Let $X, Y$ be CW complexes and the following proof (from J. P. May's "A Concise Course in Algebraic Topology"; see page 102. Here the full pdf document: https://www.maths.ed.ac.uk/~v1ranick/papers/maybook.pdf
Here the excerpt:
The goal is to show that $C_*(X \times Y) \cong C_*(X) \otimes C_*( Y)$ holds ***as CW complexes$.
By definition the of $\otimes$ for complexes, we have $( C_*(X) \otimes C_*( Y))_n = \oplus_{p +q =n} C_p(X) \otimes C_q(Y)$.
Futhermore, since each $n$ cell of$ X \times Y$ comes from $p$-cell $[i]$ of $X$ and $q$-cell $[j]$ of $Y$ we can donote the free generators of $C_n(X \times Y)$ by $[i \times j]$, we can make the idenitification $[i \times j] = [i] \otimes [j]$.
Then we get indeed a bijection between $C_*(X \times Y) $ and $ C_*(X) \otimes C_*( Y)$.
The candidate for a isomorphism of complexes is geven by $\kappa:C_*(X) \otimes C_*( Y) \to C_*(X \times Y)$ by $[i] \otimes [j] \mapsto (-1)^{pq}[i \times j]$.
The cruical point is to show that $\kappa$ respects differentials, therefore that $\kappa \circ d_{C_*(X) \otimes C_*( Y), n} = d_{C_*(X \times Y),n}\circ \kappa$ holds.
Remark: The differential $d_{C_*(X) \otimes C_*( Y)}$ for $[i] \in C_p(X)$ and $[j] \in C_q(Y)$ with $p+q=n$ is given by $ d_{C_*(X) \otimes C_*( Y), n}([i] \times [j]) = d_X([i]) \otimes [j] + (-1)^p[i] \otimes d_Y([Y])$.
The autor introduces that if we can show that the two diagrams above commutate up to homotopy then we are done.
I don't see why. Ok, that is a weird work to prove the commutativity op to homotopy, but if we assump that it holds, why does it instantly imply, that $\kappa$ respects the differentials and is therefore a iso between complexes?
BTW: I know that this famous statement (called Eilenberg-Zilber) can be proved in different other ways but the main point of my question is to understand the step used in the proof above.