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If you'll have patience with me, I understand that this is not the first time that a variation of this question is being asked on MathOverflow, but alas, I am unable to truly make sense of those answers, so pray have patience with me.

The problem I have is that I frankly find myself unable to truly understand how distinguished triangles (or exact triangles, depending on your taste) are generalizations of short exact sequences.

I suppose that what I would really need to get this through my skull would be to see an explanation that proceeds in one of two fashions:

  1. You start with a triangulated category, add some additional structure to make it into an abelian category, and show how that renders all distinguished triangles into short exact sequences.
  2. You start with an abelian category, and look at a generic short exact sequence of three objects. You remove some structure so as to make the category merely triangulated, and show how in doing so, while before you could say that the three objects were related by a short exact sequence, now you can only say that they are related by means of a distinguished triangle.

Most ideally, I'd like to see both explanations of course.

I thank you in advance. Such an explanation would make life far easier for me.

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    $\begingroup$ Definitely not an answer, only a small comment, but I always thought of exact triangles as generalizations of long exact sequences (in (co)homology) rather than short ones. I guess this might sound like only a semantic distinction, but my point is about the different emphasis. I might be biased, though, because my context for triangulated categories is derived categories (of sheaves), where you want to encode homological information in the morphisms themselves (i.e. turn quasi-isomorphisms into actual isomorphisms). $\endgroup$
    – M.G.
    Sep 5, 2022 at 19:45
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    $\begingroup$ This fits neither 1. nor 2. but a short exact sequence in an abelian category $A$ yields an exact triangle in the derived category of $A$. That's one sense in which it is a generalization (more generally, any triangulated category with a $t$-structure has an abelian heart, and triangles between objects in the heart are exactly short exact sequences) $\endgroup$ Sep 5, 2022 at 20:33
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    $\begingroup$ I think the correct style explanation is something like (3) you start with an abelian category, embed it into a triangulated category, and show that under the embedding the short exact sequences are precisely those that are sent to distinguished triangles. Would that help? $\endgroup$ Sep 7, 2022 at 4:19
  • $\begingroup$ @MikeShulman It might very well! Thanks for the tip! I'll have a look at it! $\endgroup$ Sep 7, 2022 at 14:26
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    $\begingroup$ The triangulated category you embed it in should probably be some homotopy category of unbounded chain complexes in your abelian category. I don't have time to work out or look up the details right now, or I'd post it as an Answer. $\endgroup$ Sep 7, 2022 at 18:24

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