When are quotients of homeomorphic spaces homeomorphic?

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Let $X$ and $Y$ be homeomorphic topological spaces, connected by the homeomorphism $f : X rightarrow Y$. Let $sim$ be an equivalence relation on $X$ and $approx$ be an equivalence relation on $Y$.



I would think that there would be some structure that I could place on the equivalence relations $sim$ and $approx$ that would allow me to construct, using the homeomorphism $f$, a homeomorphism $barf$ from $X/sim$ to $Y/approx$. I suspect that that relationship is $a sim b leftrightarrow f(a) approx f(b), forall a, b in X$.



That is, I suspect that "equivalent" quotients of homeomorphic spaces are homeomorphic, but I don't know exactly how to formulate this or prove it.



(This result, by the way, to me, seems to hinge on the existence of a "reverse" universal quotient property: juts as functions $f : X/sim rightarrow Y$ correspond uniquely to a subset of functions $f : X rightarrow Y$, I suspect that functions $f : X rightarrow Y/approx$ correspond uniquely to some subset of functions $f : X rightarrow Y$, but I don't know how this works either.)










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  • 1




    There is no such "reverse universal property". For instance you have many maps $S^1to S^1$ that don't factor through the quotient map $mathbbRto S^1$
    – Max
    Dec 15 at 15:59










  • Hmm, interesting. Is there a familiar subset of maps that do factor through in this fashion? I wonder if the question I'm asking here is best answered topologically or categorically.
    – Billy Smith
    Dec 15 at 17:59











  • In the specific case of $mathbbRto S^1$ the maps $Xto S^1$ that factor through it are precisely the nullhomotopic maps if $X$ is connected and locally path-connected. I don't know if there's a criterion for general quotient maps, but I don't think there is.
    – Max
    Dec 15 at 18:04















5














Let $X$ and $Y$ be homeomorphic topological spaces, connected by the homeomorphism $f : X rightarrow Y$. Let $sim$ be an equivalence relation on $X$ and $approx$ be an equivalence relation on $Y$.



I would think that there would be some structure that I could place on the equivalence relations $sim$ and $approx$ that would allow me to construct, using the homeomorphism $f$, a homeomorphism $barf$ from $X/sim$ to $Y/approx$. I suspect that that relationship is $a sim b leftrightarrow f(a) approx f(b), forall a, b in X$.



That is, I suspect that "equivalent" quotients of homeomorphic spaces are homeomorphic, but I don't know exactly how to formulate this or prove it.



(This result, by the way, to me, seems to hinge on the existence of a "reverse" universal quotient property: juts as functions $f : X/sim rightarrow Y$ correspond uniquely to a subset of functions $f : X rightarrow Y$, I suspect that functions $f : X rightarrow Y/approx$ correspond uniquely to some subset of functions $f : X rightarrow Y$, but I don't know how this works either.)










share|cite|improve this question

















  • 1




    There is no such "reverse universal property". For instance you have many maps $S^1to S^1$ that don't factor through the quotient map $mathbbRto S^1$
    – Max
    Dec 15 at 15:59










  • Hmm, interesting. Is there a familiar subset of maps that do factor through in this fashion? I wonder if the question I'm asking here is best answered topologically or categorically.
    – Billy Smith
    Dec 15 at 17:59











  • In the specific case of $mathbbRto S^1$ the maps $Xto S^1$ that factor through it are precisely the nullhomotopic maps if $X$ is connected and locally path-connected. I don't know if there's a criterion for general quotient maps, but I don't think there is.
    – Max
    Dec 15 at 18:04













5












5








5


1





Let $X$ and $Y$ be homeomorphic topological spaces, connected by the homeomorphism $f : X rightarrow Y$. Let $sim$ be an equivalence relation on $X$ and $approx$ be an equivalence relation on $Y$.



