Homotopy type of a specific discrete monoid
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is always onto).
What do we know about the homotopy type of this monoid (viewed as a
one-object category) ? In particular, about its homotopy groups ?
My background on this subject is very small. By a paper from Dusa McDuff (On the classifying space of discrete monoids), every path-connected space has the same homotopy type as the classifying space of some monoid, and the fundamental group of $BM$ is the groupification of $M$.
EDIT: I am a bit confused between the English meaning and the French meaning of nondecreasing. The monoid I am talking about is not a group because a nondecreasing map preserving extremities is not necessarily one-to-one. I hope that this clarification will be helpful.
at.algebraic-topology monoids classifying-spaces
edited Oct 1 at 18:00
John Feminella
1032
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
add a comment |Â
up vote
6
down vote
favorite
Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is always onto).
What do we know about the homotopy type of this monoid (viewed as a
one-object category) ? In particular, about its homotopy groups ?
My background on this subject is very small. By a paper from Dusa McDuff (On the classifying space of discrete monoids), every path-connected space has the same homotopy type as the classifying space of some monoid, and the fundamental group of $BM$ is the groupification of $M$.
EDIT: I am a bit confused between the English meaning and the French meaning of nondecreasing. The monoid I am talking about is not a group because a nondecreasing map preserving extremities is not necessarily one-to-one. I hope that this clarification will be helpful.
at.algebraic-topology monoids classifying-spaces
edited Oct 1 at 18:00
John Feminella
1032
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
2
This seems a complicated way of asking what's the homotopy type of $BM$. I'm not aware of any tool beyond the group-completion theorem to do this kind of analysis though
– Denis Nardin
Oct 1 at 7:16
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
Oct 1 at 7:42
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is always onto).
What do we know about the homotopy type of this monoid (viewed as a
one-object category) ? In particular, about its homotopy groups ?
My background on this subject is very small. By a paper from Dusa McDuff (On the classifying space of discrete monoids), every path-connected space has the same homotopy type as the classifying space of some monoid, and the fundamental group of $BM$ is the groupification of $M$.
EDIT: I am a bit confused between the English meaning and the French meaning of nondecreasing. The monoid I am talking about is not a group because a nondecreasing map preserving extremities is not necessarily one-to-one. I hope that this clarification will be helpful.
at.algebraic-topology monoids classifying-spaces
edited Oct 1 at 18:00
John Feminella
1032
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is always onto).
What do we know about the homotopy type of this monoid (viewed as a
one-object category) ? In particular, about its homotopy groups ?
My background on this subject is very small. By a paper from Dusa McDuff (On the classifying space of discrete monoids), every path-connected space has the same homotopy type as the classifying space of some monoid, and the fundamental group of $BM$ is the groupification of $M$.
EDIT: I am a bit confused between the English meaning and the French meaning of nondecreasing. The monoid I am talking about is not a group because a nondecreasing map preserving extremities is not necessarily one-to-one. I hope that this clarification will be helpful.
at.algebraic-topology monoids classifying-spaces
at.algebraic-topology monoids classifying-spaces
edited Oct 1 at 18:00
John Feminella
1032
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
edited Oct 1 at 18:00
John Feminella
1032
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
edited Oct 1 at 18:00
John Feminella
1032
edited Oct 1 at 18:00
John Feminella
1032
edited Oct 1 at 18:00
John Feminella
1032
1032
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
asked Oct 1 at 6:50
Philippe Gaucher
1,607919
1,607919
2
This seems a complicated way of asking what's the homotopy type of $BM$. I'm not aware of any tool beyond the group-completion theorem to do this kind of analysis though
– Denis Nardin
Oct 1 at 7:16
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
Oct 1 at 7:42
add a comment |Â
2
This seems a complicated way of asking what's the homotopy type of $BM$. I'm not aware of any tool beyond the group-completion theorem to do this kind of analysis though
– Denis Nardin
Oct 1 at 7:16
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
Oct 1 at 7:42
2
2
This seems a complicated way of asking what's the homotopy type of $BM$. I'm not aware of any tool beyond the group-completion theorem to do this kind of analysis though
– Denis Nardin
Oct 1 at 7:16
This seems a complicated way of asking what's the homotopy type of $BM$. I'm not aware of any tool beyond the group-completion theorem to do this kind of analysis though
– Denis Nardin
Oct 1 at 7:16
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
Oct 1 at 7:42
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
Oct 1 at 7:42
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
11
down vote
accepted
This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
beginalign*
A(x) &=
begincases 2x &textif x leq 1/2\1 &textif x geq 1/2endcases\
B(x) &=
begincases 0 &textif x leq 1/2\2x-1 &textif x geq 1/2endcases
endalign*
$$
Define three monoid homomorphisms $Id, U, V: M to M$ by:
$$
beginalign*
(Id(f))(x) &= f(x)\
(Uf)(x) &=
begincases tfrac12f(2x) &textif x leq 1/2\x &textif x geq 1/2endcases\
(Vf)(x) &= x
endalign*
$$
For any $f in M$, we have the following identities:
$$
beginalign*
A circ (Uf) &= f circ A\
B circ (Uf) &= (Vf) circ B
endalign*
$$
As a result, we can reinterpret this in terms of the one-object category $M$: we get three functors $Id,U,V: M to M$ and natural transformations $A: U to I$ and $B: U to V$.
