Lemma 38.36.1. Let $\mathcal{F}$ be a sheaf on a site $(\mathit{Sch}/S)_{ph}$, see Topologies, Definition 34.8.11. Then for any blow up square (38.36.0.1) in the category $(\mathit{Sch}/S)_{ph}$ the diagram
is cartesian in the category of sets.
Let $X$ be a scheme. Let $Z \subset X$ be a closed subscheme such that the inclusion morphism is of finite presentation, i.e., the quasi-coherent sheaf of ideals corresponding to $Z$ is of finite type. Let $b : X' \to X$ be the blowup of $X$ in $Z$ and let $E = b^{-1}(Z)$ be the exceptional divisor. See Divisors, Section 31.32. In this situation and in this section, let us say
is a blow up square.
Lemma 38.36.1. Let $\mathcal{F}$ be a sheaf on a site $(\mathit{Sch}/S)_{ph}$, see Topologies, Definition 34.8.11. Then for any blow up square (38.36.0.1) in the category $(\mathit{Sch}/S)_{ph}$ the diagram is cartesian in the category of sets.
Proof. Since $Z \amalg X' \to X$ is a surjective proper morphism we see that $\{ Z \amalg X' \to X\} $ is a ph covering (Topologies, Lemma 34.8.6). We have
Since $\mathcal{F}$ is a Zariski sheaf we see that $\mathcal{F}$ sends disjoint unions to products. Thus the sheaf condition for the covering $\{ Z \amalg X' \to X\} $ says that $\mathcal{F}(X) \to \mathcal{F}(Z) \times \mathcal{F}(X')$ is injective with image the set of pairs $(t, s')$ such that (a) $t|_ E = s'|_ E$ and (b) $s'$ is in the equalizer of the two maps $\mathcal{F}(X') \to \mathcal{F}(X' \times _ X X')$. Next, observe that the obvious morphism
is a surjective proper morphism as $b$ induces an isomorphism $X' \setminus E \to X \setminus Z$. We conclude that $\mathcal{F}(X' \times _ X X') \to \mathcal{F}(E \times _ Z E) \times \mathcal{F}(X')$ is injective. It follows that (a) $\Rightarrow $ (b) which means that the lemma is true. $\square$
Lemma 38.36.2. Let $\mathcal{F}$ be a sheaf on a site $(\mathit{Sch}/S)_{ph}$ as in Topologies, Definition 34.8.11. Let $X \to X'$ be a morphism of $(\mathit{Sch}/S)_{ph}$ which is a thickening. Then $\mathcal{F}(X') \to \mathcal{F}(X)$ is bijective.
Proof. Observe that $X \to X'$ is a proper surjective morphism of and $X \times _{X'} X = X$. By the sheaf property for the ph covering $\{ X \to X'\} $ (Topologies, Lemma 34.8.6) we conclude. $\square$
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