Chain homotopy of Cartan-Eilenberg resolutions induced by homotopic maps of chain complexes 
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?
 A: 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.
