# Do topos-valued sheaves form a topos?

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?

• 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. – Simon Henry Jan 16 at 23:40
• Can you elaborate/give references for these claims? – mattecapu Jan 16 at 23:41
• 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 $\mathcal S$. – Andreas Blass Jan 17 at 0:09
• 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. – Simon Henry Jan 17 at 0:16