**Proof.**
Let $Y_1, Y_2$ be finite étale over $X$ and let $\varphi : (Y_1)_ U \to (Y_2)_ U$ be a morphism over $U$. We have to show that $\varphi $ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma 58.10.1. We will prove existence by showing that we can enlarge $U$ if $U \not= X$ and using Zorn's lemma to finish the proof.

Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Lemma 58.10.2 we can extend the morphism $\varphi _ V : (Y_1)_ V \to (Y_2)_ V$ (uniquely) to a morphism $(Y_1)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})} \to (Y_2)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})}$. Choose an affine neighbourhood $W \subset X$ of $x$. Since $U \cap W$ is dense in $W$ it contains the generic points $\eta _1, \ldots , \eta _ n$ of $W$. Choose an affine open $W' \subset W \cap U$ containing $\eta _1, \ldots , \eta _ n$. Set $V' = W' \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Limits, Lemma 32.20.3 applied to $x \in W \supset W'$ we find an open $W' \subset W'' \subset W$ with $x \in W''$ and a morphism $\varphi '' : (Y_1)_{W''} \to (Y_2)_{W''}$ agreeing with $\varphi $ over $W'$. Since $W'$ is dense in $W'' \cap U$, we see by Lemma 58.10.1 that $\varphi $ and $\varphi ''$ agree over $U \cap W'$. Thus $\varphi $ and $\varphi ''$ glue to a morphism $\varphi '$ over $U' = U \cup W''$ agreeing with $\varphi $ over $U$. Observe that $x \in U'$ so that we've extended $\varphi $ to a strictly larger open.

Consider the set $\mathcal{S}$ of pairs $(U', \varphi ')$ where $U \subset U'$ and $\varphi '$ is an extension of $\varphi $. We endow $\mathcal{S}$ with a partial ordering in the obvious manner. If $(U'_ i, \varphi '_ i)$ is a totally ordered subset, then it has a maximum $(U', \varphi ')$. Just take $U' = \bigcup U'_ i$ and let $\varphi ' : (Y_1)_{U'} \to (Y_2)_{U'}$ be the morphism agreeing with $\varphi '_ i$ over $U'_ i$. Thus Zorn's lemma applies and $\mathcal{S}$ has a maximal element. By the argument above we see that this maximal element is an extension of $\varphi $ over all of $X$.
$\square$

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