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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.

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  • $\begingroup$ 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 $\mathcal{A}$. $\endgroup$ – Sasha Nov 21 '19 at 19:47
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I found the answer two 1. and 2. in Bridgeland's previous work Stability Conditions on Triangulated Categories:

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 $\mathscr{D}$.

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  • $\begingroup$ You should accept your answer :). $\endgroup$ – Slup Nov 22 '19 at 9:09
  • $\begingroup$ Sure, but one has to wait for 2 days to do so. $\endgroup$ – red_trumpet Nov 22 '19 at 9:40

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