Homotopy equivalent chains and I take the tensor product of both with the same complex?

Question

This is my first post here, so forgive me for any potential mistakes. So i have two chain complexes that are chain homotopy equivalent, say $C_1$ and $C_2$. Now if i take the tensor product of both of them with the same complex, as in $C_1 \otimes F[t,t^{-1}]$ and $C_2 \otimes F[t,t^{-1}]$ where the field is just $\mathbb{Z}_2$, is there a universal property or theorem that will give me a chain homotopy equivalence between the new complexes? Intuitively, I have thought about just considering a map of the form $\mathcal{H} \otimes id$ where $\mathcal{H}$ is the chain homotopy equivalence before, but I failed to show why this should work. Thanks a lot for any help!

Answer 1

I think you had the right ideia about the problem. Let's formalize that:

Let $f: C_1\rightarrow C_2$ and $g: C_2\rightarrow C_1$, such that $fg\sim id_{C_2}$ and $gf\sim id_{C_1}$. Let $id_F$ be the identity of $F[t, t^{-1}]$ and consider the maps $$f\otimes id_F: C_1\otimes F[t,t^{-1}] \rightarrow C_2\otimes F[t, t^{-1}]$$ $$g\otimes id_F: C_2\otimes F[t,t^{-1}] \rightarrow C_1\otimes F[t, t^{-1}]$$ Since $fg\sim id_{C_2}$ there exists $h$ such that $fg-id_{C_2}=hd+dh$, so $$(f\otimes id_F)(g\otimes id_F)-id_{C_2}\otimes id_F= (fg-id_{C_2})\otimes id_F= (hd+dh)\otimes id_F = (h\otimes id_F)(d\otimes id_F) + (d\otimes id_F)(h\otimes id_F)$$

The same argument holds to show that $(g\otimes id_F)(f\otimes id_F)\sim id_{C_1}\otimes id_F$.