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Does the category of functors from the category of finite sets to a Grothendieck category have enough projectives and injectives?

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    $\begingroup$ It is a Grothendieck category again, so has enough injectives. It has enough projectives if and only if the target does. I think. $\endgroup$ – მამუკა ჯიბლაძე Nov 21 '17 at 6:17
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    $\begingroup$ To prove the last claim: if $C$ is a nonempty small category and $\mathscr A^C$ has enough projectives then so does $\mathscr A$. Indeed, first note that $\varinjlim:\mathscr A^C\to\mathscr A$ preserves projectives (because its right adjoint $\operatorname{const}$ preserves epis, having itself a right adjoint $\varprojlim$). Given that, for an object $A$, cover $\operatorname{const}(A)$ by a projective $P$, apply $\varinjlim$ and take the composite $ \varinjlim P\twoheadrightarrow \varinjlim\operatorname{const}(A)\twoheadrightarrow A $. $\endgroup$ – მამუკა ჯიბლაძე Nov 21 '17 at 9:32
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    $\begingroup$ Yes, and if $\mathscr A$ has enough projectives then so does $\mathscr A^C$ - this is fairly standard. It follows from considering the functors $P[h^c]$ for $P$ a projective in $\mathscr A$ and $c$ an object of $C$, with $P[h^c](c')$ the direct sum of $\hom(c,c')$ many copies of $P$. Then$$\hom(P[h^c],F)\cong\hom(P,F(c))$$for any $F\in\mathscr A^C$, which implies that each $P[h^c]$ is projective. Moreover it is more or less clear from the above that each $F$ can be covered by a direct sum of $P[h^c]$'s. $\endgroup$ – მამუკა ჯიბლაძე Nov 21 '17 at 11:44
  • $\begingroup$ @მამუკაჯიბლაძე: Why not make those comments an answer? $\endgroup$ – Peter LeFanu Lumsdaine Nov 22 '17 at 8:54
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    $\begingroup$ In fact I now looked at the "Related" column to the right and this is almost duplicate of mathoverflow.net/q/162795/41291 $\endgroup$ – მამუკა ჯიბლაძე Nov 22 '17 at 10:38

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