4

Given two topological spaces $X,Y$. Is every chain map $f_{\ast}:S_{\ast}(X)\rightarrow S_{\ast}(Y)$ induced by a map (of topological spaces) $f:X\rightarrow Y$?

  • Which chain complex does $S_*(X)$ denote? – Amitai Yuval Apr 09 '17 at 21:27
  • We are in the singular homology. –  Apr 09 '17 at 21:28
  • The examples below are counterexamples because they concern singular chains or because a chain map induced by a map of spaces is an isomorphism in degree zero. What about if one uses CW cellular chains? Then is it true that any chain map (that is an isomorphism in degree 0) can be realized by a map of spaces on CW cellular chains? – user39598 Jun 10 '20 at 03:31

4 Answers4

5

Given a topological space $X$, its group of singular $n$-chains $S_n(X)$ comes equipped with a particular basis, namely the basis of singular $n$-simplices $\sigma : \Delta^n \to X$.

Given two topological spaces and a continuous map $f : X \to Y$, the induced homomorphism $f_* : S_n(X) \to S_n(Y)$ takes each basis element $\sigma$ to a basis element $f \circ \sigma$.

Thus, a necessary condition for a chain map to be induced by a continuous map is that it take each basis element of each $S_n(X)$ to a basis element of $S_n(Y)$.

Lee Mosher
  • 120,280
4

This is very very far from being true. For a simple example, there is a chain map $S_*(X)\to S_*(Y)$ that is just $0$ in every degree. This map cannot be induced by any map $X\to Y$ if $X$ is nonempty, since any map $S_*(X)\to S_*(Y)$ induced by a map $X\to Y$ is nonzero on every singular simplex in $X$ (namely, it maps it to a singular simplex in $Y$).

Eric Wofsey
  • 330,363
  • So your argument is: Since $X$ is not empty there is at least one singular $n$-simplex for every $n$ namely the constant map. The induced map sends singular $n$-simplicies to singular $n$-simplicies. But why can't this be the zero map? –  Apr 10 '17 at 07:17
  • 1
    @user372565 An $n$-simplex is a basis element. Recall that $S_*(X)$ is the free abelian group generated by them. – Aloizio Macedo Apr 10 '17 at 10:16
3

Given $X,Y$ any nonempty spaces and $f: X \to Y$ a continuous map, then the diagram below commutes

$$\require{AMScd} \begin{CD} H_0(X) @>{f_*}>> H_0(Y)\\ @A{i_*}AA @V{r_*}VV \\ H_0(\{p\}) @>{g_*}>> H_0(\{q\}), \end{CD}$$

where $i: \{p\} \to X$ is an inclusion of a point, $r: Y \to \{q\}$ is the constant map and $g:=r \circ f \circ i$. Since $g$ is an homeomorphism, $g_*$ is an isomorphism. In particular, $f_*$ can never be the zero map (which it would be, if it came from the zero chain map), since $H_0(\{q\})$ is non-trivial. It follows that the the induced chain map can never be the zero chain map.

Aloizio Macedo
  • 34,292
2

If $X$ and $Y$ are arcwise connected, then $H_0(X)=H_0(Y)=\mathbb{Z}.$ However, for every continuous map $f:X\to Y$, the induced map on $H_0$ is the identity. Hence, the answer is negative.

Edit: In the spirit of Aloizio's comment, there isn't actually a well-defined "identity" between $H_0(X)$ and $H_0(Y)$. However, the induced map will always be an isomorphism of abelian groups.

Amitai Yuval
  • 19,308
  • Thank you! This is not so clear for me. Why is every induced map on $H_0$ the identity and how do you conclude that the answer is negative? –  Apr 09 '17 at 21:40
  • @user372565 The fact that the induced $f_*$ is the identity follows immediately from the definition. The answer to the question is negative, as there are many homomorphisms $\mathbb{Z}\to\mathbb{Z}$ that are not the identity. – Amitai Yuval Apr 09 '17 at 21:49
  • There are some misconceptions here. Firstly, $H_0$ of a connected space is not necessarily $\mathbb{Z}$. $H_0$ of an arcwise connected space is. Secondly, it is not true that the induced map is the identity, since there is an underlying choice of generator which you must first specify in order for that to be true (it can also be minus the identity). – Aloizio Macedo Apr 09 '17 at 21:51
  • @AloizioMacedo Thank you. You inspired me to edit my answer. – Amitai Yuval Apr 09 '17 at 22:28