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Let $X$ be a small site and $C$ a bicomplete (i.e. complete and cocomplete) category (or rather assume enough so that associated sheaves exists; see the comments). Consider the functor $\text{PSh}(X;C) \to \text{Sh}(X;C), F \mapsto F'$, which takes a presheaf to its associated sheaf. To what extent it is natural in $C$? I.e. for which funcor $\alpha : C \to D$ is the canonical map $\alpha(F)' \to \alpha(F')$ an isomorphism for all presheaves $F$ on $X$?

It seems to me that the explicit construction of the associated sheaf shows that this is true if $\alpha$ is continuous and preserves directed colimits, right? Is there a proof which just uses universal properties? I imagine a proof which uses adjoint functors to $\alpha$ in extended categories.

Do you know of any example of a functor where the above map is not an isomorphism?

Similary we may ask these questions about morphisms in $X$. For example if $i : U \to X$ is an open immerson of topological spaces, then $i^*$ commutes with taking associated sheaves. What are other examples of such morphisms, or is there even a characterization?

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For associated sheaf existence isn't enought the "bicomplete" property, for a general discussion: P. Freyd and M. Kelly , Categories of continuous functors I. J. Pure Appl. Algebra 2 (1972) ; J.W. Gray, Sheaves with values in a category, Topology 3. ANyway if we supposing the associated sheaf, a formalization in terms of adjoint functors is the BEck_Chevalley property, but this is and simply equal to your thesis (I checked). In "J. W. Gray, Fibred and cofibred categories, Proc. Conf. Categorical Algebra", pag.63-7 find a similar discussion (but no exatly the answere you looking for ) – Buschi Sergio Jan 8 '11 at 10:57
The most common general well-behaved condition for the existence of sheafifications is that the target category is bicomplete, has exact filtered colimits, and has the IPC property: . Also see the nLab page on sheafification:… , which discusses the difficulties in finding sheafification functors in general. – Harry Gindi Jan 9 '11 at 10:05
Thank you for this information. – Martin Brandenburg Jan 9 '11 at 13:04
About the inverse Image: for definition the image invese of a sheaf if the sheafication of the presheaf inverse image. Then if you check on stalks follow that map is Iso on stalks, then is Iso. (I hope my English is understable) – Buschi Sergio Jan 9 '11 at 23:43

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