Let $\mathcal{A}$ be an abelian category, we have $D(\mathcal{A})$
For $A,C \in \mathcal{A}$, we know that $Ext^1(A,C)$ classifies extensions $0 \to C \to B \to A \to 0$
I want an analog for $Der(\mathcal{A})$.
Exact sequences should be replaced with distinguished triangles. A splitting condition is precisely that the natural element in $(A,C[1])$ vanishes. But $Hom(A,C[1])$ doesn't classify distinguished triangles, indeed if $\mathcal{A} = Vect$ then all distinguished triangles split, but $Hom(A,C[1])$ can be nonzero.
In other words there is a map $A \to C[1]$ that can't be raised to a distinguished triangle I guess?
Thus my question is, is there a way to compute the class of distinguished triangles $C \to B \to A$?