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Let $\mathcal F$ be a sheaf of a topological space $X$ and $D_{\mathcal F}(U)=\prod_{U\ni x}\mathcal F_x$.

I don't understand this morphism $\varphi_{\mathcal F}:\mathcal F\to D_{\mathcal F}$, when we apply in the open subset $U$ we have $({\varphi_{\mathcal F}})_U:\mathcal F(U)\to D_{\mathcal F}(U),\ s\mapsto (s_x)_{x\in U}$ but what exactly means $s\mapsto (s_x)_{x\in U}$?

Can I say that $s\mapsto (\overline{(s,U)},\overline{(s,U)},\ldots)$? since $s\in \mathcal F(U)$, we have $s_x=\overline{(s,U)}$ for every $x\in U$.

I think maybe I didn't fully understand the meaning of $s_x$.

I really need help.

Thanks a lot.

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It maps a section of U to the tuple of germs at all the points in U. $s_x$ denotes the germ at $x$, i.e. the equivalence class of (U, s) as you wrote. –  Adeel Sep 20 '13 at 0:41

1 Answer 1

up vote 1 down vote accepted

An element in a product over a set of indices $U$ is an arrow from the set of indices to the coproduct: so your $D_\mathcal F(U)$ is $$\prod_{x\in U}\mathcal F_x=\textrm{Maps }(U,\coprod_{x\in U}\mathcal F_x).$$

Thus, concretely, your morphism $\phi_\mathcal F:\mathcal F\to D_\mathcal F$ on an open subset $U\subset X$ sends a section $s\in\mathcal F(U)$ to the map $U\to \coprod_{x\in U}\mathcal F_x$ defined by $x\mapsto s_x$, where $s_x\in\mathcal F_x$ is the germ of $s$ at $x$, that you correctly described as an equivalence class.

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So the coordinates of the image of $s$ are equal to each other? –  user42912 Sep 20 '13 at 9:43
    
I'm not sure I understand. Every $x\in U$ defines a germ $s_x\in \mathcal F_x$, the corresponding stalk. So the notation $\phi_\mathcal F(U)(s)=(s_x)_{x\in U}$ means: take $s$ to the map $U\to \coprod_{x\in U}\mathcal F_x$ given by $x\mapsto s_x$. –  Brenin Sep 20 '13 at 9:58
    
yes, but if $x, y\in U$ and $s\in \mathcal F(U)$, don't we have $s_x=s_y=\overline{(s,U)}$? –  user42912 Sep 20 '13 at 10:25
    
$s_x$ and $s_y$ live in different stalks. The notation $\overline{(s,U)}$ might be misleading if you do not remember that you are looking at $U$ as a neighborhood of $x$ rather that $y$. –  Brenin Sep 20 '13 at 10:41
    
It's true, thank you very much for your help! –  user42912 Sep 20 '13 at 11:32

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