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Homotopy type of a specific discrete monoid
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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 fondamental 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
asked 3 hours ago
Philippe Gaucher
1,587918
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up vote
2
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 fondamental 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
asked 3 hours ago
Philippe Gaucher
1,587918
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
3 hours ago
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
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 fondamental 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
asked 3 hours ago
Philippe Gaucher
1,587918
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 fondamental 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
asked 3 hours ago
Philippe Gaucher
1,587918
asked 3 hours ago
Philippe Gaucher
1,587918
edited 3 hours ago
asked 3 hours ago
Philippe Gaucher
1,587918
asked 3 hours ago
Philippe Gaucher
1,587918
asked 3 hours ago
Philippe Gaucher
1,587918
1,587918
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
3 hours ago
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
2 hours ago
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
3 hours ago
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
2 hours ago
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
3 hours ago
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
3 hours ago
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
2 hours ago
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
2 hours ago
add a comment |Â
1 Answer
1
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oldest
votes
up vote
3
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 1 hour ago
Tyler Lawson
37.8k7129195
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
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 1 hour ago
Tyler Lawson
37.8k7129195
add a comment |Â
up vote
3
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 1 hour ago
Tyler Lawson
37.8k7129195
add a comment |Â
up vote
3
down vote
accepted
up vote
3
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 1 hour ago
Tyler Lawson
37.8k7129195
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 1 hour ago
Tyler Lawson
37.8k7129195
answered 1 hour ago
Tyler Lawson
37.8k7129195
answered 1 hour ago
Tyler Lawson
37.8k7129195
answered 1 hour ago
Tyler Lawson
37.8k7129195
37.8k7129195
add a comment |Â
add a comment |Â
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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
3 hours ago
@DenisNardin It is just a question about the "state of the art", nothing else.
– Philippe Gaucher
2 hours ago