Lemma 6.33.2. Let $X$ be a topological space. Let $X = \bigcup _{i\in I} U_ i$ be an open covering. Given any glueing data $(\mathcal{F}_ i, \varphi _{ij})$ for sheaves of sets with respect to the covering $X = \bigcup U_ i$ there exists a sheaf of sets $\mathcal{F}$ on $X$ together with isomorphisms

$\varphi _ i : \mathcal{F}|_{U_ i} \to \mathcal{F}_ i$

such that the diagrams

$\xymatrix{ \mathcal{F}|_{U_ i \cap U_ j} \ar[r]_{\varphi _ i} \ar[d]_{\text{id}} & \mathcal{F}_ i|_{U_ i \cap U_ j} \ar[d]^{\varphi _{ij}} \\ \mathcal{F}|_{U_ i \cap U_ j} \ar[r]^{\varphi _ j} & \mathcal{F}_ j|_{U_ i \cap U_ j} }$

are commutative.

Proof. First proof. In this proof we give a formula for the set of sections of $\mathcal{F}$ over an open $W \subset X$. Namely, we define

$\mathcal{F}(W) = \{ (s_ i)_{i \in I} \mid s_ i \in \mathcal{F}_ i(W \cap U_ i), \varphi _{ij}(s_ i|_{W \cap U_ i \cap U_ j}) = s_ j|_{W \cap U_ i \cap U_ j} \} .$

Restriction mappings for $W' \subset W$ are defined by the restricting each of the $s_ i$ to $W' \cap U_ i$. The sheaf condition for $\mathcal{F}$ follows immediately from the sheaf condition for each of the $\mathcal{F}_ i$.

We still have to prove that $\mathcal{F}|_{U_ i}$ maps isomorphically to $\mathcal{F}_ i$. Let $W \subset U_ i$. In this case the condition in the definition of $\mathcal{F}(W)$ implies that $s_ j = \varphi _{ij}(s_ i|_{W \cap U_ j})$. And the commutativity of the diagrams in the definition of a glueing data assures that we may start with any section $s \in \mathcal{F}_ i(W)$ and obtain a compatible collection by setting $s_ i = s$ and $s_ j = \varphi _{ij}(s_ i|_{W \cap U_ j})$.

Second proof (sketch). Let $\mathcal{B}$ be the set of opens $U \subset X$ such that $U \subset U_ i$ for somje $i \in I$. Then $\mathcal{B}$ is a base for the topology on $X$. For $U \in \mathcal{B}$ we pick $i \in I$ with $U \subset U_ i$ and we set $\mathcal{F}(U) = \mathcal{F}_ i(U)$. Using the isomorphisms $\varphi _{ij}$ we see that this prescription is “independent of the choice of $i$”. Using the restriction mappings of $\mathcal{F}_ i$ we find that $\mathcal{F}$ is a sheaf on $\mathcal{B}$. Finally, use Lemma 6.30.6 to extend $\mathcal{F}$ to a unique sheaf $\mathcal{F}$ on $X$. $\square$

Comment #3423 by Samir Canning on

Maybe a slightly shorter proof can be given as follows: define a base for the topology on $X$ consisting of all the open sets of each $U_i$ and then define$\mathcal{F}$ on the base in the obvious way. Then just use tag 009N to get a (unique) sheaf on $X$.

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