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I'm trying to understand the category-theoretic proof that sheafification preserves stalks, using adjoints, as outlined e.g. in this mathoverflow answer:

What is bugging me is that I'm having a hard time seeing why the colimit described is actually isomorphic to the stalk (as a skyscraper sheaf based at x using the stalk as the fixed set).

Specifically, in his answer, Ryan Reich claims that for a sheaf $F$ on $X$, the stalk $F_x$ is the colimit over opens $U$ containing $x$ of the sheaves $j_* j^* F$, where $j: U \rightarrow X$ is the inclusion map.

Every time I try to show that the skyscraper sheaf satisfies the universal property I end up in a seeming dead end. I feel like it shouldn't be THAT hard, and that I must be missing something or confusing myself somehow. Does anyone know a reference for this, or be willing to explain why it works? I'd be very grateful.

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That seems like a roundabout way of doing things. What's wrong with the usual definition of stalk as a filtered colimit of sections? – Zhen Lin Jul 15 '12 at 17:00
Well, I suppose because one wants to use the fact that left adjoints commute with colimits. Since sheafification is a left adjoint, one wants to describe the stalk as a certain colimit in order to immediately conclude that the stalks are preserved. But sheafification is a functor from PreSh(X) so it commutes with colimits in that category. Hence we want the stalk as a colimit of (pre)sheaves, and not just sections. At least, this is how I understand it. Maybe I am mistaken. – KristianJS Jul 15 '12 at 17:33
Right. But the obvious way of turning the filtered colimit of sections into a filtered colimit of sheaves is by the procedure you mention. The (pre)sheaf $j_* j^* F$ can be explicitly described as the (pre)sheaf where $(j_* j^* F)(V) = F(U \cap V)$. – Zhen Lin Jul 15 '12 at 18:03
Ah, ok, this is making sense now. I was always thinking of $j_* j^*F(V)$ as some awkward colimit, but in fact it simplifies like you say to $F(U \cap V)$. From there I was able to convince myself that the colimit of sheaves coincides with the skyscraper sheaf desired. Thanks! If you wish to make your comment an answer, I'll happily accept it. – KristianJS Jul 15 '12 at 21:45

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