3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I'm wondering if this is a mistake in the notes. It would make more sense if they defined exactness of a triangle to require that $\text{im}(h) \simeq \ker(f)$. Is that what is meant? $\endgroup$ – Luke Jun 26 '18 at 12:39
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you! Sorry if it was supposed to be obvious, I just didn't want to carry on with the wrong assumption. $\endgroup$ – Luke Jun 26 '18 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.