Chain maps induces by maps of topological spaces

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 $S_n(Y)$.

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 $Y$).

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.