1
$\begingroup$

If $f,g:A\to B$ are homotopic maps of chain complexes, and $\tilde{f},\tilde{g}:P\to Q$ are maps of Cartan-Eilenberg resolutions lying over them, show that $\tilde{f}$ is chain homotopic to $\tilde{g}$.

I could easily verified that: Suppose $f$ is a chain map which induces a zero map from $H_*A$ to $H_*B$. Then, the induced map $\tilde{f}$ between two Cartan-Eilenberg resolutions again induces zero maps from $H_*(\operatorname{Tot}^\oplus(P))$ to $H_*(\operatorname{Tot}^\oplus(Q))$. But I cannot proceed to the level of chain homotopy. I tried to find all $s^h:P_{pq}\to Q_{p+1,q}$ and $s^v:P_{pq}\to Q_{p,q+1}$ using the property of projective objects but this seems to be also difficult. Is there any good ideas?

$\endgroup$
2
$\begingroup$

I can only sketch a solution. First consider what $\tilde{f}$ could be if $f=0$ (not homotopic to zero but just zero). Then what would happen is that the C-E (Cartan Eilenberg) resolution over H and B (the homology and boundary projective resolutions which generate the C-E resolution) would be a chain map over zero. What are the possible chain maps over projective resolutions over zero? They must be chain homotopic to zero. There's a version of horseshoe lemma which allows you to extend these homotopies to the C-E resolution, this gives a vertical homotopy to zero over any identically zero map. Now returning to the question, we wonder what happens if we are a C-E resolution over a homotopy zero map. In this case, simply lift the horizontal homotopy to projective resolutions to get a C-E map which is horizontally homotopic to zero. Now subtract this C-E map from the one given, it must be a C-E map over zero which is vertically homotopic to zero. This shows that any C-E map over a chain map homotopic to zero must be homotopic to zero.

$\endgroup$
  • $\begingroup$ I have solved it but thank you so much! $\endgroup$ – Herace Jul 16 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.