Why is induced map on zero homology the identity and not negative the identity?
$begingroup$
Suppose we have a simplicial map $f$ on a path connected simplicial complex $X$. The answer here: Induced map on zeroth homology is zero claims that the induced map on the $0$-homology given by $f_*: H_0(X) to H_0(X)$ is the identity map. As $X$ is path connected I see why $H_0(X)$ is $mathbb{Z}$ and why the equivalence class of any vertex of $X$ is a generator of $H_0(X)$ corresponding to $1$ in $mathbb{Z}$.
The linked answer then continues to say that $f_*$ sends generators of $H_0(X)$ to generators of $H_0(X)$ which I understand. It then claims that for any vertex $x$ we have $1 leftrightarrow [x]to[f(x)] leftrightarrow 1$ so the map must be the identity. However I don't understand why we can't have $[f(x)] leftrightarrow -1$ as $-1$ is also a generator of $mathbb{Z}$. Can someone please clear up this confusion for me? Thanks!
algebraic-topology homology-cohomology homological-algebra simplicial-stuff simplicial-complex
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
$endgroup$
add a comment |
$begingroup$
Suppose we have a simplicial map $f$ on a path connected simplicial complex $X$. The answer here: Induced map on zeroth homology is zero claims that the induced map on the $0$-homology given by $f_*: H_0(X) to H_0(X)$ is the identity map. As $X$ is path connected I see why $H_0(X)$ is $mathbb{Z}$ and why the equivalence class of any vertex of $X$ is a generator of $H_0(X)$ corresponding to $1$ in $mathbb{Z}$.
The linked answer then continues to say that $f_*$ sends generators of $H_0(X)$ to generators of $H_0(X)$ which I understand. It then claims that for any vertex $x$ we have $1 leftrightarrow [x]to[f(x)] leftrightarrow 1$ so the map must be the identity. However I don't understand why we can't have $[f(x)] leftrightarrow -1$ as $-1$ is also a generator of $mathbb{Z}$. Can someone please clear up this confusion for me? Thanks!
algebraic-topology homology-cohomology homological-algebra simplicial-stuff simplicial-complex
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
$endgroup$
3
$begingroup$
$H_0$ has a distinguished choice of generator given by any point (as opposed to the negative of a point), and any map $f : X to X$ maps points to points (as opposed to negatives of points).
$endgroup$
– Qiaochu Yuan
Dec 25 '18 at 9:35
$begingroup$
You say that $f$ is a simplicial map on a simplicial complex $X$ Do you consider the simplicial homology groups of $X$ or the singular homology groups of the polyhedron determined by $X$ (i.e. the geometric realization of $X$)?
$endgroup$
– Paul Frost
Dec 25 '18 at 11:50
$begingroup$
The simplicial homology, although the comment above yours showed me where I was going wrong. Feel free to add a more detailed answer though in case someone else has the same misunderstanding as me in the future.
$endgroup$
– Abdul Hadi Khan
Dec 25 '18 at 13:07
add a comment |
$begingroup$
Suppose we have a simplicial map $f$ on a path connected simplicial complex $X$. The answer here: Induced map on zeroth homology is zero claims that the induced map on the $0$-homology given by $f_*: H_0(X) to H_0(X)$ is the identity map. As $X$ is path connected I see why $H_0(X)$ is $mathbb{Z}$ and why the equivalence class of any vertex of $X$ is a generator of $H_0(X)$ corresponding to $1$ in $mathbb{Z}$.
The linked answer then continues to say that $f_*$ sends generators of $H_0(X)$ to generators of $H_0(X)$ which I understand. It then claims that for any vertex $x$ we have $1 leftrightarrow [x]to[f(x)] leftrightarrow 1$ so the map must be the identity. However I don't understand why we can't have $[f(x)] leftrightarrow -1$ as $-1$ is also a generator of $mathbb{Z}$. Can someone please clear up this confusion for me? Thanks!
algebraic-topology homology-cohomology homological-algebra simplicial-stuff simplicial-complex
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
$endgroup$
Suppose we have a simplicial map $f$ on a path connected simplicial complex $X$. The answer here: Induced map on zeroth homology is zero claims that the induced map on the $0$-homology given by $f_*: H_0(X) to H_0(X)$ is the identity map. As $X$ is path connected I see why $H_0(X)$ is $mathbb{Z}$ and why the equivalence class of any vertex of $X$ is a generator of $H_0(X)$ corresponding to $1$ in $mathbb{Z}$.
