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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