Proof.
For (1), we consider the more complete diagram, without the sign change on c:
\xymatrix{ x\ar@<0.5ex>[r]^{\alpha } \ar[d] & y\ar@<0.5ex>[l]^{\pi } \ar@<0.5ex>[r]^{b}\ar[d] & C(\alpha )\ar@<0.5ex>[l]^{p} \ar@<0.5ex>[r]^{c}\ar@{.>}@<0.5ex>[d]^{e} & x[1]\ar@<0.5ex>[l]^{\sigma } \ar[d]\ar@<0.5ex>[r]^{\alpha } & y[1]\ar@<0.5ex>[l]^{\pi } \\ x\ar@<0.5ex>[r]^{\alpha } & y\ar@<0.5ex>[r]^{\beta } \ar@<0.5ex>[l]^{\pi } & z\ar[r]^{\delta }\ar@<0.5ex>[l]^{s} \ar@{.>}@<0.5ex>[u]^{f} & x[1] }
where the admissible short exact sequence x\xrightarrow {\alpha } y\xrightarrow {\beta } z is given the splitting \pi , s, and the admissible short exact sequence y\xrightarrow {b}C(\alpha )\xrightarrow {c}x[1] is given the splitting p, \sigma . Note that (identifying hom-sets under shifting)
\alpha = pd(\sigma ) = -d(p)\sigma ,\quad \delta = \pi d(s) = -d(\pi )s
by the construction in Lemma 22.27.1.
We define e=\beta p and f=bs-\sigma \delta . We first check that they are morphisms in \text{Comp}(\mathcal{A}). To show that d(e)=\beta d(p) vanishes, it suffices to show that \beta d(p)b and \beta d(p)\sigma both vanish, whereas
\beta d(p)b = \beta d(pb) = \beta d(1_ y) = 0,\quad \beta d(p)\sigma = -\beta \alpha = 0
Similarly, to check that d(f)=bd(s)-d(\sigma )\delta vanishes, it suffices to check the post-compositions by p and c both vanish, whereas
\begin{align*} pbd(s) - pd(\sigma )\delta = & d(s)-\alpha \delta = d(s)-\alpha \pi d(s) = 0 \\ cbd(s)-cd(\sigma )\delta = & -cd(\sigma )\delta = -d(c\sigma )\delta = 0 \end{align*}
The commutativity of left two squares of the diagram 22.27.13.1 follows directly from definition. Before we prove the commutativity of the right square (up to homotopy), we first check that e is a homotopy equivalence. Clearly,
ef=\beta p (bs-\sigma \delta )=\beta s=1_ z
To check that fe is homotopic to 1_{C(\alpha )}, we first observe
b\alpha = bpd(\alpha ) = d(\sigma ),\quad \alpha c = -d(p)\sigma c = -d(p),\quad d(\pi )p = d(\pi )s\beta p = -\delta \beta p
Using these identities, we compute
\begin{align*} 1_{C(\alpha )} = & bp + \sigma c \quad (\text{from }y \xrightarrow {b} C(\alpha ) \xrightarrow {c} x[1]) \\ = & b(\alpha \pi + s\beta )p + \sigma (\pi \alpha )c \quad (\text{from }x \xrightarrow {\alpha } y \xrightarrow {\beta } z) \\ = & d(\sigma )\pi p + bs\beta p - \sigma \pi d(p) \quad (\text{by the first two identities above}) \\ = & d(\sigma )\pi p + bs\beta p - \sigma \delta \beta p + \sigma \delta \beta p - \sigma \pi d(p) \\ = & (bs - \sigma \delta )\beta p + d(\sigma )\pi p - \sigma d(\pi )p - \sigma \pi d(p)\quad (\text{by the third identity above}) \\ = & fe + d(\sigma \pi p) \end{align*}
since \sigma \in \mathop{\mathrm{Hom}}\nolimits ^{-1}(x, C(\alpha )) (cf. proof of Lemma 22.27.4). Hence e and f are homotopy inverses. Finally, to check that the right square of diagram 22.27.13.1 commutes up to homotopy, it suffices to check that -cf=\delta . This follows from
-cf = -c(bs-\sigma \delta ) = c\sigma \delta = \delta
since cb=0.
For (2), consider the factorization x\xrightarrow {\tilde{\alpha }}\tilde{y}\to y given as in Lemma 22.27.6, so the second morphism is a homotopy equivalence. By Lemmas 22.27.3 and 22.27.12, there exists an isomorphism of triangles between
x \xrightarrow {\alpha } y \to C(\alpha ) \to x[1] \quad \text{and}\quad x \xrightarrow {\tilde{\alpha }} \tilde{y} \to C(\tilde{\alpha }) \to x[1]
Since we can compose isomorphisms of triangles, by replacing \alpha by \tilde{\alpha }, y by \tilde{y}, and C(\alpha ) by C(\tilde{\alpha }), we may assume \alpha is an admissible monomorphism. In this case, the result follows from (1).
\square
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