# About "strict" short exact sequences in quasi-abelian subcategory of a derived category

I'm reading Bridgeland's Stability conditions on K3 surfaces. In Lemma 4.4 there appears a full quasi-abelian subcategory $$\mathscr{A} \subset \mathscr{D}$$ of a triangulated category $$\mathscr{D} = \mathscr{D}(X)$$ of a smooth variety $$X$$. Then he considers a "strict" short exact sequence $$0 \to A \to B \to C \to 0, \tag{*}$$ with $$A, B, C \in \mathscr{A}$$.

1. Question: How exactly does one define (short) exact sequences in a quasi-abelian category?

The definition of abelian categories $$\operatorname{Ker}(f_i) = \operatorname{Im}(f_{i+1})$$ seems problematic, because I don't know how to define the image. In general it seems that $$\operatorname{coker ker} f \neq\operatorname{ker coker} f$$, see Wikipedia.

2. Question: What does "strict" mean in that context? Does it have anything to do with the first question, or does it just mean that $$A \neq 0 \neq B$$, i.e. those are non-trivial subobjects / quotients?

After that Bridgeland goes on to argue that $$f(B) = f(A) + f(C)$$, where $$f: K(\mathscr{D}) \to \mathbb{R}$$ is an additive function.

3. Question: Why does a short exact sequence $$(*)$$ in $$\mathscr{A}$$ induce a triangle in $$\mathscr{D}$$?

I know that this is true if $$\mathscr{A}$$ is abelian, and $$\mathscr{D} = \mathscr{D}(\mathscr{A})$$ is the derived category of $$\mathscr{A}$$, but I see no reason why this is true in our case.

• I don't really know, but just a possibility: Maybe one should define a short exact sequence as a distinguished triangle all of whose elements lie in $\mathcal{A}$? somewhat analogously, it is a map $A \to B$ of objects in $\mathcal{A}$ whose cone also lies in $\mathcal{A}$. Nov 21, 2019 at 19:47

Let $$\mathscr{A}$$ be an additive category with kernels and cokernels. A morphism $$f: A \to B$$ is called strict, if the canonical map $$\operatorname{coker ker} f \to \operatorname{ker coker} f$$ is an isomorphism.
$$\mathscr{A}$$ is called quasi-abelian if the pullback of every strict epi is a strict epi, and the pushout of every strict mono is a strict mono. Then a strict short exact sequence is a diagram $$0 \to A \xrightarrow{i} B \xrightarrow{j} C \to 0$$ in which $$i$$ is the kernel of $$j$$ and $$j$$ is the cokernel of $$i$$. In particular $$i$$ is mono and $$j$$ is epi, so $$\operatorname{ker} i = 0$$ and $$\operatorname{coker} j = 0$$. Hence \begin{align} \operatorname{ker coker} i = \operatorname{ker}j & = i = \operatorname{coker ker} i \\ \operatorname{coker ker} j = \operatorname{coker} i & = j = \operatorname{ker coker} j, \end{align} so both $$i$$ and $$j$$ are strict. In Lemma 4.3 Bridgeland then proceeds to prove that in exactly the situation which appears in the K3 surface paper, strict short exact sequences in $$\mathscr{A}$$ are in one-to-one correspondence to exact triangles in $$\mathscr{D}$$.