I would think that there would be some structure that I could place on the equivalence relations $sim$ and $approx$ that would allow me to construct, using the homeomorphism $f$, a homeomorphism $barf$ from $X/sim$ to $Y/approx$. I suspect that that relationship is $a sim b leftrightarrow f(a) approx f(b), forall a, b in X$.



That is, I suspect that "equivalent" quotients of homeomorphic spaces are homeomorphic, but I don't know exactly how to formulate this or prove it.



(This result, by the way, to me, seems to hinge on the existence of a "reverse" universal quotient property: juts as functions $f : X/sim rightarrow Y$ correspond uniquely to a subset of functions $f : X rightarrow Y$, I suspect that functions $f : X rightarrow Y/approx$ correspond uniquely to some subset of functions $f : X rightarrow Y$, but I don't know how this works either.)










share|cite|improve this question













Let $X$ and $Y$ be homeomorphic topological spaces, connected by the homeomorphism $f : X rightarrow Y$. Let $sim$ be an equivalence relation on $X$ and $approx$ be an equivalence relation on $Y$.



I would think that there would be some structure that I could place on the equivalence relations $sim$ and $approx$ that would allow me to construct, using the homeomorphism $f$, a homeomorphism $barf$ from $X/sim$ to $Y/approx$. I suspect that that relationship is $a sim b leftrightarrow f(a) approx f(b), forall a, b in X$.



That is, I suspect that "equivalent" quotients of homeomorphic spaces are homeomorphic, but I don't know exactly how to formulate this or prove it.



(This result, by the way, to me, seems to hinge on the existence of a "reverse" universal quotient property: juts as functions $f : X/sim rightarrow Y$ correspond uniquely to a subset of functions $f : X rightarrow Y$, I suspect that functions $f : X rightarrow Y/approx$ correspond uniquely to some subset of functions $f : X rightarrow Y$, but I don't know how this works either.)







general-topology category-theory






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asked Dec 15 at 14:27









Billy Smith

685




685







  • 1




    There is no such "reverse universal property". For instance you have many maps $S^1to S^1$ that don't factor through the quotient map $mathbbRto S^1$
    – Max
    Dec 15 at 15:59










  • Hmm, interesting. Is there a familiar subset of maps that do factor through in this fashion? I wonder if the question I'm asking here is best answered topologically or categorically.
    – Billy Smith
    Dec 15 at 17:59











  • In the specific case of $mathbbRto S^1$ the maps $Xto S^1$ that factor through it are precisely the nullhomotopic maps if $X$ is connected and locally path-connected. I don't know if there's a criterion for general quotient maps, but I don't think there is.
    – Max
    Dec 15 at 18:04












  • 1




    There is no such "reverse universal property". For instance you have many maps $S^1to S^1$ that don't factor through the quotient map $mathbbRto S^1$
    – Max
    Dec 15 at 15:59










  • Hmm, interesting. Is there a familiar subset of maps that do factor through in this fashion? I wonder if the question I'm asking here is best answered topologically or categorically.
    – Billy Smith
    Dec 15 at 17:59











  • In the specific case of $mathbbRto S^1$ the maps $Xto S^1$ that factor through it are precisely the nullhomotopic maps if $X$ is connected and locally path-connected. I don't know if there's a criterion for general quotient maps, but I don't think there is.
    – Max
    Dec 15 at 18:04







1




1




There is no such "reverse universal property". For instance you have many maps $S^1to S^1$ that don't factor through the quotient map $mathbbRto S^1$
– Max
Dec 15 at 15:59




There is no such "reverse universal property". For instance you have many maps $S^1to S^1$ that don't factor through the quotient map $mathbbRto S^1$
– Max
Dec 15 at 15:59












Hmm, interesting. Is there a familiar subset of maps that do factor through in this fashion? I wonder if the question I'm asking here is best answered topologically or categorically.
– Billy Smith
Dec 15 at 17:59





Hmm, interesting. Is there a familiar subset of maps that do factor through in this fashion? I wonder if the question I'm asking here is best answered topologically or categorically.
– Billy Smith
Dec 15 at 17:59













