Lemma 59.82.2. Let $X$ be a quasi-compact and quasi-separated scheme. Let $i : Z \to X$ be a closed immersion. Assume that
for any sheaf $\mathcal{F}$ on $X_{Zar}$ the map $\Gamma (X, \mathcal{F}) \to \Gamma (Z, i^{-1}\mathcal{F})$ is bijective, and
for any finite morphism $X' \to X$ assumption (1) holds for $Z \times _ X X' \to X'$.
Then for any sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $\Gamma (X, \mathcal{F}) = \Gamma (Z, i^{-1}_{small}\mathcal{F})$.
Proof.
Let $\mathcal{F}$ be a sheaf on $X_{\acute{e}tale}$. There is a canonical (base change) map
\[ i^{-1}(\mathcal{F}|_{X_{Zar}}) \longrightarrow (i_{small}^{-1}\mathcal{F})|_{Z_{Zar}} \]
of sheaves on $Z_{Zar}$. We will show this map is injective by looking at stalks. The stalk on the left hand side at $z \in Z$ is the stalk of $\mathcal{F}|_{X_{Zar}}$ at $z$. The stalk on the right hand side is the colimit over all elementary étale neighbourhoods $(U, u) \to (X, z)$ such that $U \times _ X Z \to Z$ has a section over a neighbourhood of $z$. As étale morphisms are open, the image of $U \to X$ is an open neighbourhood $U_0$ of $z$ in $X$. The map $\mathcal{F}(U_0) \to \mathcal{F}(U)$ is injective by the sheaf condition for $\mathcal{F}$ with respect to the étale covering $U \to U_0$. Taking the colimit over all $U$ and $U_0$ we obtain injectivity on stalks.
It follows from this and assumption (1) that the map $\Gamma (X, \mathcal{F}) \to \Gamma (Z, i^{-1}_{small}\mathcal{F})$ is injective. By (2) the same thing is true on all $X'$ finite over $X$.
Let $s \in \Gamma (Z, i^{-1}_{small}\mathcal{F})$. By construction of $i^{-1}_{small}\mathcal{F}$ there exists an étale covering $\{ V_ j \to Z\} $, étale morphisms $U_ j \to X$, sections $s_ j \in \mathcal{F}(U_ j)$ and morphisms $V_ j \to U_ j$ over $X$ such that $s|_{V_ j}$ is the pullback of $s_ j$. Observe that every nonempty closed subscheme $T \subset X$ meets $Z$ by assumption (1) applied to the sheaf $(T \to X)_*\underline{\mathbf{Z}}$ for example. Thus we see that $\coprod U_ j \to X$ is surjective. By More on Morphisms, Lemma 37.45.7 we can find a finite surjective morphism $X' \to X$ such that $X' \to X$ Zariski locally factors through $\coprod U_ j \to X$. It follows that $s|_{Z'}$ Zariski locally comes from a section of $\mathcal{F}|_{X'}$. In other words, $s|_{Z'}$ comes from $t' \in \Gamma (X', \mathcal{F}|_{X'})$ by assumption (2). By injectivity we conclude that the two pullbacks of $t'$ to $X' \times _ X X'$ are the same (after all this is true for the pullbacks of $s$ to $Z' \times _ Z Z'$). Hence we conclude $t'$ comes from a section of $\mathcal{F}$ over $X$ by Remark 59.55.6.
$\square$
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