The Stacks project

Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $B$. 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 on Spaces, Lemma 73.9.3. 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 $\mathcal{O}_{X_ T, \overline{x}} \to \mathcal{O}_{X_ i, \overline{x_ i}}$ is flat, hence faithfully flat (see Morphisms of Spaces, Section 67.30). 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. The lemma follows. $\square$

Proof. let $v : \mathcal{H} \to \mathcal{E}$ be a map of quasi-coherent modules on an algebraic space $Y$ and let $\varphi : Y' \to Y$ be a surjective flat morphism of algebraic spaces, then $v$ is an isomorphism, injective, surjective, or zero if and only if $\varphi ^*v$ is an isomorphism, injective, surjective, or zero. Namely, for every $y \in |Y|$ there exists a $y' \in |Y'|$ and the map of local rings $\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{Y', \overline{y'}}$ is faithfully flat (see Morphisms of Spaces, Section 67.30). Of course, to check for injectivity or being zero it suffices to look at the points in the support of $\mathcal{H}$, and to check for surjectivity it suffices to look at points in the support of $\mathcal{E}$. Moreover, under the finite type assumptions as in the statement of the lemma, taking the supports commutes with base change, see Morphisms of Spaces, Lemma 67.15.2. Thus the lemma is clear. $\square$

Proof. Proof of (1). Let $i : Z \to X$ be the scheme theoretic support of $\mathcal{G}$ and think of $\mathcal{G}$ as a finite type quasi-coherent module on $Z$. We may replace $X$ by $Z$ and $u$ by the map $i^*\mathcal{F} \to \mathcal{G}$ (details omitted). Hence we may assume $f$ is 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 $B$-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 (1) we have to show that $u_ i$ is surjective for some $i$. Pick $0 \in I$ and replace $I$ by $\{ i \mid i \geq 0\} $. Since $f$ is quasi-compact we see $X_0$ is quasi-compact. Hence we may choose a surjective étale morphism $\varphi _0 : W_0 \to X_0$ where $W_0$ is an affine scheme. Set $W = W_0 \times _{T_0} T$ and $W_ i = W_0 \times _{T_0} T_ i$ for $i \geq 0$. These are affine schemes endowed with a surjective étale morphisms $\varphi : W \to X_ T$ and $\varphi _ i : W_ i \to X_ i$. Note that $W = \mathop{\mathrm{lim}}\nolimits W_ i$. Hence $\varphi ^*u_ T$ is surjective and it suffices to prove that $\varphi _ i^*u_ i$ is surjective for some $i$. Thus we have reduced the problem to the affine case which is Algebra, Lemma 10.127.5 part (2).

Proof of (2). Assume $\mathcal{F}$ is of finite type with scheme theoretic support $Z \subset B$ quasi-compact over $B$. Let $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ be a directed limit of affine $B$-schemes and assume that $u_ T$ is zero. Set $X_ i = T_ i \times _ B X$ and denote $u_ i : \mathcal{F}_ i \to \mathcal{G}_ i$ the pullback. Choose $0 \in I$ and replace $I$ by $\{ i \mid i \geq 0\} $. Set $Z_0 = Z \times _ X X_0$. By Morphisms of Spaces, Lemma 67.15.2 the support of $\mathcal{F}_ i$ is $|Z_0|$. Since $|Z_0|$ is quasi-compact we can find an affine scheme $W_0$ and an étale morphism $W_0 \to X_0$ such that $|Z_0| \subset \mathop{\mathrm{Im}}(|W_0| \to |X_0|)$. Set $W = W_0 \times _{T_0} T$ and $W_ i = W_0 \times _{T_0} T_ i$ for $i \geq 0$. These are affine schemes endowed with étale morphisms $\varphi : W \to X_ T$ and $\varphi _ i : W_ i \to X_ i$. Note that $W = \mathop{\mathrm{lim}}\nolimits W_ i$ and that the support of $\mathcal{F}_ T$ and $\mathcal{F}_ i$ is contained in the image of $|W| \to |X_ T|$ and $|W_ i| \to |X_ i|$. Now $\varphi ^*u_ T$ is injective and it suffices to prove that $\varphi _ i^*u_ i$ is injective for some $i$. Thus we have reduced the problem to the affine case which is Algebra, Lemma 10.127.5 part (1).

