Lemma 22.20.13. In Situation 22.20.2.

1. Given an admissible short exact sequence $x\xrightarrow {\alpha } y\xrightarrow {\beta } z$. Then there exists a homotopy equivalence $e:C(\alpha )\to z$ such that the diagram

22.20.13.1
\begin{equation} \label{dga-equation-cone-isom-triangle} \vcenter { \xymatrix{ x\ar[r]^{\alpha }\ar[d] & y\ar[r]^{b}\ar[d] & C(\alpha )\ar[r]^{-c}\ar@{.>}[d]^{e} & x\ar[d] \\ x\ar[r]^{\alpha } & y\ar[r]^{\beta } & z\ar[r]^{\delta } & x } } \end{equation}

defines an isomorphism of triangles in $K(\mathcal{A})$. Here $y\xrightarrow {b}C(\alpha )\xrightarrow {c}x$ is the admissible short exact sequence given as in axiom (C).

2. Given a morphism $\alpha : x \to y$ in $\text{Comp}(\mathcal{A})$, let $x \xrightarrow {\tilde{\alpha }} \tilde{y} \to y$ be the factorization given as in Lemma 22.20.6, where the admissible monomorphism $x \xrightarrow {\tilde{\alpha }} y$ extends to the admissible short exact sequence

$\xymatrix{ x \ar[r]^{\tilde{\alpha }} & \tilde{y} \ar[r] & z }$

Then there exists an isomorphism of triangles

$\xymatrix{ x \ar[r]^{\tilde{\alpha }} \ar[d] & \tilde{y} \ar[r] \ar[d] & z \ar[r]^{\delta } \ar@{.>}[d]^{e} & x \ar[d] \\ x \ar[r]^{\alpha } & y \ar[r] & C(\alpha ) \ar[r]^{-c} & x }$

where the upper triangle is the triangle associated to the sequence $x \xrightarrow {\tilde{\alpha }} \tilde{y} \to z$.

Proof. For (1), we consider the more complete diagram, without the sign change on $c$:

$\xymatrix{ x\ar@<0.5ex>[r]^{\alpha } \ar[d] & y\ar@<0.5ex>[l]^{\pi } \ar@<0.5ex>[r]^{b}\ar[d] & C(\alpha )\ar@<0.5ex>[l]^{p} \ar@<0.5ex>[r]^{c}\ar@{.>}@<0.5ex>[d]^{e} & x\ar@<0.5ex>[l]^{\sigma } \ar[d]\ar@<0.5ex>[r]^{\alpha } & y\ar@<0.5ex>[l]^{\pi } \\ x\ar@<0.5ex>[r]^{\alpha } & y\ar@<0.5ex>[r]^{\beta } \ar@<0.5ex>[l]^{\pi } & z\ar[r]^{\delta }\ar@<0.5ex>[l]^{s} \ar@{.>}@<0.5ex>[u]^{f} & x }$

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 $y\xrightarrow {b}C(\alpha )\xrightarrow {c}x$ 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 22.20.1.

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 22.20.13.1 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) \\ = & 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 22.20.4). Hence $e$ and $f$ are homotopy inverses. Finally, to check that the right square of diagram 22.20.13.1 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 22.20.6, so the second morphism is a homotopy equivalence. By Lemmas 22.20.3 and 22.20.12, there exists an isomorphism of triangles between

$x \xrightarrow {\alpha } y \to C(\alpha ) \to x \quad \text{and}\quad x \xrightarrow {\tilde{\alpha }} \tilde{y} \to C(\tilde{\alpha }) \to x$

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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