Proof.
Proof of (1). We have $C(\alpha ) = L \oplus K$ and we simply define $C(\alpha ) \to M$ via the projection onto $L$ followed by $\beta $. This defines a morphism of differential graded modules because the compositions $K^{n + 1} \to L^{n + 1} \to M^{n + 1}$ are zero. Choose splittings $s : M \to L$ and $\pi : L \to K$ with $\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{Im}}(s)$ and set $\delta = \pi \circ \text{d}_ L \circ s$ as usual. To get a homotopy inverse we take $M \to C(\alpha )$ given by $(s , -\delta )$. This is compatible with differentials because $\delta ^ n$ can be characterized as the unique map $M^ n \to K^{n + 1}$ such that $\text{d} \circ s^ n - s^{n + 1} \circ \text{d} = \alpha \circ \delta ^ n$, see proof of Homology, Lemma 12.14.10. The composition $M \to C(f) \to M$ is the identity. The composition $C(f) \to M \to C(f)$ is equal to the morphism
\[ \left( \begin{matrix} s \circ \beta
& 0
\\ -\delta \circ \beta
& 0
\end{matrix} \right) \]
To see that this is homotopic to the identity map use the homotopy $h : C(\alpha ) \to C(\alpha )$ given by the matrix
\[ \left( \begin{matrix} 0
& 0
\\ \pi
& 0
\end{matrix} \right) : C(\alpha ) = L \oplus K \to L \oplus K = C(\alpha ) \]
It is trivial to verify that
\[ \left( \begin{matrix} 1
& 0
\\ 0
& 1
\end{matrix} \right) - \left( \begin{matrix} s
\\ -\delta
\end{matrix} \right) \left( \begin{matrix} \beta
& 0
\end{matrix} \right) = \left( \begin{matrix} \text{d}
& \alpha
\\ 0
& -\text{d}
\end{matrix} \right) \left( \begin{matrix} 0
& 0
\\ \pi
& 0
\end{matrix} \right) + \left( \begin{matrix} 0
& 0
\\ \pi
& 0
\end{matrix} \right) \left( \begin{matrix} \text{d}
& \alpha
\\ 0
& -\text{d}
\end{matrix} \right) \]
To finish the proof of (1) we have to show that the morphisms $-p : C(\alpha ) \to K[1]$ (see Definition 22.6.1) and $C(\alpha ) \to M \to K[1]$ agree up to homotopy. This is clear from the above. Namely, we can use the homotopy inverse $(s, -\delta ) : M \to C(\alpha )$ and check instead that the two maps $M \to K[1]$ agree. And note that $p \circ (s, -\delta ) = -\delta $ as desired.
Proof of (2). We let $\tilde f : K \to \tilde L$, $s : L \to \tilde L$ and $\pi : L \to L$ be as in Lemma 22.7.4. By Lemmas 22.6.2 and 22.9.3 the triangles $(K, L, C(f), i, p)$ and $(K, \tilde L, C(\tilde f), \tilde i, \tilde p)$ are isomorphic. Note that we can compose isomorphisms of triangles. Thus we may replace $L$ by $\tilde L$ and $f$ by $\tilde f$. In other words we may assume that $f$ is an admissible monomorphism. In this case the result follows from part (1).
$\square$
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