Chain maps induces by maps of topological spaces 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) $f:X\rightarrow Y$?
 A: 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 $S_n(Y)$.
A: 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 $Y$).
A: 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.
A: 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.
