12 at 1:57












Sounds like univalent foundations to me.
– Nik Weaver
Dec 12 at 5:29




Sounds like univalent foundations to me.
– Nik Weaver
Dec 12 at 5:29

















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