Lemma 6.33.1. Let $X$ be a topological space. Let $X = \bigcup U_ i$ be an open covering. Let $\mathcal{F}$, $\mathcal{G}$ be sheaves of sets on $X$. Given a collection

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

of maps of sheaves such that for all $i, j \in I$ the maps $\varphi _ i, \varphi _ j$ restrict to the same map $\mathcal{F}|_{U_ i \cap U_ j} \to \mathcal{G}|_{U_ i \cap U_ j}$ then there exists a unique map of sheaves

$\varphi : \mathcal{F} \longrightarrow \mathcal{G}$

whose restriction to each $U_ i$ agrees with $\varphi _ i$.

Proof. Omitted. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).