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Hello,

I have terminological question. Consider the following properties of a full subcategory $B \subset A$, where $A$ is an abelian category, and we assume $B$ to be closed under finite direct sums.

(1) $B$ is closed under subs, quotients and extensions, i.e. for a short exact sequence $0 \to X \to Y \to Z \to 0$ in $A$, $Y$ is in $B$ if and only if $X,Z$ are in $B$.

(2) $B$ is closed under kernels, cokernels and extensions.

(3) $B$ is closed under kernels and cokernels.

I know that if $B$ satisfies (1) it is called Serre subcategory. Do the other two have a name?

I ask this because it seems that for derived category of $A$ with cohomologies in $B$ to be a triangulated sub-category of derived category of $A$, it is enough to have property (2). For example, it seems constructible sheaves satisfy (2) but not (1).

Thank you, Sasha

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  • $\begingroup$ (2) is just an abelian subcategory and (3) is an exact subcategory if you assume that $B$ contains a zero object $\endgroup$
    – Fernando Muro
    Commented Jan 19, 2012 at 11:22
  • $\begingroup$ @Martin: I would say a thick subcategory is one which is closed under summands and has the two out of three property for short exact sequences rather than being the same as a Serre subcategory. $\endgroup$
    – Greg Stevenson
    Commented Jan 19, 2012 at 12:38
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    $\begingroup$ I am sorry, I changed (3); I wrote by mistake "closed under extensions". I wanted (3) as it is now; Then, as far as I understand, $B$ satisfying (3) is Abelian by itself. $\endgroup$
    – Sasha
    Commented Jan 19, 2012 at 13:28
  • $\begingroup$ I correct myself, I think that (3) is an abelian subcategory and (2) a thick abelian subcategory, but I can still be mistaken $\endgroup$
    – Fernando Muro
    Commented Jan 19, 2012 at 14:11

1 Answer 1

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Since $A$ is an abelian category, it is in particular additive, i.e. has a zero object, a product functor: $A\times A\rightarrow A$, and is Ab-enriched, i.e. hom sets are abelian groups. Since $B$ is a full subcategory, it is also Ab-enriched. If $B$ is closed under finite direct sums, then $B$ also contains the zero object (via an empty direct sum). According to page 5 of Weibel's An Introduction to Homological Algebra, finite direct sums are the same as finite products in $A$. Hence, if $B$ is closed under finite direct sums then it also has a product functor: $B\times B\rightarrow B$. Thus, before we even place (1)-(3) we know $B$ is additive.

Now, if $B$ satisfies (3) then it's an abelian subcategory because it's abelian and exact sequences in $B$ are still exact in $A$ (see page 7 of Weibel). Of course, it's enough to verify that short exact sequences in $B$ are still exact in $A$, and this comes down to $B$ being closed under kernels and cokernels.

A category is exact if it possesses a class of short exact sequences. See wikipedia's article for details. Every abelian category is exact and is closed under extensions. Thus, if $B$ satisfies (3) then it must also satisfy (2), under your initial assumptions on $B$.

A final note of caution: when someone says thick subcategory they usually mean (in my experience) that the ambient category is triangulated, which $A$ need not be. According to nLab, some people use the term thick subcategory if the ambient category is abelian, but there seems to be disagreement about the meaning of the term when so applied. The majority seem to use it to mean that the subcategory is closed under extensions (but not necessarily under kernels and cokernels). So...this paragraph is answering a question of terminology which you didn't ask, but which Fernando Muro's second comment brought up.

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    $\begingroup$ The full subcategory of projectives in the category of modules over most rings is closed under finite direct sums but is not abelian. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 19, 2012 at 20:31
  • $\begingroup$ (Your last paragraph somehow makes me think that triangulated is something more than abelian («...is only abelian.») but very few triangulated categories are abelian!) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 19, 2012 at 20:33
  • $\begingroup$ @Mariano: Whoops, you have caught a typo. I meant to say "B is additive" and then with (3) you get B to be abelian. $\endgroup$
    – David White
    Commented Jan 19, 2012 at 21:04
  • $\begingroup$ @David: I am not sure what do you mean by (3) implies (2). In (2) I mean that if 0->X->Y->Z->0 is a sequence in A, and X,Z are in B, then Y is in B. Then (3) certainly doesn't imply (2); For example, semisimple modules as a subcategory of all modules satisfy (3) but not (2). Am I correct? $\endgroup$
    – Sasha
    Commented Jan 23, 2012 at 18:20
  • $\begingroup$ @Sasha: The claim that (3) implies (2) was the one I felt least sure about, because it seems like you're getting something for free. I'm basing it on wikipedia's claim that every abelian category is exact, and exact means you get $Y\in B$ in the notation of your comment. Does your example satisfy the standing hypotheses on $B$, i.e. full subcategory and closed under finite direct sums? If so, can you come up with an explicit short exact sequence to show $B$ is not closed under extensions? $\endgroup$
    – David White
    Commented Jan 23, 2012 at 18:31

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