Lemma 59.104.7. Let $f : X' \to X$ be a proper morphism of schemes. Let $i : Z \to X$ be a closed immersion. Set $E = Z \times _ X X'$. Picture
\[ \xymatrix{ E \ar[d]_ g \ar[r]_ j & X' \ar[d]^ f \\ Z \ar[r]^ i & X } \]
If $f$ is an isomorphism over $X \setminus Z$, then the functor
\[ \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}) \longrightarrow \mathop{\mathit{Sh}}\nolimits (X'_{\acute{e}tale}) \times _{\mathop{\mathit{Sh}}\nolimits (E_{\acute{e}tale})} \mathop{\mathit{Sh}}\nolimits (Z_{\acute{e}tale}) \]
is an equivalence of categories.
Proof.
We will work with the $2$-fibre product category as constructed in Categories, Example 4.31.3. The functor sends $\mathcal{F}$ to the triple $(f^{-1}\mathcal{F}, i^{-1}\mathcal{F}, c)$ where $c : j^{-1}f^{-1}\mathcal{F} \to g^{-1}i^{-1}\mathcal{F}$ is the canonical isomorphism. We will construct a quasi-inverse functor. Let $(\mathcal{F}', \mathcal{G}, \alpha )$ be an object of the right hand side of the arrow. We obtain an isomorphism
\[ i^{-1}f_*\mathcal{F}' = g_*j^{-1}\mathcal{F}' \xrightarrow {g_*\alpha } g_*g^{-1}\mathcal{G} \]
The first equality is Lemma 59.91.5. Using this we obtain maps $i_*\mathcal{G} \to i_*g_*g^{-1}\mathcal{G}$ and $f'_*\mathcal{F}' \to i_*g_*g^{-1}\mathcal{G}$. We set
\[ \mathcal{F} = f_*\mathcal{F}' \times _{i_*g_*g^{-1}\mathcal{G}} i_*\mathcal{G} \]
and we claim that $\mathcal{F}$ is an object of the left hand side of the arrow whose image in the right hand side is isomorphic to the triple we started out with. Let us compute the stalk of $\mathcal{F}$ at a geometric point $\overline{x}$ of $X$.
If $\overline{x}$ is not in $Z$, then on the one hand $\overline{x}$ comes from a unique geometric point $\overline{x}'$ of $X'$ and $\mathcal{F}'_{\overline{x}'} = (f_*\mathcal{F}')_{\overline{x}}$ and on the other hand we have $(i_*\mathcal{G})_{\overline{x}}$ and $(i_*g_*g^{-1}\mathcal{G})_{\overline{x}}$ are singletons. Hence we see that $\mathcal{F}_{\overline{x}}$ equals $\mathcal{F}'_{\overline{x}'}$.
If $\overline{x}$ is in $Z$, i.e., $\overline{x}$ is the image of a geometric point $\overline{z}$ of $Z$, then we obtain $(i_*\mathcal{G})_{\overline{x}} = \mathcal{G}_{\overline{z}}$ and
\[ (i_*g_*g^{-1}\mathcal{G})_{\overline{x}} = (g_*g^{-1}\mathcal{G})_{\overline{z}} = \Gamma (E_{\overline{z}}, g^{-1}\mathcal{G}|_{E_{\overline{z}}}) \]
(by the proper base change for pushforward used above) and similarly
\[ (f_*\mathcal{F}')_{\overline{x}} = \Gamma (X'_{\overline{x}}, \mathcal{F}'|_{X'_{\overline{x}}}) \]
Since we have the identification $E_{\overline{z}} = X'_{\overline{x}}$ and since $\alpha $ defines an isomorphism between the sheaves $\mathcal{F}'|_{X'_{\overline{x}}}$ and $g^{-1}\mathcal{G}|_{E_{\overline{z}}}$ we conclude that we get
\[ \mathcal{F}_{\overline{x}} = \mathcal{G}_{\overline{z}} \]
in this case.
To finish the proof, we observe that there are canonical maps $i^{-1}\mathcal{F} \to \mathcal{G}$ and $f^{-1}\mathcal{F} \to \mathcal{F}'$ compatible with $\alpha $ which on stalks produce the isomorphisms we saw above. We omit the careful construction of these maps.
$\square$
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