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Let $\mathcal F$ be a sheaf of sets on a site. Fix an object $X$ of the underlying category of the site, which is assumed to contain a final object and have products. Define a presheaf $\mathcal G$ by setting $\mathcal G(U) := \mathcal F(U \times X)$. Is $\mathcal G$ a sheaf?

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2 Answers 2

If $(U_i)$ cover $U$, then $\mathcal{G}(U)$ is supposed to be a certain limit of a diagram of the $\mathcal{G}(U_i)$ and $\mathcal{G}(U_i \times_U U_j)$. The $U_i \times X$ also cover $U \times X$ (this is the base change axiom for the site), and $(U_i \times_U U_j) \times X = (U_i \times X) \times_{U \times X} (U_j \times X)$, so the analogous diagram for $\mathcal{F}(U \times X)$ is indeed a limit diagram, by the sheaf condition on $\mathcal{F}$. Thus the answer is yes.

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Your operation can be defined intrinsically without reference to sites: if $\mathcal{E}$ is your topos and $X$ is an object in $\mathcal{E}$, the slice category $\mathcal{E}_{/X}$ is again a topos, and there is a geometric morphism $p^* \dashv p_* : \mathcal{E}_{/X} \to \mathcal{E}$ where the inverse image functor $p^*$ is the functor $- \times X$, and $p^*$ itself has a left adjoint $p_! : \mathcal{E}_{/X} \to \mathcal{E}$, which is just the evident projection functor.

I claim the operation you describe is the endofunctor $p_* p^* : \mathcal{E} \to \mathcal{E}$. Indeed, by the Yoneda lemma, if $U$ and $X$ are representable sheaves in $\mathcal{E}$ and the site has products, then, $$p_* p^* \mathscr{F} (U) = \mathcal{E}(U, p_* p^* \mathscr{F}) \cong \mathcal{E}_{/X}(p^* U, p^* \mathscr{F}) \cong \mathcal{E} (p_! p^* U, \mathscr{F})$$ but $p_! p^* U = U \times X$, so $$p_* p^* \mathscr{F} (U) \cong \mathscr{F} (U \times X)$$ as claimed.

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This may be a dumb question, but isn't $X$ supposed to be an object of the site corresponding to $\mathcal{E}$? –  Martin Brandenburg Mar 20 '13 at 14:07
Yes, but if the site has a subcanonical topology you can just embed the site as a full subcategory of $\mathcal{E}$ via Yoneda. If not then you can take the sheafification. –  Zhen Lin Mar 20 '13 at 16:18

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