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 \ref{topologies-lemma-fpqc}. 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 \ref{algebra-lemma-local-flat-ff}) 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. \medskip\noindent Proof of (2). Assume $f$ quasi-compact and $\mathcal{G}$ of finite type. Let $T = \lim_{i \in I} T_i$ 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 $U_j = \lim_i U_{j, i}$ 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 \ref{properties-lemma-finite-type-module} 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 $R = \colim_{i \in I} R_i$ 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 \ref{algebra-lemma-module-map-property-in-colimit} part (2). \medskip\noindent 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 $R = \colim_{i \in I} R_i$ 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 \ref{algebra-lemma-module-map-property-in-colimit} part (1). \medskip\noindent 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 \ref{properties-lemma-finite-presentation-module}) 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 $R = \colim_{i \in I} R_i$ 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 \ref{algebra-lemma-module-map-property-in-colimit} part (3).