Upon taking geometric realization, we get a space $BM$, these functors turn into continuous maps $Id, U, V:BM to BM$ and homotopies from $U$ to $Id$ and from $U$ to $V$. However, $V$ is a constant map, and so this says that the homotopy type of $BM$ is contractible.
(I believe that I learned this from somewhere in a paper of Lurie's, but I can't find it currently.)
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
4
Counterexample 4.4.5.19, HTT?
– Charles Rezk
Oct 1 at 13:32
@Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it...
– Tyler Lawson
Oct 1 at 17:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
beginalign*
A(x) &=
begincases 2x &textif x leq 1/2\1 &textif x geq 1/2endcases\
B(x) &=
begincases 0 &textif x leq 1/2\2x-1 &textif x geq 1/2endcases
endalign*
$$
Define three monoid homomorphisms $Id, U, V: M to M$ by:
$$
beginalign*
(Id(f))(x) &= f(x)\
(Uf)(x) &=
begincases tfrac12f(2x) &textif x leq 1/2\x &textif x geq 1/2endcases\
(Vf)(x) &= x
endalign*
$$
For any $f in M$, we have the following identities:
$$
beginalign*
A circ (Uf) &= f circ A\
B circ (Uf) &= (Vf) circ B
endalign*
$$
As a result, we can reinterpret this in terms of the one-object category $M$: we get three functors $Id,U,V: M to M$ and natural transformations $A: U to I$ and $B: U to V$.
Upon taking geometric realization, we get a space $BM$, these functors turn into continuous maps $Id, U, V:BM to BM$ and homotopies from $U$ to $Id$ and from $U$ to $V$. However, $V$ is a constant map, and so this says that the homotopy type of $BM$ is contractible.
(I believe that I learned this from somewhere in a paper of Lurie's, but I can't find it currently.)
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
4
Counterexample 4.4.5.19, HTT?
– Charles Rezk
Oct 1 at 13:32
@Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it...
– Tyler Lawson
Oct 1 at 17:08
add a comment |Â
up vote
11
down vote
accepted
This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
beginalign*
A(x) &=
begincases 2x &textif x leq 1/2\1 &textif x geq 1/2endcases\
B(x) &=
begincases 0 &textif x leq 1/2\2x-1 &textif x geq 1/2endcases
endalign*
$$
Define three monoid homomorphisms $Id, U, V: M to M$ by:
$$
beginalign*
(Id(f))(x) &= f(x)\
(Uf)(x) &=
begincases tfrac12f(2x) &textif x leq 1/2\x &textif x geq 1/2endcases\
(Vf)(x) &= x
endalign*
$$
For any $f in M$, we have the following identities:
$$
beginalign*
A circ (Uf) &= f circ A\
B circ (Uf) &= (Vf) circ B
endalign*
$$
As a result, we can reinterpret this in terms of the one-object category $M$: we get three functors $Id,U,V: M to M$ and natural transformations $A: U to I$ and $B: U to V$.
