The Stacks project

I think this result could be expanded/rephrased in this way:

Lemma. Let $\mathcal{D}$ be a pre-triangulated category.

• For any objects $X$, $Z$ of $\mathcal{D}$ the triangle $(X,X\oplus Z,Z,(1,0),(0,1),0)$ is distinguished.

• Let $(X,Y,Z,f,g,h)$ be a distinguished triangle in $\mathcal{D}$. The following are equivalent:

  1. $h=0$,
  2. $f$ is a monomorphism,
  3. $f$ has a left inverse,
  4. $g$ is an epimorphism, and
  5. $g$ has a right inverse.

Moreover, if these equivalent conditions hold, then for any right inverse $s:Z\to Y$ of $g$ the map $f\oplus s:X\oplus Z\to Y$ is an isomorphism.

Proof. The proof of the last assertion is written in the already existing proof. The first point follows from TR1 and Lemma 13.4.10. We prove the second point. Let $W$ be an object of $\mathcal{D}$. The sequences $$\begin{gather*} \operatorname{Hom}(W,Z)\xrightarrow{h_*} \operatorname{Hom}(W,X[1])\xrightarrow{f[1]_*} \operatorname{Hom}(W,Y[1])\\ \operatorname{Hom}(X[1],W)\xrightarrow{h^*} \operatorname{Hom}(Z,W)\xrightarrow{g^*} \operatorname{Hom}(Y,W) \end{gather*}$$ are exact. By Yoneda, $h=0$ if and only $h_*=0$ for all $W$ and/or $h^*=0$ for all $W$. That is, $h=0$ if and only if $f[1]_*$ is injective for all $W$ and/or $g^*$ is injective for all $W$. In other words, $h=0$ if and only if $f[1]$ is monic and/or $g$ is epic. Since $f$ is monic iff $f[1]$ is, this shows 1$\Leftrightarrow$2$\Leftrightarrow$4. It is clear that 3$\Rightarrow$2 and $5\Rightarrow$4. Now assume 1. Then plugging $(X,Y,Z,f,g,h)$ into $\operatorname{Hom}(-,X)$ and $\operatorname{Hom}(Z,-)$ gives exactness of $$\begin{gather*} \operatorname{Hom}(Y,X)\xrightarrow{f^*} \operatorname{Hom}(X,X)\xrightarrow{h^*=0} 0\\ \operatorname{Hom}(Z,Y)\xrightarrow{g_*} \operatorname{Hom}(Z,Z)\xrightarrow{h_*=0} 0 \end{gather*}$$ Thus there are lifts of $1_X\in\operatorname{Hom}(X,X)$ and $1_Z\in\operatorname{Hom}(Z,Z)$ along these maps, i.e., a left inverse for $f$ and a right inverse for $g$, respectively. This finishes the proof of the second point. $\square$

Actually, one could add another a sixth point into the list of equivalences:

1. There is an isomorphism $\beta:X\oplus Z\cong Y$ such that $(1_X,\beta,1_Z):(X,X\oplus Z,Z,(1,0),(0,1),0)\to(X,Y,Z,f,h,g)$ is a triangle isomorphism.

Implication 6$\Rightarrow$1 is trivial. Implication (1-5)$\Rightarrow$6 is what is the current proof does to show that for any right inverse $s:Z\to Y$ of $g$ the map $f\oplus s:X\oplus Z\to Y$ is an isomorphism.