In the specific case of $mathbbRto S^1$ the maps $Xto S^1$ that factor through it are precisely the nullhomotopic maps if $X$ is connected and locally path-connected. I don't know if there's a criterion for general quotient maps, but I don't think there is.
– Max
Dec 15 at 18:04




In the specific case of $mathbbRto S^1$ the maps $Xto S^1$ that factor through it are precisely the nullhomotopic maps if $X$ is connected and locally path-connected. I don't know if there's a criterion for general quotient maps, but I don't think there is.
– Max
Dec 15 at 18:04










3 Answers
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Your suspicion is correct. If $X$ and $Y$ are topological spaces with equivalence relations $sim$ and $approx$ and $f:Xto Y$ is a homeomorphism such that $xsim y$ if and only if $f(x)approx f(y)$, then $f$ passes to the quotient as $X/_sim cong Y/_approx$.



Proof goes roughly like this: define $widetildef([x]_sim)doteq [f(x)]_approx$. We have that $widetildef$ is well-defined and injective because of the condition relating $sim$ and $approx$, while surjectivity follows from the surjectivity of $f$. Continuity of $widetildef$ and $widetildef^-1$ follow from the universal property of the quotient topologies applied to the relations $widetildefcircpi_sim=pi_approxcirc f$ and $widetildef^-1circ pi_approx=pi_simcirc f^-1$ with the continuity of $f$ and $f^-1$.






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    3














    I think you are correct. Here is a sketch of the argument:



    $phi:=pi_Ycirc f$ is a continuous surjection: $Xto Y/approx$ and it induces a bijection: $X^*=left phi^-1([y]): [y]in Y/approx right oversetoverlinephirightarrow Y/approx , $ which is a homeomorphism if and only if $phi$ is a quotient map.



    As $phi $ is indeed a quotient map, the induced map is a homeomorphism. So, $X/sim$ and $Y/approx$ will be homoeomorphic if and only if $X^*$ and $X/sim$ are.



    This means that the map $[x]mapsto f^-1circ pi^-1_Y([y])$ must be a homeomorphism. That is, $f([x])mapsto pi^-1_Y([y])$ must be a homeomorphism.



    But the above map is well-defined if and only if $xsim x'Rightarrow f(x)approx f(x'),$ and in this case, it is obviously continuous, and it should be easy to check that it is bijective with continuous inverse.






    share|cite|improve this answer






























      3














      For a categorical argument, a topological space equipped with an equivalence relation can be viewed as a parallel pair of arrows $Rrightrightarrows X$, where $Rsubset Xtimes X$ is the subspace consisting of the related pairs of points. Then the claim follows from reinterpreting your condition that $f$ preserve and reflect the equivalence relation as asking that $f$ extend to a natural isomorphism between the diagrams $Rrightrightarrows X$ and $Srightrightarrows Y$. Then your claim is a special case of the fact that naturally isomorphic diagrams have naturally isomorphic colimits, which follows from the fact that colimits are functorial.



      This perspective clarifies that this really has nothing to do with equivalence relations, nor with topological spaces. As far as the technical details of the proof, any map $f:Xto Y$ induces a natural transformation between the diagrams $Xtimes Xrightrightarrows X$ and $Ytimes Yrightrightarrows Y$, so one is reduced to proving that $f$ induces a bijection $Rtimes Rto Stimes S$, which is exactly your assumption. There's no need to say anything more nitty-gritty.






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        3 Answers
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        3 Answers
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        Your suspicion is correct. If $X$ and $Y$ are topological spaces with equivalence relations $sim$ and $approx$ and $f:Xto Y$ is a homeomorphism such that $xsim y$ if and only if $f(x)approx f(y)$, then $f$ passes to the quotient as $X/_sim cong Y/_approx$.



