Let $X$ and $Y$ be topological spaces and $h_*:S_*(X\times Y)\to S_*(X\times Y)$ a homomorphism of natural chain complexes. If $h_0=Id$, show that $h_*$ is homotopic to $Id$.
To solve this problem I have thought about using naturalness. So first I proof this result for $ X=\Delta_n$ and $Y=\Delta_n$, where $h_*':S_*(\Delta_n\times \Delta_n)\to S_*(\Delta_n\times \Delta_n)$ is homotopic to $Id$. I also have the following diagram:
$$\require{AMScd} \begin{CD} S_*(\Delta_n\times \Delta_n) @> h_*' >> S_*(\Delta_n\times \Delta_n) \\ @V f_1 VV @V f_0 VV @. \\ S_*(X\times Y) @> h_* >> S_*(X\times Y) \end{CD} $$
But I don't know what functions to put on the vertical lines. It follows that $h_*$ is homotopic to $Id$? Why? Thank you.