Upon taking geometric realization, we get a space $BM$, these functors turn into continuous maps $Id, U, V:BM to BM$ and homotopies from $U$ to $Id$ and from $U$ to $V$. However, $V$ is a constant map, and so this says that the homotopy type of $BM$ is contractible.
(I believe that I learned this from somewhere in a paper of Lurie's, but I can't find it currently.)
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
4
Counterexample 4.4.5.19, HTT?
– Charles Rezk
Oct 1 at 13:32
@Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it...
– Tyler Lawson
Oct 1 at 17:08
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
beginalign*
A(x) &=
begincases 2x &textif x leq 1/2\1 &textif x geq 1/2endcases\
B(x) &=
begincases 0 &textif x leq 1/2\2x-1 &textif x geq 1/2endcases
endalign*
$$
Define three monoid homomorphisms $Id, U, V: M to M$ by:
$$
beginalign*
(Id(f))(x) &= f(x)\
(Uf)(x) &=
begincases tfrac12f(2x) &textif x leq 1/2\x &textif x geq 1/2endcases\
(Vf)(x) &= x
endalign*
$$
For any $f in M$, we have the following identities:
$$
beginalign*
A circ (Uf) &= f circ A\
B circ (Uf) &= (Vf) circ B
endalign*
$$
As a result, we can reinterpret this in terms of the one-object category $M$: we get three functors $Id,U,V: M to M$ and natural transformations $A: U to I$ and $B: U to V$.
Upon taking geometric realization, we get a space $BM$, these functors turn into continuous maps $Id, U, V:BM to BM$ and homotopies from $U$ to $Id$ and from $U$ to $V$. However, $V$ is a constant map, and so this says that the homotopy type of $BM$ is contractible.
(I believe that I learned this from somewhere in a paper of Lurie's, but I can't find it currently.)
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
beginalign*
A(x) &=
begincases 2x &textif x leq 1/2\1 &textif x geq 1/2endcases\
B(x) &=
begincases 0 &textif x leq 1/2\2x-1 &textif x geq 1/2endcases
endalign*
$$
Define three monoid homomorphisms $Id, U, V: M to M$ by:
$$
beginalign*
(Id(f))(x) &= f(x)\
(Uf)(x) &=
begincases tfrac12f(2x) &textif x leq 1/2\x &textif x geq 1/2endcases\
(Vf)(x) &= x
endalign*
$$
For any $f in M$, we have the following identities:
$$
beginalign*
A circ (Uf) &= f circ A\
B circ (Uf) &= (Vf) circ B
endalign*
$$
As a result, we can reinterpret this in terms of the one-object category $M$: we get three functors $Id,U,V: M to M$ and natural transformations $A: U to I$ and $B: U to V$.
Upon taking geometric realization, we get a space $BM$, these functors turn into continuous maps $Id, U, V:BM to BM$ and homotopies from $U$ to $Id$ and from $U$ to $V$. However, $V$ is a constant map, and so this says that the homotopy type of $BM$ is contractible.
(I believe that I learned this from somewhere in a paper of Lurie's, but I can't find it currently.)
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
answered Oct 1 at 9:20
Tyler Lawson
37.9k7130196
37.9k7130196
4
Counterexample 4.4.5.19, HTT?
– Charles Rezk
Oct 1 at 13:32
@Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it...
– Tyler Lawson
Oct 1 at 17:08
add a comment |Â
4
Counterexample 4.4.5.19, HTT?
– Charles Rezk
Oct 1 at 13:32
@Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it...
– Tyler Lawson
Oct 1 at 17:08
4
4
Counterexample 4.4.5.19, HTT?
– Charles Rezk
Oct 1 at 13:32
Counterexample 4.4.5.19, HTT?
– Charles Rezk
Oct 1 at 13:32
@Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it...
– Tyler Lawson
Oct 1 at 17:08
@Charles That looks like it... although I'm reasonable certain that I've never read the version of HTT that has this in it...
– Tyler Lawson
Oct 1 at 17:08
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f311781%2fhomotopy-type-of-a-specific-discrete-monoid%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
This seems a complicated way of asking what's the homotopy type of $BM$. I'm not aware of any tool beyond the group-completion theorem to do this kind of analysis though
– Denis Nardin
Oct 1 at 7:16
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
Oct 1 at 7:42