The linked answer then continues to say that $f_*$ sends generators of $H_0(X)$ to generators of $H_0(X)$ which I understand. It then claims that for any vertex $x$ we have $1 leftrightarrow [x]to[f(x)] leftrightarrow 1$ so the map must be the identity. However I don't understand why we can't have $[f(x)] leftrightarrow -1$ as $-1$ is also a generator of $mathbb{Z}$. Can someone please clear up this confusion for me? Thanks!
algebraic-topology homology-cohomology homological-algebra simplicial-stuff simplicial-complex
algebraic-topology homology-cohomology homological-algebra simplicial-stuff simplicial-complex
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
asked Dec 25 '18 at 9:29
Abdul Hadi KhanAbdul Hadi Khan
518419
518419
3
$begingroup$
$H_0$ has a distinguished choice of generator given by any point (as opposed to the negative of a point), and any map $f : X to X$ maps points to points (as opposed to negatives of points).
$endgroup$
– Qiaochu Yuan
Dec 25 '18 at 9:35
$begingroup$
You say that $f$ is a simplicial map on a simplicial complex $X$ Do you consider the simplicial homology groups of $X$ or the singular homology groups of the polyhedron determined by $X$ (i.e. the geometric realization of $X$)?
$endgroup$
– Paul Frost
Dec 25 '18 at 11:50
$begingroup$
The simplicial homology, although the comment above yours showed me where I was going wrong. Feel free to add a more detailed answer though in case someone else has the same misunderstanding as me in the future.
$endgroup$
– Abdul Hadi Khan
Dec 25 '18 at 13:07
add a comment |
3
$begingroup$
$H_0$ has a distinguished choice of generator given by any point (as opposed to the negative of a point), and any map $f : X to X$ maps points to points (as opposed to negatives of points).
$endgroup$
– Qiaochu Yuan
Dec 25 '18 at 9:35
$begingroup$
You say that $f$ is a simplicial map on a simplicial complex $X$ Do you consider the simplicial homology groups of $X$ or the singular homology groups of the polyhedron determined by $X$ (i.e. the geometric realization of $X$)?
$endgroup$
– Paul Frost
Dec 25 '18 at 11:50
$begingroup$
The simplicial homology, although the comment above yours showed me where I was going wrong. Feel free to add a more detailed answer though in case someone else has the same misunderstanding as me in the future.
$endgroup$
– Abdul Hadi Khan
Dec 25 '18 at 13:07
3
3
$begingroup$
$H_0$ has a distinguished choice of generator given by any point (as opposed to the negative of a point), and any map $f : X to X$ maps points to points (as opposed to negatives of points).
$endgroup$
– Qiaochu Yuan
Dec 25 '18 at 9:35
$begingroup$
$H_0$ has a distinguished choice of generator given by any point (as opposed to the negative of a point), and any map $f : X to X$ maps points to points (as opposed to negatives of points).
$endgroup$
– Qiaochu Yuan
Dec 25 '18 at 9:35
$begingroup$
You say that $f$ is a simplicial map on a simplicial complex $X$ Do you consider the simplicial homology groups of $X$ or the singular homology groups of the polyhedron determined by $X$ (i.e. the geometric realization of $X$)?
$endgroup$
– Paul Frost
Dec 25 '18 at 11:50
$begingroup$
You say that $f$ is a simplicial map on a simplicial complex $X$ Do you consider the simplicial homology groups of $X$ or the singular homology groups of the polyhedron determined by $X$ (i.e. the geometric realization of $X$)?
$endgroup$
– Paul Frost
Dec 25 '18 at 11:50
$begingroup$
The simplicial homology, although the comment above yours showed me where I was going wrong. Feel free to add a more detailed answer though in case someone else has the same misunderstanding as me in the future.
