**Proof.**
Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $S$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and $u_ i = u_{T_ i}$. Note that $\{ X_ i \to X_ T\} _{i \in I}$ is an fpqc covering of $X_ T$, see Topologies, Lemma 34.9.7. In particular, for every $x \in X_ T$ there exists an $i \in I$ and an $x_ i \in X_ i$ mapping to $x$. Since $\mathcal{O}_{X_ T, x} \to \mathcal{O}_{X_ i, x_ i}$ is flat, hence faithfully flat (see Algebra, Lemma 10.39.17) we conclude that $(u_ i)_{x_ i}$ is injective, surjective, bijective, or zero if and only if $(u_ T)_ x$ is injective, surjective, bijective, or zero. Whence part (1) of the lemma.

Proof of (2). Assume $f$ quasi-compact and $\mathcal{G}$ of finite type. Let $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ be a directed limit of affine $S$-schemes and assume that $u_ T$ is surjective. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and $u_ i = u_{T_ i} : \mathcal{F}_ i = \mathcal{F}_{T_ i} \to \mathcal{G}_ i = \mathcal{G}_{T_ i}$. To prove part (2) we have to show that $u_ i$ is surjective for some $i$. Pick $i_0 \in I$ and replace $I$ by $\{ i \mid i \geq i_0\} $. Since $f$ is quasi-compact the scheme $X_{i_0}$ is quasi-compact. Hence we may choose affine opens $W_1, \ldots , W_ m \subset X$ and an affine open covering $X_{i_0} = U_{1, i_0} \cup \ldots \cup U_{m, i_0}$ such that $U_{j, i_0}$ maps into $W_ j$ under the projection morphism $X_{i_0} \to X$. For any $i \in I$ let $U_{j, i}$ be the inverse image of $U_{j, i_0}$. Setting $U_ j = \mathop{\mathrm{lim}}\nolimits _ i U_{j, i}$ we see that $X_ T = U_1 \cup \ldots \cup U_ m$ is an affine open covering of $X_ T$. Now it suffices to show, for a given $j \in \{ 1, \ldots , m\} $ that $u_ i|_{U_{j, i}}$ is surjective for some $i = i(j) \in I$. Using Properties, Lemma 28.16.1 this translates into the following algebra problem: Let $A$ be a ring and let $u : M \to N$ be an $A$-module map. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras. If $N$ is a finite $A$-module and if $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is surjective, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is surjective. This is Algebra, Lemma 10.127.5 part (2).

Proof of (3). Exactly the same arguments as given in the proof of (2) reduces this to the following algebra problem: Let $A$ be a ring and let $u : M \to N$ be an $A$-module map. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras. If $M$ is a finite $A$-module and if $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is zero, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is zero. This is Algebra, Lemma 10.127.5 part (1).

Proof of (4). Assume $f$ quasi-compact and $\mathcal{F}, \mathcal{G}$ of finite presentation. Arguing in exactly the same manner as in the previous paragraph (using in addition also Properties, Lemma 28.16.2) part (3) translates into the following algebra statement: Let $A$ be a ring and let $u : M \to N$ be an $A$-module map. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras. Assume $M$ is a finite $A$-module, $N$ is a finitely presented $A$-module, and $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is an isomorphism. Then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is an isomorphism. This is Algebra, Lemma 10.127.5 part (3).
$\square$

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