Proof of (3). This can be proven in exactly the same manner as in the previous two paragraphs using Algebra, Lemma 10.127.5 part (3). We can also deduce it from (1) and (2) as follows. Let $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ be a directed limit of affine $B$-schemes and assume that $u_ T$ is an isomorphism. By part (1) there exists an $0 \in I$ such that $u_{T_0}$ is surjective. Set $\mathcal{K} = \mathop{\mathrm{Ker}}(u_{T_0})$ and consider the map of quasi-coherent modules $v : \mathcal{K} \to \mathcal{F}_{T_0}$. For $i \geq 0$ the base change $v_{T_ i}$ is zero if and only if $u_ i$ is an isomorphism. Moreover, $v_ T$ is zero. Since $\mathcal{G}_{T_0}$ is of finite presentation, $\mathcal{F}_{T_0}$ is of finite type, and $u_{T_0}$ is surjective we conclude that $\mathcal{K}$ is of finite type (Modules on Sites, Lemma 18.24.1). It is clear that the support of $\mathcal{K}$ is contained in the support of $\mathcal{F}_{T_0}$ which is quasi-compact over $T_0$. Hence we can apply part (2) to see that $v_{T_ i}$ is zero for some $i$. $\square$

Proof. Since pullback is right exact we see that $\mathcal{F}_ T \to \mathcal{G}_ T \to \mathcal{H}_ T \to 0$ is exact for every scheme $T$ over $B$. Hence $u_ T$ is surjective if and only if $v_ T$ is an isomorphism. $\square$

Proof. Let $T$ be a scheme over $B$. Denote $i_ T : Z_ T \to X_ T$ the base change of $i$ and $\mathcal{H}_ T$ the pullback of $\mathcal{H}$ to $Z_ T$. Observe that $(i^*\mathcal{F})_ T = i_ T^*\mathcal{F}_ T$ and $i_{T, *}\mathcal{H}_ T = (i_*\mathcal{H})_ T$. The first statement follows from commutativity of pullbacks and the second from Cohomology of Spaces, Lemma 69.11.1. Hence we see that $u_ T$ and $v_ T$ are adjoint maps as well. Thus $u_ T = 0$ if and only if $v_ T = 0$. This proves (1). In case (2) we see that $u_ T$ is surjective if and only if $v_ T$ is surjective because $u_ T$ factors as

and the fact that $i_{T, *}$ is an exact functor fully faithfully embedding the category of quasi-coherent modules on $Z_ T$ into the category of quasi-coherent $\mathcal{O}_{X_ T}$-modules. See Morphisms of Spaces, Lemma 67.14.1. $\square$

Proof. By Cohomology of Spaces, Lemma 69.11.1 we have $g_{T, *}u_ T = u'_ T$. Moreover, $g_{T, *} : \mathit{QCoh}(\mathcal{O}_{X_ T}) \to \mathit{QCoh}(\mathcal{O}_ X)$ is a faithful, exact functor reflecting isomorphisms, injective maps, and surjective maps. $\square$

Proof. Part (1) follows from the following statement: If $T' \to T$ is a surjective flat morphism of algebraic spaces over $Y$, then $\mathcal{F}_{T'}$ is flat over $T'$ if and only if $\mathcal{F}_ T$ is flat over $T$, see Morphisms of Spaces, Lemma 67.31.3. Part (2) follows from Limits of Spaces, Lemma 70.6.12 if $f$ is also quasi-separated (i.e., $f$ is of finite presentation). For the general case, first reduce to the case where the base is affine and then cover $X$ by finitely many affines to reduce to the quasi-separated case. Details omitted. $\square$