$endgroup$
– Abdul Hadi Khan
Dec 25 '18 at 13:07
$begingroup$
The simplicial homology, although the comment above yours showed me where I was going wrong. Feel free to add a more detailed answer though in case someone else has the same misunderstanding as me in the future.
$endgroup$
– Abdul Hadi Khan
Dec 25 '18 at 13:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $X$ be an abstract simplicial complex and $lvert X rvert$ its geometric realization which is a topological space representable as the union of geometric simplices. It is well-known that there exist natural isomorphisms $phi_n^X : H_n(X) to H_n(lvert X rvert)$ between the simplicial homology groups $H_n(X)$ of $X$ and the singular homology groups $H_n(lvert X rvert)$ of $lvert X rvert$. In particular, if $f : Xto Y$ is a simplicial map, then $lvert f rvert_* circ phi_n^X = phi_n^Y circ f_*$.
Using this, Qiaochu Yuan's comment answers your question. However, we can also see this without changing to topological spaces.
The simplicial chain complex of $X$ is based on oriented simplices. An orientation of an $n$-simplex $s^n = { v_0,dots,v_n }$ is an equivalence class $[v_{pi(0)},dots,v_{pi(n)}]$, where $pi$ ranges over all permutations of ${ 0,dots,n }$. For $n > 0$ there exist two orientations, for $n = 0$ only one. The simplicial chain map $C_*(f) : C_*(X) to C_*(Y)$ is given on the simplicial generators $sigma^n = [v_0,dots,v_n]$ of the free abelian group $C_n(X)$ by
$$C_n(f)([v_0,dots,v_n]) = [f(v_0),dots,f(v_n)].$$
In dimension $0$ we simply have
$$(*) phantom{xx} C_0(f)(v) = f(v)$$
with the vertices $v$.
Now you consider a simplicial map $f : X to X$ on a path connected $X$. In my understanding of your question you know that all vertices $v$ of $X$ (i.e. all simplicial generators of $C_0(X)$) are in the same homology class $g in H_0(X)$ and that $H_0(X)$ is free abelian with $g$ as a generator. Let us call $g$ the canonical simplicial generator of $H_0(X)$. This provides a canonical isomorphism $H_0(X) approx mathbb{Z}$ (mapping $g$ to $1$). $H_0(X)$ of course has an alternative generator: This is $-g$.
But now it obvious from $(*)$ that $f_*(g) = g$ since both $v$ and $f(v)$ are vertices and therefore in the same homology class.
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
$endgroup$
add a comment |
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$begingroup$
Let $X$ be an abstract simplicial complex and $lvert X rvert$ its geometric realization which is a topological space representable as the union of geometric simplices. It is well-known that there exist natural isomorphisms $phi_n^X : H_n(X) to H_n(lvert X rvert)$ between the simplicial homology groups $H_n(X)$ of $X$ and the singular homology groups $H_n(lvert X rvert)$ of $lvert X rvert$. In particular, if $f : Xto Y$ is a simplicial map, then $lvert f rvert_* circ phi_n^X = phi_n^Y circ f_*$.
Using this, Qiaochu Yuan's comment answers your question. However, we can also see this without changing to topological spaces.
The simplicial chain complex of $X$ is based on oriented simplices. An orientation of an $n$-simplex $s^n = { v_0,dots,v_n }$ is an equivalence class $[v_{pi(0)},dots,v_{pi(n)}]$, where $pi$ ranges over all permutations of ${ 0,dots,n }$. For $n > 0$ there exist two orientations, for $n = 0$ only one. The simplicial chain map $C_*(f) : C_*(X) to C_*(Y)$ is given on the simplicial generators $sigma^n = [v_0,dots,v_n]$ of the free abelian group $C_n(X)$ by
$$C_n(f)([v_0,dots,v_n]) = [f(v_0),dots,f(v_n)].$$
In dimension $0$ we simply have
$$(*) phantom{xx} C_0(f)(v) = f(v)$$
with the vertices $v$.
