Lemma 14.29.2. Let \mathcal{A} be an abelian category. Let
0 \to A \oplus A \to B \to C \to 0
be a short exact sequence of chain complexes of \mathcal{A}. Suppose given in addition morphisms s_ n : C_ n \to B_ n splitting the associated short exact sequence in degree n. Let \delta (s) : C \to (A \oplus A)[-1] = A[-1] \oplus A[-1] be the associated morphism of complexes, see Homology, Lemma 12.14.4. If \delta (s) factors through the morphism (1, -1) : A[-1] \to A[-1] \oplus A[-1], then there is a unique morphism B \to \diamond A fitting into a commutative diagram
\xymatrix{ 0 \ar[r] & A \oplus A \ar[d] \ar[r] & B \ar[r] \ar[d] & C \ar[d] \ar[r] & 0 \\ 0 \ar[r] & A \oplus A \ar[r] & \diamond A \ar[r] & A[-1] \ar[r] & 0 }
where the vertical maps are compatible with the splittings s_ n and the splittings of \diamond A_ n \to A[-1]_ n as well.
Proof.
Denote (p_ n, q_ n) : B_ n \to A_ n \oplus A_ n the morphism \pi _ n of Homology, Lemma 12.14.4. Also write (a, b) : A \oplus A \to B, and r : B \to C for the maps in the short exact sequence. Write the factorization of \delta (s) as \delta (s) = (1, -1) \circ f. This means that p_{n - 1} \circ d_{B, n} \circ s_ n = f_ n, and q_{n - 1} \circ d_{B, n} \circ s_ n = - f_ n, and Set B_ n \to \diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1} equal to (p_ n, q_ n, f_ n \circ r_ n).
Now we have to check that this actually defines a morphism of complexes. We will only do this in the case of abelian groups. Pick x \in B_ n. Then x = a_ n(x_1) + b_ n(x_2) + s_ n(x_3) and it suffices to show that our definition commutes with differential for each term separately. For the term a_ n(x_1) we have (p_ n, q_ n, f_ n \circ r_ n)(a_ n(x_1)) = (x_1, 0, 0) and the result is obvious. Similarly for the term b_ n(x_2). For the term s_ n(x_3) we have
\begin{eqnarray*} (p_ n, q_ n, f_ n \circ r_ n)(d_ n(s_ n(x_3))) & = & (p_ n, q_ n, f_ n \circ r_ n)( \\ & & \ \ \ \ \ a_ n(f_ n(x_3)) - b_ n(f_ n(x_3)) + s_ n(d_ n(x_3))) \\ & = & (f_ n(x_3), -f_ n(x_3), f_ n(d_ n(x_3))) \end{eqnarray*}
by definition of f_ n. And
\begin{eqnarray*} d_ n(p_ n, q_ n, f_ n \circ r_ n)(s_ n(x_3)) & = & d_ n(0, 0, f_ n(x_3)) \\ & = & (f_ n(x_3), - f_ n(x_3), d_{A[-1], n}(f_ n(x_3))) \end{eqnarray*}
The result follows as f is a morphism of complexes.
\square
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