## 38.36 Blow up squares and the ph topology

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

38.36.0.1
$$\label{flat-equation-blow-up-square} \vcenter { \xymatrix{ E \ar[d] \ar[r] & X' \ar[d]^ b \\ Z \ar[r] & X } }$$

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

$\xymatrix{ \mathcal{F}(E) & \mathcal{F}(X') \ar[l] \\ \mathcal{F}(Z) \ar[u] & \mathcal{F}(X) \ar[u] \ar[l] }$

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

$(Z \amalg X') \times _ X (Z \amalg X') = Z \amalg E \amalg E \amalg X' \times _ X X'$

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

$E \times _ Z E \amalg X' \longrightarrow X' \times _ X X'$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).