
Lemma 6.30.5. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$. Let $U \in \mathcal{B}$. Let $\mathcal{F}$ be a sheaf of sets on $\mathcal{B}$. The map

$\mathcal{F}(U) \to \prod \nolimits _{x \in U} \mathcal{F}_ x$

identifies $\mathcal{F}(U)$ with the elements $(s_ x)_{x\in U}$ with the property

• For any $x \in U$ there exists a $V \in \mathcal{B}$, with $x \in V \subset U$ and a section $\sigma \in \mathcal{F}(V)$ such that for all $y \in V$ we have $s_ y = (V, \sigma )$ in $\mathcal{F}_ y$.

Proof. First note that the map $\mathcal{F}(U) \to \prod \nolimits _{x \in U} \mathcal{F}_ x$ is injective by the uniqueness in the sheaf condition of Definition 6.30.2. Let $(s_ x)$ be any element on the right hand side which satisfies $(*)$. Clearly this means we can find a covering $U = \bigcup U_ i$, $U_ i \in \mathcal{B}$ such that $(s_ x)_{x \in U_ i}$ comes from certain $\sigma _ i \in \mathcal{F}(U_ i)$. For every $y \in U_ i \cap U_ j$ the sections $\sigma _ i$ and $\sigma _ j$ agree in the stalk $\mathcal{F}_ y$. Hence there exists an element $V_{ijy} \in \mathcal{B}$, $y \in V_{ijy}$ such that $\sigma _ i|_{V_{ijy}} = \sigma _ j|_{V_{ijy}}$. Thus the sheaf condition $(**)$ of Definition 6.30.2 applies to the system of $\sigma _ i$ and we obtain a section $s \in \mathcal{F}(U)$ with the desired property. $\square$

Comment #3233 by on

Xena says "Maybe forall y in U intersect V'' would be better in (*)".

Comment #3332 by on

OK, I think the more natural change is to require $V \subset U$ which I clearly just forgot to type. Thanks Xena! Fix is here.

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