$K_\star$ and $L_\star$ are homotopic as chain complex. Is it true that $H_\star(\operatorname{Hom}(K_\star,L_\star))$ trivial?

Question

Consider $K_\star,L_\star$ 2 chain complexes and suppose they are homotopic say $f:K\to L$.

$\textbf{Q:}$ Consider $\operatorname{Hom}(K_\star, L_\star)$ complex and take its homology. It is clear that $[f]=0\in H_0(\operatorname{Hom}_\star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $\operatorname{Hom}_1(K_\star,L_\star)$). However, do I even know or can say $H_\star(\operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_\star(\operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(\operatorname{Hom}(K,L))=0$ in general as it might be very large.

Answer 1

It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps: $$H_0(\underline{Hom}(L,K)) = [L,K]$$ So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $L\simeq K$, then the following are equivalent:

1. $H_*(\underline{Hom}(L,K)) = 0$,
2. $H_0(\underline{Hom}(L,K)) = 0$,
3. $L$ is contractible, i.e. $\text{id}_L \simeq 0$.

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In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.