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I'm reading about sheaves on sites and I have a question about a particular example in these notes:

Let $\mathcal{C}$ be a small category with a terminal object $T$ and suppose that every fibered product exists in $\mathcal{C}$. Let $K$ be a pretopology in $\mathcal{C}$.

If we fix an object $S\in \mathcal{C}$, we can define the comma category $\mathcal{C}_{/S}$. Its objects are morphisms $f:X\to S$ and the morphisms between $(f:X\to S)\in \mathcal{C}_{/S} $ and $(g:Y\to S) \in \mathcal{C}_{/S}$ are $h:X\to Y$ such that $f=gh$.

If for each $(f:X\to S)\in \mathcal{C}_{/S} $ we define $K_{/S}(f)=K(X)$, then it defines a pretopology in $\mathcal{C}_{/S}$ (more or less by definition, I guess). Here finishes the example in the notes.

Now I have a concrete question: if $F:\mathcal{C}_{/S}^{op}\to \mathbf{Set}$ is a functor, we can define the functor $\widehat{F}:\mathcal{C}^{op} \to \mathbf{Set}$ by setting

$$\widehat{F}(X)=\coprod_{f\in \mbox{Hom}(X,S)}F(f) $$

I want to show that $F$ is a sheaf with respect to $K_{/S}$ if and only if $\widehat{F}$ is a sheaf with respect to $K$.

Following the notes, $\widehat{F}$ is a sheaf with respect to the pretopology $K$ if for every $X\in \mathcal{C}$ and $R\in K(X)$, the natural morphism

$$\widehat{F}(X)\to \varprojlim \left(\prod_{(W\to X)\in R} \widehat{F}(W) \rightrightarrows \prod_{(U\to X)\in R}\prod_{(V\to X)\in R}\widehat{F}(U\times_X V) \right) $$ is an isomorphism.

I guess that by the definition of disjoint union, if $F$ is a sheaf then $\widehat{F}$ is a sheaf. Because we consider the same natural morphism but considering $F$ instead of $\widehat{F}$, so if is it an isomorphism for each $(f:X\to S)\in \mathcal{C}_{/S} $ its induces an isomorphism of the disjoint union of the $F(f)$. Is this right?

But I don't see why it should be true the converse.

Thank you very much !

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I would not expect $\hat{F}$ to be a sheaf. Why do you think it is? – Zhen Lin Mar 11 '14 at 16:19

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