Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.