Tensor product of two chain homotopic maps are again chain homotopic Let $C$, $C'$, $D$, $D'$ be chain complexes, $f$, $f'\colon C\to C'$ and $g$, $g'\colon D \to D'$ two pairs of homotopic chain maps. How to show $f \otimes g$ and $f' \otimes g' \colon C\otimes D\to C'\otimes D'$ are homotopic? I have tried many formulas for such a homotopy, but none of them worked.
 A: Here is an approach that avoids some computations (really they're hidden under the hood and I'm pretty sure that if you followed the advice given in t.b.'s comment you would find the same thing). I'm implicitly using the enriched $\operatorname{Hom}(C,D)$ functor of chain complex, where a cycle in degree zero is a chain map $C \to D$, and a boundary of degree zero is a nullhomotopic chain map.
Let $F = f - f'$ and $G = g - g'$. Then both $F$ and $G$ are null homotopic, i.e. there exist homotopies $h$ and $k$ such that $F = \partial(h)$ and $G = \partial(k)$ (where $\partial(h) = dh + hd$). We also know that $f$, $f'$, $g$, and $g'$ are chain maps, i.e. $\partial(f) = \partial(f') = 0$ and $\partial(g) = \partial(g') = 0$. What we want to do is express $f \otimes g - f' \otimes g'$ as the boundary of some homotopy. So let's write:
$$\begin{align}
f \otimes g - f' \otimes g'
& = f \otimes g - f' \otimes g + f' \otimes g - f' \otimes g' \\
& = F \otimes g + f' \otimes G \\
& = \partial(h) \otimes g + f' \otimes \partial(k) \\
& = \partial(h \otimes g + f' \otimes k)
\end{align}$$
Where I used the fact that $\partial(h \otimes g) = \partial(h) \otimes g - h \otimes \partial(g)$ and the second term is zero.
So to conclude, if $h$ is the homotopy between $f$ and $f'$, and $k$ is the homotopy between $g$ and $g'$, the homotopy between $f \otimes g$ and $f' \otimes g'$ is given by
$$c \otimes d \mapsto h(c) \otimes g(d) + (-1)^{|c|} f'(c) \otimes k(d),$$
where the sign comes from the Koszul rule of signs
$$(f' \otimes k)(c \otimes d) = (-1)^{|k| \cdot |c|} f'(c) \otimes k(d) = (-1)^{|c|} f'(c) \otimes k(d).$$
