3
$\begingroup$

Let $A$ be an abelian category and $I$ some arbitrary category. It follows, that the functor category $A^I$ is also an abelian category.

Is there a general characterization of the projective objects of $A^I$ in terms of the projective objects of $A$? If not, how much structure for $I$ do we have to require to obtain nice results?

For example, the complex category $\text{Comp}(A)$ can be understood as a category of additive functors according to this answer and there is a characterization of projective chain complexes in terms of their underlying objects, cycles, boundaries and homology (they must all be projective).

$\endgroup$
1
  • 1
    $\begingroup$ I think that's a very hard question. For instance, if you take $A=\mathbf{Ab}$, you have special projectives that are $\mathbb Z[\hom(i,-)], i\in I$, and these are generators so any projective functor is a summand of one of these - but said summands can be wild. Another point of view on why it's a hard question is to take $I=BG$, for $G$ a group. Then you're looking at the category of $G$-representations in $A$, and projectives in that one can be weird too (although in some cases you can determine them) $\endgroup$ Commented Apr 1, 2020 at 17:36

1 Answer 1

3
$\begingroup$

Here is a partial answer when the target category is the category $\textbf{Ab}$ of abelian groups.

Lemma. If $\mathcal{C}$ is a small preadditive category then the finitely generated projective objects in $(\mathcal{C},\textbf{Ab})$ are the finite direct summands of direct sums of representable functors. If $\mathcal{C}$ has split idempotents and finite direct sums then these are precisely the representable functors.

Recall that a functor is representable if it is of the form $\text{Hom}(C,-)$ for some $C\in\mathcal{C}$.

For example, if $R$ is a ring then the finitely generated projective objects in $(R\text{-mod},\textbf{Ab})$ are of the form $\text{Hom}_{R}(A,-)$ for $A\in R\text{-mod}$.

Proofs and more info can be found here.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .