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.


1 Answer 1


There is such a sequence, but it's not very interesting.

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.


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