**Proof.**
Proof of (1). We have $C(\alpha )^ n = B^ n \oplus A^{n + 1}$ and we simply define $C(\alpha )^ n \to C^ n$ via the projection onto $B^ n$ followed by $\beta ^ n$. This defines a morphism of complexes because the compositions $A^{n + 1} \to B^{n + 1} \to C^{n + 1}$ are zero. To get a homotopy inverse we take $C^\bullet \to C(\alpha )^\bullet $ given by $(s^ n , -\delta ^ n)$ in degree $n$. This is a morphism of complexes because the morphism $\delta ^ n$ can be characterized as the unique morphism $C^ n \to A^{n + 1}$ such that $d \circ s^ n - s^{n + 1} \circ d = \alpha \circ \delta ^ n$, see proof of Homology, Lemma 12.14.10. The composition $C^\bullet \to C(\alpha )^\bullet \to C^\bullet $ is the identity. The composition $C(\alpha )^\bullet \to C^\bullet \to C(\alpha )^\bullet $ is equal to the morphism

\[ \left( \begin{matrix} s^ n \circ \beta ^ n
& 0
\\ -\delta ^ n \circ \beta ^ n
& 0
\end{matrix} \right) \]

To see that this is homotopic to the identity map use the homotopy $h^ n : C(\alpha )^ n \to C(\alpha )^{n - 1}$ given by the matrix

\[ \left( \begin{matrix} 0
& 0
\\ \pi ^ n
& 0
\end{matrix} \right) : C(\alpha )^ n = B^ n \oplus A^{n + 1} \to B^{n - 1} \oplus A^ n = C(\alpha )^{n - 1} \]

It is trivial to verify that

\[ \left( \begin{matrix} 1
& 0
\\ 0
& 1
\end{matrix} \right) - \left( \begin{matrix} s^ n
\\ -\delta ^ n
\end{matrix} \right) \left( \begin{matrix} \beta ^ n
& 0
\end{matrix} \right) = \left( \begin{matrix} d
& \alpha ^ n
\\ 0
& -d
\end{matrix} \right) \left( \begin{matrix} 0
& 0
\\ \pi ^ n
& 0
\end{matrix} \right) + \left( \begin{matrix} 0
& 0
\\ \pi ^{n + 1}
& 0
\end{matrix} \right) \left( \begin{matrix} d
& \alpha ^{n + 1}
\\ 0
& -d
\end{matrix} \right) \]

To finish the proof of (1) we have to show that the morphisms $-p : C(\alpha )^\bullet \to A^\bullet [1]$ (see Definition 13.9.1) and $C(\alpha )^\bullet \to C^\bullet \to A^\bullet [1]$ agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse $(s, -\delta ) : C^\bullet \to C(\alpha )^\bullet $ and check instead that the two maps $C^\bullet \to A^\bullet [1]$ agree. And note that $p \circ (s, -\delta ) = -\delta $ as desired.

Proof of (2). We let $\tilde f : K^\bullet \to \tilde L^\bullet $, $s : L^\bullet \to \tilde L^\bullet $ and $\pi : \tilde L^\bullet \to L^\bullet $ be as in Lemma 13.9.6. By Lemmas 13.9.2 and 13.9.13 the triangles $(K^\bullet , L^\bullet , C(f), i, p)$ and $(K^\bullet , \tilde L^\bullet , C(\tilde f), \tilde i, \tilde p)$ are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace $L^\bullet $ by $\tilde L^\bullet $ and $f$ by $\tilde f$. In other words we may assume that $f$ is a termwise split injection. In this case the result follows from part (1).
$\square$

## Comments (5)

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