Let $f:A^\bullet\to B^\bullet$ be a morphism of chain complexes (of any give abelian category). We know that if $f$ is homotopic to the zero map, then $f$ will induce zero map on cohomology. I want to know if the converse true, i.e. if $f$ induces zero map on cohomology, is it true that $f$ is homotopic to zero map?
If not, under what conditions it can be true?
Best Answer
Let both $A^*$ and $B^*$ be the chain complex of abelian groups $\cdots\to0\to\mathbb Z\to\mathbb Z\to\mathbb Z/2\to0\to\cdots$ where the map $\mathbb Z\to\mathbb Z$ is multiplication by 2 and the $\cdots$ represents just $0$'s. Let $f:A^*\to B^*$ be the identity map; it induces $0$ in cohomology, because the complex is an exact sequence, so its cohomology is zero. But $f$ is not null-homotopic. To see that, notice that the only homomorphism from $\mathbb Z/2$ to $\mathbb Z$ is zero, so a homotopy would consist entirely of zero maps except for one map from the second $\mathbb Z$ in $A^*$ to the first $\mathbb Z$ in $B^*$. That's not enough to provide the desired null-homotopy.