Lemma 12.14.10. Let $\mathcal{A}$ be an additive category. Let

$0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$

be a complex (!) of complexes. Suppose that we are given splittings $B^ n = A^ n \oplus C^ n$ compatible with the maps in the displayed sequence. Let $s^ n : C^ n \to B^ n$ and $\pi ^ n : B^ n \to A^ n$ be the corresponding maps. Then the family of morphisms

$\pi ^{n + 1} \circ d_ B^ n \circ s^ n : C^ n \to A^{n + 1}$

define a morphism of complexes $\delta : C^\bullet \to A[1]^\bullet$.

Proof. Denote $i : A^\bullet \to B^\bullet$ and $q : B^\bullet \to C^\bullet$ the maps of complexes in the short exact sequence. Then $i^{n + 1} \circ \pi ^{n + 1} \circ d_ B^ n \circ s^ n = d_ B^ n \circ s^ n - s^{n + 1} \circ d_ C^ n$. Hence $i^{n + 2} \circ d_ A^{n + 1} \circ \pi ^{n + 1} \circ d_ B^ n \circ s^ n = d_ B^{n + 1} \circ (d_ B^ n \circ s^ n - s^{n + 1} \circ d_ C^ n) = - d_ B^{n + 1} \circ s^{n + 1} \circ d_ C^ n$ as desired. $\square$

Comment #5076 by Remy on

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.

Comment #5081 by Laurent Moret-Bailly on

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).

Comment #5289 by on

@#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 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.

Comment #7427 by Elías Guisado on

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

Comment #7428 by Elías Guisado on

not *before

Comment #8369 by on

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 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 commutes for all $n$, then the following diagram commutes: 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.

Comment #8974 by on

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?

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• 2 comment(s) on Section 12.14: Homotopy and the shift functor

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