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Given a topological space $X$ the Godement construction for a sheaf $F$ returns a sheaf $G^0(F)$ constructed as follows. For each point $x\in X$, define $$ G^0(F)(U):=\prod_{x\in U} F_x.$$

This clearly induces a functor, which is also easily seen to be exact.

In the related Wikipedia page i read:

Another way to view $G^0$ is as follows. Let $Y = \coprod_{x \in X} \{x\}$ be the disjoint union of the points of $X$. There is a continuous map $p : Y \to X$. This induces adjoint pushforward and pullback functors $p^\star$ and $p_\star$.

The canonical map $F\to G^0F$ turns out to be the endofuntor of a monad on $Sh(X)$ (as far as I understood the same argument explained in Weibel, one takes the monad induced by the adjunction between skyscraper and stalk functors and then exploit the fact that the category $\mathbf{Mnd}(\mathbf C)$ of monads over $\mathbf C$ admits products), so the possibility to find a resolution for $F$ in terms of $G^0(-)$ turns out to be a merely categorical fact. Wiki is not at all clear in explaining how it's done, and Weibel leaves the much part of the argument as an exercise.

Can you help me to better understand this construction? I'm also stuck with tihs conflict of notation with the functor $p^\star$: the same Wiki defines $p^\star := p^{-1}(-) \otimes_{p^{-1}\mathcal{O}_Y} \mathcal{O}_X$, which is not left adjoint to $p_\star$ (its left adjoint is $p^{-1}$).

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Is this what you're looking for? See also here –  t.b. Jan 2 '12 at 16:34
    
@t.b.: Precisely something like this. But I'm in trouble in visualizing LaTeX code, when the page is rendered FF somehow "forgets" all the backslashes –  tetrapharmakon Jan 2 '12 at 16:39
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@tetrapharmakon: The functor $p^{-1} (-) \mathbin{\otimes_{p^{-1} \mathscr{O}_X}} \mathscr{O}_Y$ is left adjoint to $p_*$, provided you interpret the first one as a functor from $\mathscr{O}_X \textbf{-Mod}$ to $\mathscr{O}_Y \textbf{-Mod}$. So it really depends on the categories you're working with. For your purposes I think the correct left adjoint is the plain $p^{-1}$. –  Zhen Lin Jan 2 '12 at 16:54

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