        Proof goes roughly like this: define $widetildef([x]_sim)doteq [f(x)]_approx$. We have that $widetildef$ is well-defined and injective because of the condition relating $sim$ and $approx$, while surjectivity follows from the surjectivity of $f$. Continuity of $widetildef$ and $widetildef^-1$ follow from the universal property of the quotient topologies applied to the relations $widetildefcircpi_sim=pi_approxcirc f$ and $widetildef^-1circ pi_approx=pi_simcirc f^-1$ with the continuity of $f$ and $f^-1$.






        share|cite|improve this answer

























          4














          Your suspicion is correct. If $X$ and $Y$ are topological spaces with equivalence relations $sim$ and $approx$ and $f:Xto Y$ is a homeomorphism such that $xsim y$ if and only if $f(x)approx f(y)$, then $f$ passes to the quotient as $X/_sim cong Y/_approx$.



          Proof goes roughly like this: define $widetildef([x]_sim)doteq [f(x)]_approx$. We have that $widetildef$ is well-defined and injective because of the condition relating $sim$ and $approx$, while surjectivity follows from the surjectivity of $f$. Continuity of $widetildef$ and $widetildef^-1$ follow from the universal property of the quotient topologies applied to the relations $widetildefcircpi_sim=pi_approxcirc f$ and $widetildef^-1circ pi_approx=pi_simcirc f^-1$ with the continuity of $f$ and $f^-1$.






          share|cite|improve this answer























            4












            4








            4






            Your suspicion is correct. If $X$ and $Y$ are topological spaces with equivalence relations $sim$ and $approx$ and $f:Xto Y$ is a homeomorphism such that $xsim y$ if and only if $f(x)approx f(y)$, then $f$ passes to the quotient as $X/_sim cong Y/_approx$.



            Proof goes roughly like this: define $widetildef([x]_sim)doteq [f(x)]_approx$. We have that $widetildef$ is well-defined and injective because of the condition relating $sim$ and $approx$, while surjectivity follows from the surjectivity of $f$. Continuity of $widetildef$ and $widetildef^-1$ follow from the universal property of the quotient topologies applied to the relations $widetildefcircpi_sim=pi_approxcirc f$ and $widetildef^-1circ pi_approx=pi_simcirc f^-1$ with the continuity of $f$ and $f^-1$.






            share|cite|improve this answer












            Your suspicion is correct. If $X$ and $Y$ are topological spaces with equivalence relations $sim$ and $approx$ and $f:Xto Y$ is a homeomorphism such that $xsim y$ if and only if $f(x)approx f(y)$, then $f$ passes to the quotient as $X/_sim cong Y/_approx$.



            Proof goes roughly like this: define $widetildef([x]_sim)doteq [f(x)]_approx$. We have that $widetildef$ is well-defined and injective because of the condition relating $sim$ and $approx$, while surjectivity follows from the surjectivity of $f$. Continuity of $widetildef$ and $widetildef^-1$ follow from the universal property of the quotient topologies applied to the relations $widetildefcircpi_sim=pi_approxcirc f$ and $widetildef^-1circ pi_approx=pi_simcirc f^-1$ with the continuity of $f$ and $f^-1$.







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            answered Dec 15 at 15:31









            Ivo Terek

            45.2k951139




            45.2k951139





















                3














                I think you are correct. Here is a sketch of the argument:



                $phi:=pi_Ycirc f$ is a continuous surjection: $Xto Y/approx$ and it induces a bijection: $X^*=left phi^-1([y]): [y]in Y/approx right oversetoverlinephirightarrow Y/approx , $ which is a homeomorphism if and only if $phi$ is a quotient map.



                As $phi $ is indeed a quotient map, the induced map is a homeomorphism. So, $X/sim$ and $Y/approx$ will be homoeomorphic if and only if $X^*$ and $X/sim$ are.



                This means that the map $[x]mapsto f^-1circ pi^-1_Y([y])$ must be a homeomorphism. That is, $f([x])mapsto pi^-1_Y([y])$ must be a homeomorphism.



