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Although you write "complex of complexes", the further assumption implies it is in fact a short exact sequence of complexes, right? The current wording is confusing.

Maybe this is a bad day, but checking the first formula in the proof was not completely immediate to me. There could be a hint such as "use $i^{n+1}\circ\pi^{n+1}=\mathrm{Id_{B^{n+1}}}-s^{n+1}\circ q^{n+1}$".

Similarly, the formula to be proved would be $d_A^{n+1}\circ \pi^{n+1}\circ d_B^{n}\circ s^{n}= -\pi^{n+2}\circ d_B^{n+1}\circ s^{n+1}\circ d_C^n$, but this is done (tacitly) after composition with $i^{n+2}$ (which is OK since $i^{n+2}$ is left invertible).

@#5076: yes because later we say we have the direct sum decompositions

@#5081: yes you have to work to get it. In fact I think personally the best way is to think of the differential on $B$ as given by matrices $$d_B= \left( \begin{matrix} d_A & \delta \\\ 0 & d_C \end{matrix} \right)$$ and then look what it means that $d_B^2 = 0$. Of course this is a completely standard thing that everybody should do once for themselves. Another things is that this chapter is in the prerequisites.

I agree with @#5081: not after I did read Laurent's comments I could understand well the proof.

not *before

Suggestion: Maybe one could add to the statement "Moreover, the morphism $\delta$ is natural in the following way: If we have a commutative diagram of complexes $$\require{AMScd} \begin{CD} 0@>>> A^\bullet@>>> B^\bullet@>>> C^\bullet@>>> 0\\ @.@VV{\alpha}V@VV{\beta}V@VV{\gamma}V@.\\ 0@>>>\tilde{A}^\bullet@>>> \tilde{B}^\bullet@>>> \tilde{C}@>>> 0 \end{CD}$$ where the rows are termwise split short exact sequences and we are given termwise splittings $\widetilde{s}^n:\widetilde{C}^n\to\widetilde{B}^n$ and $\widetilde{\pi}^n:\widetilde{B}^n\to\widetilde{A}^n$ such that the diagram $$\require{AMScd} \begin{CD} A^n@<{\pi^n}<< B^n@<{s^n}<< C^n\\ @V{\alpha^n}VV@V{\beta^n}VV@VV{\gamma^n}V\\ \widetilde{A}^n@<<{\widetilde{\pi}^n}< \widetilde{B}^n@<<{\widetilde{s}^n}< \widetilde{C}^n \end{CD}$$ commutes for all $n$, then the following diagram commutes: $$\require{AMScd} \begin{CD} C^\bullet@>{\delta}>> A[1]^\bullet\\ @V{\gamma}VV@VV{\alpha[1]}V\\ \widetilde{C}^\bullet@>>{\widetilde{\delta}}>\widetilde{A}[1]^\bullet \end{CD}$$ where $\widetilde{\delta}^n=\widetilde{\pi}^n\circ d^n_{\widetilde{B}}\circ\widetilde{s}^n$."

The proof is immediate from commutativity of the second-to-last diagram plus the fact that $\beta$ is a cochain map.

OK, I think that this is the sort of "easy" case. Even if we cannot choose splittings compatible with the maps, then there is a relation between the two deltas. In order words, we should upgrade Lemma 12.14.12 to the case of a map between termwise split short exact sequences (not compatible with splittings) whose conclusion is some homotopy between two maps involving deltas. Maybe this is what commenter R in #5086 had in mind?