I know the usual proof of the fact that if a morphism between chain complexes $f$ is homotopic to zero then it induces the $0$ map on cohomology.

I was wondering if there is an easy proof of this fact only using category theory. I was trying to see how the restriction of $f$ to the kernel looks when we have an homotopy but I don't know how to proceed.

  • $\begingroup$ The map induced on homology by a map of chain complexes is invariant up to chain homotopy, so a nullhomotopic map induces the same map on homology as the zero map. $\endgroup$ – Qiaochu Yuan Apr 8 '16 at 16:37
  • $\begingroup$ I want to prove exactly what you said using my question. I guess I'm going in the wrong direction. Could you give me some hints? $\endgroup$ – Abellan Apr 8 '16 at 16:39

Given $A^{\bullet} \xrightarrow{f}B^{\bullet}$ such that $f$ is nullhomotopic we have by hypothesis

$$f=d^{B}s+sd^{A}$$ in particular $$f_n=d_{n-1}^{B}s+sd_n^{A}$$ Applying the universal property of the kernel we can induce a map $f^{k}_n:\operatorname{Ker}(d^{A}_n) \to \operatorname{Ker}(d^{B}_n) $ with the property

$$ f_n \circ k^{A}_n= k^{B}_n \circ f^{k}_n$$ where $k^{A}_n$ is the kernel monomorphism. Using the description of $f_n$ we can get: $$(d_{n-1}^{B}s+sd_n^{A}) \circ k^{A}_n= d_{n-1}^{B}s \circ k^{A}_n=k^{B}_n \circ f^{k}_n$$

We can decompose $d_{n-1}^{B}=k_{n}^{B} \circ \alpha \circ \pi$ with $\pi$ the cannonical image factorization and $\alpha$ the induce map from the image of $d_{n-1}^{B}$ into the kernel of $d_{n}^{B}$. Thus we can see

$$d_{n-1}^{B}s \circ k^{A}_n=k_{n}^{B} \circ \alpha \circ \pi \circ s \circ k^{A}_n=k^{B}_n \circ f^k_n \implies f^{k}_n= \alpha \circ \pi \circ s \circ k^{A}_n$$

Finally denoting $\pi^{A}_n:\operatorname{Ker}(d_n^{A}) \to H^{n}(A^{\bullet})$ (respect $B$) it is easy to see that

$$H^{n}(f) \circ \pi^{A}_n =\pi^{B}_n \circ f^{k}_n=0$$

But $\pi^{A}_n$ is an epimorphism so $H^n(f)=0$. This shows that the induced maps on cohomology only depend on the homotopy class of $f$.


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