                But the above map is well-defined if and only if $xsim x'Rightarrow f(x)approx f(x'),$ and in this case, it is obviously continuous, and it should be easy to check that it is bijective with continuous inverse.






                share|cite|improve this answer



























                  3














                  I think you are correct. Here is a sketch of the argument:



                  $phi:=pi_Ycirc f$ is a continuous surjection: $Xto Y/approx$ and it induces a bijection: $X^*=left phi^-1([y]): [y]in Y/approx right oversetoverlinephirightarrow Y/approx , $ which is a homeomorphism if and only if $phi$ is a quotient map.



                  As $phi $ is indeed a quotient map, the induced map is a homeomorphism. So, $X/sim$ and $Y/approx$ will be homoeomorphic if and only if $X^*$ and $X/sim$ are.



                  This means that the map $[x]mapsto f^-1circ pi^-1_Y([y])$ must be a homeomorphism. That is, $f([x])mapsto pi^-1_Y([y])$ must be a homeomorphism.



                  But the above map is well-defined if and only if $xsim x'Rightarrow f(x)approx f(x'),$ and in this case, it is obviously continuous, and it should be easy to check that it is bijective with continuous inverse.






                  share|cite|improve this answer

























                    3












                    3








                    3






                    I think you are correct. Here is a sketch of the argument:



                    $phi:=pi_Ycirc f$ is a continuous surjection: $Xto Y/approx$ and it induces a bijection: $X^*=left phi^-1([y]): [y]in Y/approx right oversetoverlinephirightarrow Y/approx , $ which is a homeomorphism if and only if $phi$ is a quotient map.



                    As $phi $ is indeed a quotient map, the induced map is a homeomorphism. So, $X/sim$ and $Y/approx$ will be homoeomorphic if and only if $X^*$ and $X/sim$ are.



                    This means that the map $[x]mapsto f^-1circ pi^-1_Y([y])$ must be a homeomorphism. That is, $f([x])mapsto pi^-1_Y([y])$ must be a homeomorphism.



                    But the above map is well-defined if and only if $xsim x'Rightarrow f(x)approx f(x'),$ and in this case, it is obviously continuous, and it should be easy to check that it is bijective with continuous inverse.






                    share|cite|improve this answer














                    I think you are correct. Here is a sketch of the argument:



                    $phi:=pi_Ycirc f$ is a continuous surjection: $Xto Y/approx$ and it induces a bijection: $X^*=left phi^-1([y]): [y]in Y/approx right oversetoverlinephirightarrow Y/approx , $ which is a homeomorphism if and only if $phi$ is a quotient map.



                    As $phi $ is indeed a quotient map, the induced map is a homeomorphism. So, $X/sim$ and $Y/approx$ will be homoeomorphic if and only if $X^*$ and $X/sim$ are.



                    This means that the map $[x]mapsto f^-1circ pi^-1_Y([y])$ must be a homeomorphism. That is, $f([x])mapsto pi^-1_Y([y])$ must be a homeomorphism.



                    But the above map is well-defined if and only if $xsim x'Rightarrow f(x)approx f(x'),$ and in this case, it is obviously continuous, and it should be easy to check that it is bijective with continuous inverse.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 15 at 17:36

























                    answered Dec 15 at 16:33









                    Matematleta

                    9,8722918




                    9,8722918





















                        3














                        For a categorical argument, a topological space equipped with an equivalence relation can be viewed as a parallel pair of arrows $Rrightrightarrows X$, where $Rsubset Xtimes X$ is the subspace consisting of the related pairs of points. Then the claim follows from reinterpreting your condition that $f$ preserve and reflect the equivalence relation as asking that $f$ extend to a natural isomorphism between the diagrams $Rrightrightarrows X$ and $Srightrightarrows Y$. Then your claim is a special case of the fact that naturally isomorphic diagrams have naturally isomorphic colimits, which follows from the fact that colimits are functorial.