Now you consider a simplicial map $f : X to X$ on a path connected $X$. In my understanding of your question you know that all vertices $v$ of $X$ (i.e. all simplicial generators of $C_0(X)$) are in the same homology class $g in H_0(X)$ and that $H_0(X)$ is free abelian with $g$ as a generator. Let us call $g$ the canonical simplicial generator of $H_0(X)$. This provides a canonical isomorphism $H_0(X) approx mathbb{Z}$ (mapping $g$ to $1$). $H_0(X)$ of course has an alternative generator: This is $-g$.
But now it obvious from $(*)$ that $f_*(g) = g$ since both $v$ and $f(v)$ are vertices and therefore in the same homology class.
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
$endgroup$
add a comment |
$begingroup$
Let $X$ be an abstract simplicial complex and $lvert X rvert$ its geometric realization which is a topological space representable as the union of geometric simplices. It is well-known that there exist natural isomorphisms $phi_n^X : H_n(X) to H_n(lvert X rvert)$ between the simplicial homology groups $H_n(X)$ of $X$ and the singular homology groups $H_n(lvert X rvert)$ of $lvert X rvert$. In particular, if $f : Xto Y$ is a simplicial map, then $lvert f rvert_* circ phi_n^X = phi_n^Y circ f_*$.
Using this, Qiaochu Yuan's comment answers your question. However, we can also see this without changing to topological spaces.
The simplicial chain complex of $X$ is based on oriented simplices. An orientation of an $n$-simplex $s^n = { v_0,dots,v_n }$ is an equivalence class $[v_{pi(0)},dots,v_{pi(n)}]$, where $pi$ ranges over all permutations of ${ 0,dots,n }$. For $n > 0$ there exist two orientations, for $n = 0$ only one. The simplicial chain map $C_*(f) : C_*(X) to C_*(Y)$ is given on the simplicial generators $sigma^n = [v_0,dots,v_n]$ of the free abelian group $C_n(X)$ by
$$C_n(f)([v_0,dots,v_n]) = [f(v_0),dots,f(v_n)].$$
In dimension $0$ we simply have
$$(*) phantom{xx} C_0(f)(v) = f(v)$$
with the vertices $v$.
Now you consider a simplicial map $f : X to X$ on a path connected $X$. In my understanding of your question you know that all vertices $v$ of $X$ (i.e. all simplicial generators of $C_0(X)$) are in the same homology class $g in H_0(X)$ and that $H_0(X)$ is free abelian with $g$ as a generator. Let us call $g$ the canonical simplicial generator of $H_0(X)$. This provides a canonical isomorphism $H_0(X) approx mathbb{Z}$ (mapping $g$ to $1$). $H_0(X)$ of course has an alternative generator: This is $-g$.
But now it obvious from $(*)$ that $f_*(g) = g$ since both $v$ and $f(v)$ are vertices and therefore in the same homology class.
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
$endgroup$
add a comment |
$begingroup$
Let $X$ be an abstract simplicial complex and $lvert X rvert$ its geometric realization which is a topological space representable as the union of geometric simplices. It is well-known that there exist natural isomorphisms $phi_n^X : H_n(X) to H_n(lvert X rvert)$ between the simplicial homology groups $H_n(X)$ of $X$ and the singular homology groups $H_n(lvert X rvert)$ of $lvert X rvert$. In particular, if $f : Xto Y$ is a simplicial map, then $lvert f rvert_* circ phi_n^X = phi_n^Y circ f_*$.
Using this, Qiaochu Yuan's comment answers your question. However, we can also see this without changing to topological spaces.
The simplicial chain complex of $X$ is based on oriented simplices. An orientation of an $n$-simplex $s^n = { v_0,dots,v_n }$ is an equivalence class $[v_{pi(0)},dots,v_{pi(n)}]$, where $pi$ ranges over all permutations of ${ 0,dots,n }$. For $n > 0$ there exist two orientations, for $n = 0$ only one. The simplicial chain map $C_*(f) : C_*(X) to C_*(Y)$ is given on the simplicial generators $sigma^n = [v_0,dots,v_n]$ of the free abelian group $C_n(X)$ by
$$C_n(f)([v_0,dots,v_n]) = [f(v_0),dots,f(v_n)].$$
In dimension $0$ we simply have
$$(*) phantom{xx} C_0(f)(v) = f(v)$$
with the vertices $v$.
