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Let $\mathcal A$ be an abelian category with enough injectives. Let $\mathcal I$ be the collection of injective objects. Let $\mathcal A/\mathcal I$ be the quotient category whose objects are same as those of $\mathcal A$ and morphisms are $\operatorname{Hom}_{\mathcal A/\mathcal I}(X,Y)/R_{X,Y}$, where $R_{X,Y}$, where $R_{X,Y}$ is the collection of all morphisms factoring through some object of $\mathcal I$. It is well-known that $\mathcal A/\mathcal I$ has a natural right triangulated structure, see Theorem 2.12(ii) of Beligiannis–Marmaridis 1994 (DOI link). Consider the stabilization $S(\mathcal A/\mathcal I)$ as defined in Remark 3.2 of Beligiannis 2000 (DOI link).

My question is: I wonder if there is an analogue of Theorem 3.8 of Beligiannis 2000 (previous link) in the above setting? In particular, I wonder if $S(\mathcal A/\mathcal I) \cong D^b(\mathcal A)/K^b(\mathcal I)$ ?

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  • $\begingroup$ I guess the answer is no because $D^b(\mathcal A)/K^b(\mathcal I)$ is triangulated while $S(\mathcal A/\mathcal I)$ is merely right triangulated. $\endgroup$
    – Leo Alonso
    Commented Jan 11 at 22:10
  • $\begingroup$ @LeoAlonso: The stabilization of any right/left triangulated category is triangulated ... $\endgroup$
    – Snake Eyes
    Commented Jan 12 at 0:06
  • $\begingroup$ Do you assume your category $\mathcal{A}$ is Frobenius? $\endgroup$
    – Leo Alonso
    Commented Jan 12 at 17:48
  • $\begingroup$ @LeoAlonso: No I do not .... If $\mathcal A$ were Frobenius, then $S(\mathcal A/\mathcal I)\cong \mathcal A/\mathcal I$ which is then just $D_{sg}(\mathcal A)$ I think ... $\endgroup$
    – Snake Eyes
    Commented Jan 12 at 23:04
  • $\begingroup$ Oh, I see your $S$ denotes stabilization, sorry for the misunderstanding. I am afraid I do not have intuition about this. $\endgroup$
    – Leo Alonso
    Commented Jan 13 at 21:10

1 Answer 1

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This follows by applying Theorem 3.8 of Beligiannis' 2000 paper to the opposite categories.

$\mathcal{I}$ is a full additive subcategory of $\mathcal{A}$, closed under direct summands. It is covariantly finite, since $\mathcal{A}$ has enough injectives, and every $\mathcal{I}$-monic has a cokernel in $\mathcal{A}$, since $\mathcal{A}$ is abelian.

So $\mathcal{I}^{op}$ is a full contravariantly finite additive subcategory of $\mathcal{A}^{op}$, closed under direct summands, and every $\mathcal{I}^{op}$-epic has a kernel in $\mathcal{A}^{op}$.

But these are precisely the hypotheses of Theorem 3.8, so that theorem gives a triangulated equivalence $$S(\mathcal{A}^{op}/\mathcal{I}^{op})\simeq K^{-,b}(\mathcal{I}^{op})/K^{b}(\mathcal{I}^{op}),$$ where the left hand side is the stabilization of a left triangulated category.

But the left hand side is just $S(\mathcal{A}/\mathcal{I})^{op}$ (the stabilization of a right triangulated category), and the right hand side is just $\left(K^{+,b}(\mathcal{I})/K^{b}(\mathcal{I})\right)^{op}$, so there is a triangulated equivalence $$S(\mathcal{A}/\mathcal{I})\simeq K^{+,b}(\mathcal{I})/K^{b}(\mathcal{I}) \simeq D^{b}(\mathcal{A})/K^{b}(\mathcal{I}).$$

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  • $\begingroup$ I guess the only non-trivial part for me is why $S(\mathcal A^{op}/\mathcal I^{op})\cong S(\mathcal A/\mathcal I)^{op}$ ... will think about it $\endgroup$
    – Snake Eyes
    Commented Jan 16 at 9:20
  • $\begingroup$ @SnakeEyes If it’s not clear from the construction of the stabilization of a left/right triangulated category $\mathcal{T}$, it should be clear from the universal property of the functor $\mathcal{T}\to S(\mathcal{T})$. $\endgroup$
    – Jeremy Rickard
    Commented Jan 16 at 23:04

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