                        This perspective clarifies that this really has nothing to do with equivalence relations, nor with topological spaces. As far as the technical details of the proof, any map $f:Xto Y$ induces a natural transformation between the diagrams $Xtimes Xrightrightarrows X$ and $Ytimes Yrightrightarrows Y$, so one is reduced to proving that $f$ induces a bijection $Rtimes Rto Stimes S$, which is exactly your assumption. There's no need to say anything more nitty-gritty.






                        share|cite|improve this answer

























                          3














                          For a categorical argument, a topological space equipped with an equivalence relation can be viewed as a parallel pair of arrows $Rrightrightarrows X$, where $Rsubset Xtimes X$ is the subspace consisting of the related pairs of points. Then the claim follows from reinterpreting your condition that $f$ preserve and reflect the equivalence relation as asking that $f$ extend to a natural isomorphism between the diagrams $Rrightrightarrows X$ and $Srightrightarrows Y$. Then your claim is a special case of the fact that naturally isomorphic diagrams have naturally isomorphic colimits, which follows from the fact that colimits are functorial.



                          This perspective clarifies that this really has nothing to do with equivalence relations, nor with topological spaces. As far as the technical details of the proof, any map $f:Xto Y$ induces a natural transformation between the diagrams $Xtimes Xrightrightarrows X$ and $Ytimes Yrightrightarrows Y$, so one is reduced to proving that $f$ induces a bijection $Rtimes Rto Stimes S$, which is exactly your assumption. There's no need to say anything more nitty-gritty.






                          share|cite|improve this answer























                            3












                            3








                            3






                            For a categorical argument, a topological space equipped with an equivalence relation can be viewed as a parallel pair of arrows $Rrightrightarrows X$, where $Rsubset Xtimes X$ is the subspace consisting of the related pairs of points. Then the claim follows from reinterpreting your condition that $f$ preserve and reflect the equivalence relation as asking that $f$ extend to a natural isomorphism between the diagrams $Rrightrightarrows X$ and $Srightrightarrows Y$. Then your claim is a special case of the fact that naturally isomorphic diagrams have naturally isomorphic colimits, which follows from the fact that colimits are functorial.



                            This perspective clarifies that this really has nothing to do with equivalence relations, nor with topological spaces. As far as the technical details of the proof, any map $f:Xto Y$ induces a natural transformation between the diagrams $Xtimes Xrightrightarrows X$ and $Ytimes Yrightrightarrows Y$, so one is reduced to proving that $f$ induces a bijection $Rtimes Rto Stimes S$, which is exactly your assumption. There's no need to say anything more nitty-gritty.






                            share|cite|improve this answer












                            For a categorical argument, a topological space equipped with an equivalence relation can be viewed as a parallel pair of arrows $Rrightrightarrows X$, where $Rsubset Xtimes X$ is the subspace consisting of the related pairs of points. Then the claim follows from reinterpreting your condition that $f$ preserve and reflect the equivalence relation as asking that $f$ extend to a natural isomorphism between the diagrams $Rrightrightarrows X$ and $Srightrightarrows Y$. Then your claim is a special case of the fact that naturally isomorphic diagrams have naturally isomorphic colimits, which follows from the fact that colimits are functorial.



                            This perspective clarifies that this really has nothing to do with equivalence relations, nor with topological spaces. As far as the technical details of the proof, any map $f:Xto Y$ induces a natural transformation between the diagrams $Xtimes Xrightrightarrows X$ and $Ytimes Yrightrightarrows Y$, so one is reduced to proving that $f$ induces a bijection $Rtimes Rto Stimes S$, which is exactly your assumption. There's no need to say anything more nitty-gritty.







                            share|cite|improve this answer












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                            share|cite|improve this answer










                            answered Dec 15 at 22:56









                            Kevin Carlson

                            32.4k23271




                            32.4k23271



























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