For (1), we consider the more complete diagram, \emph{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 \ref{lemma-get-triangle}. \medskip\noindent 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 \ref{equation-cone-isom-triangle} 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 $\sigma \in \Hom^{-1}(x, C(\alpha))$ (cf. proof of Lemma \ref{lemma-id-cone-null}). Hence $e$ and $f$ are homotopy inverses. Finally, to check that the right square of diagram \ref{equation-cone-isom-triangle} commutes up to homotopy, it suffices to check that $-cf=\delta$. This follows from $$ -cf = -c(bs-\sigma\delta) = c\sigma\delta = \delta $$ since $cb=0$. \medskip\noindent For (2), consider the factorization $x\xrightarrow{\tilde{\alpha}}\tilde{y}\to y$ given as in Lemma \ref{lemma-factor}, so the second morphism is a homotopy equivalence. By Lemmas \ref{lemma-cone} and \ref{lemma-analogue-third-isomorphism}, 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).