# Yoneda extensions in derived categories

If given an abelian category $$\mathcal{A}$$, we can consider the bounded derived category $$D^b(\mathcal{A})$$. For two objects $$A,B \in \mathcal{A}$$, we know that there is a natural identification between $$\text{Ext}_{\mathcal{A}}^i(A,B)$$ and $$\text{Hom}_{D^b(\mathcal{A})}(A,B[i])$$ using Yoneda extensions. This is proven in Verdier's thesis (see also group of Yoneda extensions and the EXT groups defined via derived category).

I wondered whether one in general can identify the morphisms in the derived category with Yoneda extensions in the following way:

Given two complexes $$E^\bullet, F^\bullet \in D^b(\mathcal{A})$$, can we identify the set of morphisms $$\text{Hom}_{D^b(\mathcal{A})}(E^\bullet, F^\bullet[i])$$ with sequences of morphisms in the derived category $$D^b(\mathcal{A})$$ $$F^\bullet \to Z_{i-1}^\bullet \to \dots \to Z_0^\bullet \to E^\bullet$$ such that the above sequence breaks into distinguished triangles, i.e. there exist distinguished triangles $$F^\bullet \to Z_{i-1}^\bullet \to G_{i-1}^\bullet, \quad G_{i-1}^\bullet \to Z_{i-2}^\bullet \to G_{i-2}^\bullet, \dots$$ In the case that $$E^\bullet,F^\bullet$$ are complexes concentrated in degree 0 (i.e. objects in $$\mathcal{A}$$), this is precisely the Yoneda extension with the $$Z_j^\bullet, G_j^\bullet$$ also complexes concentrated in degree 0.

For $$i=1$$ and $$E^\bullet, F^\bullet$$ arbitrary, this also holds true since a morphism $$E^\bullet \to F^\bullet[1]$$ can be completed to a distinguished triangle $$G^\bullet \to E^\bullet \to F^\bullet[1]$$ which we can rotate to obtain $$F^\bullet \to G^\bullet \to E^\bullet.$$ I am curious about the general case.

Given an element of $$\text{Hom}_{D^b(\mathcal{A})}(E,F[i])$$, then in the same way you describe, this gives a distinguished triangle $$F\to Z_{i-1}\to E[-i+1].$$
Then you can just take the distinguished triangles $$E[-i+1]\to0\to E[-i+2]$$, $$E[-i+2]\to0\to E[-i+3]$$, $$\dots$$, up to $$E[-1]\to0\to E$$, which can be spliced to give a sequence $$F\to Z_{i-1}\to Z_{i-2}\to\dots\to Z_0\to E,$$ where $$Z_{i-2}=\dots= Z_0=0$$.
This might seem like cheating, but I don't think it is. In the case of (a nonzero element of) $$\text{Ext}_{\mathcal{A}}^i(A,B)$$, the only reason you can't have zero objects in your sequence
$$B \to Z_{i-1} \to Z_{i-2}\to\dots \to Z_0 \to A$$ is that you insist that the $$Z_j$$ are not just objects of $$D^b(\mathcal{A})$$, but of $$\mathcal{A}$$. If you drop this requirement, then you just have too much freedom.