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If one Googles sufficiently hard one finds the statement that Roger Godement gives the first example of a comonad, used to compute flabby resolutions of sheaves, in his monograph "Topologie algébrique et théorie des faisceaux". For example, see here. I am interested in comonads in the category $\mathsf{Sh}(X)$ of sheaves of abelian groups on a topological space $X$, so I looked out this paper and attempted some translation. I think that the relevant part is, given a sheaf $\mathscr{F}$, we define a new sheaf $C \mathscr{F}$ where $$ C\mathscr{F}(U) = \{ \text{functions } s \colon U \longrightarrow \mathscr{E}_\mathscr{F} \mid s(x) \in \mathscr{F}_x \forall x \in U \} $$ where $\mathscr{E}_\mathscr{F}$ is the étalé space of $\mathscr{F}$, and $\mathscr{F}_x$ is the stalk of $\mathscr{F} $ at $x \in X$.

My question is, is this really a comonad? I have found it hard to define a natural transformation $\varepsilon \colon C \longrightarrow 1$ that would turn $C$ into a comonad. Is there another functor $\mathsf{Sh}(X) \longrightarrow \mathsf{Sh}(X)$ which is the comonad which is being referred to? Is the original statement incorrect?

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$F_x$ is the stalk, not the skyscraper sheaf. Are you sure that you have given the correct definition of $CF$? There is no free access to the pdf you have linked. – Martin Brandenburg Feb 11 '14 at 2:50
Since the canonical map $X^\mathrm{disc} \to X$ is continuous, we get a geometric morphism $\mathbf{Set}^X \to \mathbf{Sh}(X)$, and the adjunction induces a monad on $\mathbf{Sh}(X)$. This is the Godement standard resolution. Notably it's not a comonad – comonads are for building projective resolutions! – Zhen Lin Feb 11 '14 at 8:19
Martin, apologies I meant to write stalk, I posted the question late. Also the statement I was referring to was in the preview at the bottom of the page, not in the actual pdf. – Paul Slevin Feb 11 '14 at 9:04
Zhen Lin, thanks for confirming my suspicions. Do you know of any comonads in $ \mathsf {Sh} (X) $? – Paul Slevin Feb 11 '14 at 9:06
There are plenty. Take, for instance, a continuous map $X \to Y$; then we get a geometric morphism $\mathbf{Sh}(X) \to \mathbf{Sh}(Y)$, inducing a comonad on $\mathbf{Sh}(X)$. – Zhen Lin Feb 11 '14 at 9:13

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