# In triangulated category, why do we use triangles, but not exact sequences?

Studying the theory of derived category, I came up with a little question.

To define trianglated categories, first we choose a collection of sextuples, called the distinguished triangles.
I know that, for the homotopy category $$K(\mathscr{A})$$ of an abelian category $$\mathscr{A}$$, its distinguished triangles are almost the same to the short exact sequences. (see this post.)
And I think that in order to do the homological argument it's easy to use exact sequences rather than distinguished triangles.

So why do we use distinguished triangles in the definition of triangulated categories?
Can't we define the triangulated categories using the word "choose a collection of maps $$X \to Y \to Z$$, and call it an exact sequence"?
(E.g., for the homotopy category of an abelian category, we choose the (classical) exact sequences as the "exact sequences" of the triangulated category.)

Thank you very much!

• The latter sounds more like an exact category then, which is an additive category together with a class of short exaxt sequences that fullfills some conditions. Maybe that concept is interesting for you.
– Con
Commented Oct 20, 2019 at 11:16
• Is your question only about terminology? Do you just suggest to use "exact sequences" instead of "distinguished triangles"? Commented Oct 20, 2019 at 11:35
• @Sasha Not about terminology. Thank you for pointed out.
– k.j.
Commented Oct 20, 2019 at 11:50
• Well if it's not about terminology, the answer is that in the actual derived category, exact triangles don't behave like exact sequences Commented Oct 20, 2019 at 13:50
• @Max So what is triangles? It seems for me to be the analogues of exact sequences. For example, a cohomological functor (such as the "taking $0$-th cohomology" functor) takes a triangle to a long exact sequence.
– k.j.
Commented Oct 20, 2019 at 17:36

The standard definition of an exact sequence $$0 \to A \to B \to C \to 0$$ includes that $$A \to B$$ is a monomorphism and $$B \to C$$ is an epimorphism. It is a nice exercise to check that in a triangulated category any monomorphism is an embedding of a direct summand, and any epimorphism is a projection onto a direct summand. Thus, any exact sequence in triangulated category is split, i.e., isomorphic to $$0 \to A \to A \oplus B \to B \to 0.$$ In particular, exact sequences in a triangulated category don't give any interesting structure, while distinguished triangles do.