## Tag `009M`

Chapter 6: Sheaves on Spaces > Section 6.30: Bases and sheaves

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 $V \in \mathcal{B}$, $x \in V$ 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. $\square$

The code snippet corresponding to this tag is a part of the file `sheaves.tex` and is located in lines 3818–3835 (see updates for more information).

```
\begin{lemma}
\label{lemma-condition-star-sections}
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
\begin{itemize}
\item[$(*)$] For any $x \in U$ there exists a $V \in \mathcal{B}$,
$x \in V$ and a section $\sigma \in \mathcal{F}(V)$ such that
for all $y \in V$ we have $s_y = (V, \sigma)$ in $\mathcal{F}_y$.
\end{itemize}
\end{lemma}
\begin{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 \ref{definition-sheaf-basis}. 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 \ref{definition-sheaf-basis}
applies to the system of $\sigma_i$ and we obtain a section
$s \in \mathcal{F}(U)$ with the desired property.
\end{proof}
```

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