Why are semi-simple abelian categories pre-triangulated?

Question

The question says it all really. Page 38 of the notes here claim that a semi-simple abelian category is pre-triangulated. Here semi-simple just means that all monos (equiv. all epis) (equiv. all short exact sequences) are split. But I am stuck on how to show that an arbitrary morphism in such a category extends to a triangle. The triangles are supposed to be sequences of the form $$ M \stackrel{f}{\longrightarrow} N \stackrel{g}{\longrightarrow} P \stackrel{h}{\longrightarrow} M $$ which are exact at $N$ and $P$, and for which $\ker(f) \simeq \text{coker} (h)$. Given a morphism $f: M \rightarrow N$, the obvious choice for a triangle (which seems to also be the choice they are taking) is $$ M \stackrel{f}{\longrightarrow} N \stackrel{g}{\longrightarrow} \ker(f) \oplus \text{coker}(f) \stackrel{h}{\longrightarrow} M $$ where $g$ is given by the canonical epimorphism from $N$ to $\text{coker}(f)$ and $h$ is given by the canonical inclusion from $\ker(f)$ to $M$. But then we certainly don't have $\ker(f) \simeq \text{coker(h)}$. Indeed, \begin{align} \text{coker}(h) & \simeq M / \text{im}(h) \\ & \simeq M / \ker(f) \\ & \simeq \text{im}(f) \end{align} Is anyone able to give me an idea of how the second triangulated category axiom is satisfied, that is, how an arbitrary morphism can be extended to a triangle?

Answer 1

As you guessed, this is just an error and it should instead say $\text{im}(h) \simeq \ker(f)$. Note that here $\simeq$ should mean equivalence as subobjects of $M$, not just an abstract isomorphism.