If the homotopic category is abelian, then it is semi-simple. Let $\sf{A}$ be an abelian category and consider its homotopic category $\sf{K}(\sf{A})$. I am writing some notes about homological algebra and, since the reader usually wonders why this isn't generally an abelian category, I want to prove right after its definition that, if it is abelian, then it is semi-simple. (This means that, at this stage, I haven't yet proved that it is triangulated.)
Consider a short exact sequence in $\sf{K}(\sf{A})$:

The proof (which uses the triangulated structure of $\sf{K}(\sf{A})$) in Gelfand-Manin shows that $M^\bullet$ is homotopy equivalent to $L^\bullet\oplus \operatorname{cone}(\varphi^\bullet)$. This means that a splitting $M^\bullet\to L^\bullet$ of the exact sequence above would be given by the composition
$$M^\bullet\to L^\bullet\oplus \operatorname{cone}(\varphi^\bullet) \to L^\bullet,$$
where the first morphism is the (homotopy) inverse of
\begin{align*}
L^i\oplus M^i\oplus L^{i+1} &\to M^i \\
(l,m,l') &\mapsto \varphi^i(l)+m
\end{align*}
and the second is just the usual projection. Now, a morphism $M^\bullet\to L^\bullet\oplus \operatorname{cone}(\varphi^\bullet)$ in $\sf{K}(\sf{A})$ would be an equivalence class of morphisms $M^i\to L^i$, $M^i\to M^i$ and $M^i\to L^{i+1}$. The only reasonable choice seems to take the morphism on the middle to be the identity and the other two to be zero. This would imply that our splitting $M^\bullet\to L^\bullet$ is the zero morphism. But the zero morphism is not usually homotopic to $\operatorname{id}_{L^\bullet}$.
What am I doing wrong and/or how can we prove this result without using the triangulated structure of $\sf{K}(\sf{A})$?
 A: As JHF said in the comments, you shouldn't take a "reasonable" choice. Indeed, if you expect to construct a retraction $M^\bullet\to L^\bullet$ with only a "reasonable" map, the map would not use the fact that the category is semi-simple and would work in any category.
I would like to give a stronger statement : any triangulated category (not just the homotopy category) which is also an abelian category is semi-simple. (Here I use Gelfand-Manin definition of semi-simple : that is that any short exact sequence splits).
To see this, we have the following lemmas
Lemma 1 : Let $A\to B\to C\to A[1]$ be a distinguished triangle, then the following are equivalent :

*

*$C\to A[1]$ is the zero map

*$A\to B$ has a retraction

*$B\to C$ has a section

*There is an isomorphism $B\simeq A\oplus C$ such that $A\to B$ corresponds to the canonical inclusion $A\to A\oplus C$ and $B\to C$ correspond to the canonical projection $A\oplus C\to C$.

Proof :
$2.\Leftrightarrow 3. \Leftrightarrow 4.$ hold in any additive category.
$1.\Rightarrow 2.$ : Apply the functor $Hom(.,A)$, you get a long exact sequence
$$ ...\to Hom(B,A)\to Hom(A,A)\to Hom(C[-1],A)\to ...$$
The map $Hom(A,A)\to Hom(C[-1],A)$ is zero since this is the precomposition with $C[-1]\to A$ which is zero by hypothesis. It follows from exactness that $Hom(B,A)\to Hom(A,A)$ is onto. So $id_A$ has a preimage $r$ which is thus a retraction of $A\to B$.
$2.\Rightarrow 1.$ Since there is a map $B\to A$ such that the composition $A\to B\to A$ is the identity, the map $C[-1]\to A$ can be written as the composition $C[-1]\to A\to B\to A$, but $C[-1]\to A\to B$ is zero, so $C[-1]\to A$ is zero.
Lemma 2 : In a triangulated category, every mono and every epi are split.
Proof : The proof for epi is dual of the proof for mono, so let us prove only the mono case. Consider a monomorphism $i:A\to B$. From the axioms of triangulated categories you can form a distinguished triangle
$$A\rightarrow B\to C\rightarrow A[1]$$
We have $C[-1]\to A\to B$ is the zero map, but since $A\to B$ is mono, $C[-1]\to A$ is zero, and by lemma 1, $A\to B$ has a retraction.

The claim is now straightforward : consider a short exact sequence in a category which is both triangulated and abelian. Consider a short exact sequence :
$$0\to A\to B\to C\to 0$$
Since the category is abelian, $A\to B$ is mono. Since the category is triangulated, by lemma 2, $A\to B$ splits.
