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Let $\bf C$ be a category, $\mathcal S$ an (elementary) topos.

If $\mathcal S$ is a presheaf category over $\bf D$, then it's easy to see $[\bf C^{\rm op},\, \mathcal S] \cong [(\bf C \times \bf D)^{\rm op},\, \cal{Sets}]$ is still a topos. In more general situations I struggle to see an easy reason for it to be true as well. Thus:

When is the category $[\bf C^{\rm op},\, \mathcal S]$ of contravariant functors from $\bf C$ to $\mathcal S$ a topos?

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    $\begingroup$ If $\mathcal{S}$ is a Grothendieck topos, or more generally has large enough disjoint universal coproduct you are always fine. But If $\mathcal{S}$ is an elementary topos in general it is not going to work, but I do not know if there are nice conditions under which it works. $\endgroup$ – Simon Henry Jan 16 at 23:40
  • $\begingroup$ Can you elaborate/give references for these claims? $\endgroup$ – mattecapu Jan 16 at 23:41
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    $\begingroup$ If $\mathbf C$ were an internal category in $\mathcal S$, then the category of internal $\mathcal S$-valued presheaves on $\mathbf C$ would be a topos. (This is surely in Johnstone's "Topos Theory".) I think the point of @SimonHenry's comment is that good coproducts let you regard any genuine small category as an internal category in $\mathcal S$. $\endgroup$ – Andreas Blass Jan 17 at 0:09
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    $\begingroup$ That was indeed where I was going. But the question seem much more interesting that this observation: For example if the category C is a groupoid I think that S-valued presheaf will be an elementary topos without needing any assumption on S, so the general question definitely do not reduce to the case I was refering too. $\endgroup$ – Simon Henry Jan 17 at 0:16

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