Lemma 76.7.3. In Situation 76.7.1 let $X' \to X$ be a flat morphism of algebraic spaces. Denote $u' : \mathcal{F}' \to \mathcal{G}'$ the pullback of $u$ to $X'$. Denote $F'_{iso}$, $F'_{inj}$, $F'_{surj}$, $F'_{zero}$ the functors on $\mathit{Sch}/B$ associated to $u'$.

If $\mathcal{G}$ is of finite type and the image of $|X'| \to |X|$ contains the support of $\mathcal{G}$, then $F_{surj} = F'_{surj}$ and $F_{zero} = F'_{zero}$.

If $\mathcal{F}$ is of finite type and the image of $|X'| \to |X|$ contains the support of $\mathcal{F}$, then $F_{inj} = F'_{inj}$ and $F_{zero} = F'_{zero}$.

If $\mathcal{F}$ and $\mathcal{G}$ are of finite type and the image of $|X'| \to |X|$ contains the supports of $\mathcal{F}$ and $\mathcal{G}$, then $F_{iso} = F'_{iso}$.

**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 66.30). 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 66.15.2. Thus the lemma is clear.
$\square$

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