Let $\mathscr{D}$ be a triangulated category. Let $$X \longrightarrow Y \longrightarrow Z \longrightarrow X[1]$$ be a triangle (not necessarily distinguished). We call it special if for each $E \in \mathscr{D}$ the induced long sequence of abelian groups $$\cdots Hom(E,X) \rightarrow Hom(E,Y) \rightarrow Hom(E,Z) \rightarrow Hom(E,X[1]) \cdots $$ is exact. Since for each $E \in \mathscr{D}$ the functor $Hom(E,)$ is homological, distinguished triangles are special. But is it true that special triangles are distinguished? Thank you very much.
1 Answer
No, for instance if you multiply one of the maps in a distinguished triangle by minus one, then the resulting triangle will still be special but need no longer be distinguished.

$\begingroup$ See also here: mathoverflow.net/questions/10187/… $\endgroup$– RasmusFeb 3, 2017 at 8:55