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Suppose $C$ is a small category with a monoidal structure. Then by the special case of the Day convolution theorem for presheaves, $\operatorname{Psh}(C)$ is equipped with a corresponding biclosed monoidal structure. If $C$ is equipped with a Grothendieck topology, is there any useful condition for when the biclosed monoidal structure on presheaves descends to a biclosed monoidal structure on the category of sheaves for that Grothendieck topology?

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  • $\begingroup$ Are those categories symmetric ? $\endgroup$
    – Sov
    Commented Aug 2, 2019 at 15:43

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There is a more general form of Day's theorem that does pretty much that, at least for sub-canonical topologies:

Theorem (Day): Let $C$ be a complete and co-complete Category, and $D \subset C$ a full subcategory of $C$ endowed with monoidal structure which contains a full subcategory dense in $C$. Then if there exists functors:

$H',H: D^{op} \times C \rightarrow C$ and natural isomorphisms:

$Hom(S,H(S',X)) = Hom(S \otimes S',X) = Hom(S',H'(S,X))$

for $S,S' \in D$ and $X \in C$.

Then there is a unique biclosed monoidal closed structure on C such that the inclusion $D \rightarrow C$ extend into a monoidal functor.

The theorem as stated above appears and is proved (in french) as proposition 6.3 of this paper (Ara, Maltsiniotis), a variant of it which seems to relax the fullness of $D$ in $C$ is stated without proof as proposition 9 in this paper (Street). In both case there are references to two papers of Day which I havn't look at yet.

You can applies it to your question as follow, take $D$ to the the subcategory of representable pre-sheaves and $C$ the category of sheaves. The condition for the existence of $H$ and $H'$ is then just that given a sheaf $X$ and a representable $S$ then the presheaves $S' \mapsto Hom(S \otimes S',X)$ and $S' \mapsto Hom(S' \otimes S,X)$ are sheaves, i.e. a the condition is that "tensoring a covering by a fixed element gives you a covering". I'm convinced that this extend well to non-sub-canonical topologies, but I'm lacking of time to check it today (if someone do it, I'll be interested to know the answer)

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  • $\begingroup$ Great! Thanks! The condition in terms of coverings seems to be exactly what I'm looking for! $\endgroup$ Commented Nov 18, 2017 at 13:50

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