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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?

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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.

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  • $\begingroup$ Nice example! Btw do you know any condition to make it true? $\endgroup$ – User X Nov 13 '18 at 9:32
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Let $\mathcal A$ be an abelian category. Your condition (zero map on cohomology implies null-homotopy) is equivalent to the condition that $\mathcal A$ be semisimple (i.e. any exact triple $0\to X\to Z\to Y\to 0$ in $\mathcal A$ splits, i.e. is isomorphic to $0\to X\to X\oplus Y\to Y\to 0$).

On the one hand, your condition implies all Yoneda Exts are zero, so that the homological dimension of $\mathcal A$ is zero; i.e. $\mathcal A$ is semisimple.

On the other hand, if $\mathcal A$ is semisimple, the functor $h:(K^n,d^n)\mapsto (H^n(K),0)$ which sends a complex $K$ to its complex of cohomology objects with zero differentials in all degrees, induces an equivalence between $D(\mathcal A)$ and the category of cyclic complexes of objects of $\mathcal A$ (complexes with zero differentials in all degrees).

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