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 $X$.
Before we prove the surjectivity, let us introduce some notation. For any object $U$ of $X_{\acute{e}tale}$ and $s \in \mathcal{F}(U)$ we denote $i^{-1}s \in (i_{small}^{-1}\mathcal{F})(U \times _ X Z)$ the pullback (more precisely this is the image of $s$ under the map of Sites, Lemma 7.5.3 combined with the map from the presheaf pullback to the sheaf pullback). This construction is compatible with pullback by morphisms in $X_{\acute{e}tale}$ and by (finite) morphisms $X' \to X$ in a manner which we leave to the reader to clarify.
Let $t \in \Gamma (Z, i^{-1}_{small}\mathcal{F})$. By construction of $i^{-1}_{small}\mathcal{F}$ there exists an étale covering $\{ V_ j \to Z\} _{j \in J}$, 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 $t|_{V_ j}$ is the pullback of $i^{-1}s_ j$ by $V_ j \to U_ j \times _ X Z$. Since $V_ j \to U_ j \times _ X Z$ is étale, the image is an open subscheme. Since $U_ j \times _ X Z \subset U_ j$ is closed, we may, after replacing $U_ j$ by an open subscheme, assume that $V_ j \to U_ j \times _ X Z$ is surjective. Then $\{ V_ j \to U_ j \times _ X Z\} $ is an étale covering and we conclude that $t|_{U_ j \times _ X Z} = i^{-1}s_ j$. 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 $\coprod U_ j \to X$. Say $X' = \bigcup _{\alpha \in A} W_\alpha $ is an open covering and $j : A \to J$ is a map, such that there exist morphisms $g_\alpha : W_\alpha \to U_{j(\alpha )}$ over $X$. Denote $\mathcal{F}'$ the pullback of $\mathcal{F}$ to $X'_{\acute{e}tale}$. Denote $s'_\alpha \in \mathcal{F}'(W_\alpha )$ the pullback of $s_{j(\alpha )}$ by $g_\alpha $. Set $Z' = X' \times _ X Z$. Denote $i' : Z' \to X'$ the base change of $i$. Then $(i'_{small})^{-1}\mathcal{F}' = (Z' \to Z)_{small}^{-1}i_{small}^{-1}\mathcal{F}$. Via this identification we may denote $t' \in \Gamma (Z', (i')^{-1}_{small}\mathcal{F}')$ the pullback of $t$ to $Z'$. Then $t'|_{W_\alpha \times _ X Z}$ is equal to $(i')^{-1}s'_\alpha $ by construction. We conclude that $t'$ is a section of the subsheaf
\[ (i')^{-1}(\mathcal{F}'|_{X'_{Zar}}) \subset (i'_{small})^{-1}\mathcal{F}')|_{Z'_{Zar}} \]
By assumption (2) the section $t'$ comes from a section $s' \in \Gamma (X', \mathcal{F}'|_{X'_{Zar}}) = \Gamma (X', \mathcal{F}')$. By the injectivity proved in the second paragraph, we conclude that the two pullbacks of $s'$ to $X' \times _ X X'$ are the same (after all this is true for the pullbacks of $t'$ to $Z' \times _ Z Z'$). Hence we conclude $s'$ comes from a section of $\mathcal{F}$ over $X$ by Remark 59.55.6.
$\square$
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