Now you consider a simplicial map $f : X to X$ on a path connected $X$. In my understanding of your question you know that all vertices $v$ of $X$ (i.e. all simplicial generators of $C_0(X)$) are in the same homology class $g in H_0(X)$ and that $H_0(X)$ is free abelian with $g$ as a generator. Let us call $g$ the canonical simplicial generator of $H_0(X)$. This provides a canonical isomorphism $H_0(X) approx mathbb{Z}$ (mapping $g$ to $1$). $H_0(X)$ of course has an alternative generator: This is $-g$.
But now it obvious from $(*)$ that $f_*(g) = g$ since both $v$ and $f(v)$ are vertices and therefore in the same homology class.
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
$endgroup$
Let $X$ be an abstract simplicial complex and $lvert X rvert$ its geometric realization which is a topological space representable as the union of geometric simplices. It is well-known that there exist natural isomorphisms $phi_n^X : H_n(X) to H_n(lvert X rvert)$ between the simplicial homology groups $H_n(X)$ of $X$ and the singular homology groups $H_n(lvert X rvert)$ of $lvert X rvert$. In particular, if $f : Xto Y$ is a simplicial map, then $lvert f rvert_* circ phi_n^X = phi_n^Y circ f_*$.
Using this, Qiaochu Yuan's comment answers your question. However, we can also see this without changing to topological spaces.
The simplicial chain complex of $X$ is based on oriented simplices. An orientation of an $n$-simplex $s^n = { v_0,dots,v_n }$ is an equivalence class $[v_{pi(0)},dots,v_{pi(n)}]$, where $pi$ ranges over all permutations of ${ 0,dots,n }$. For $n > 0$ there exist two orientations, for $n = 0$ only one. The simplicial chain map $C_*(f) : C_*(X) to C_*(Y)$ is given on the simplicial generators $sigma^n = [v_0,dots,v_n]$ of the free abelian group $C_n(X)$ by
$$C_n(f)([v_0,dots,v_n]) = [f(v_0),dots,f(v_n)].$$
In dimension $0$ we simply have
$$(*) phantom{xx} C_0(f)(v) = f(v)$$
with the vertices $v$.
Now you consider a simplicial map $f : X to X$ on a path connected $X$. In my understanding of your question you know that all vertices $v$ of $X$ (i.e. all simplicial generators of $C_0(X)$) are in the same homology class $g in H_0(X)$ and that $H_0(X)$ is free abelian with $g$ as a generator. Let us call $g$ the canonical simplicial generator of $H_0(X)$. This provides a canonical isomorphism $H_0(X) approx mathbb{Z}$ (mapping $g$ to $1$). $H_0(X)$ of course has an alternative generator: This is $-g$.
But now it obvious from $(*)$ that $f_*(g) = g$ since both $v$ and $f(v)$ are vertices and therefore in the same homology class.
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
answered Dec 27 '18 at 17:22
Paul FrostPaul Frost
13.2k31035
13.2k31035
add a comment |
add a comment |
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$begingroup$
$H_0$ has a distinguished choice of generator given by any point (as opposed to the negative of a point), and any map $f : X to X$ maps points to points (as opposed to negatives of points).
$endgroup$
– Qiaochu Yuan
Dec 25 '18 at 9:35
$begingroup$
You say that $f$ is a simplicial map on a simplicial complex $X$ Do you consider the simplicial homology groups of $X$ or the singular homology groups of the polyhedron determined by $X$ (i.e. the geometric realization of $X$)?
$endgroup$
– Paul Frost
Dec 25 '18 at 11:50
$begingroup$
The simplicial homology, although the comment above yours showed me where I was going wrong. Feel free to add a more detailed answer though in case someone else has the same misunderstanding as me in the future.
$endgroup$
– Abdul Hadi Khan
Dec 25 '18 at 13:07