Lemma 59.55.3. Consider a cartesian square

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

of schemes with $f$ a finite morphism. For any sheaf of sets $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $f'_*(g')^{-1}\mathcal{F} = g^{-1}f_*\mathcal{F}$.

Proof. In great generality there is a pullback map $g^{-1}f_*\mathcal{F} \to f'_*(g')^{-1}\mathcal{F}$, see Sites, Section 7.45. It suffices to check on stalks (Theorem 59.29.10). Let $\overline{y}' : \mathop{\mathrm{Spec}}(k) \to Y'$ be a geometric point. We have

\begin{align*} (f'_*(g')^{-1}\mathcal{F})_{\overline{y}'} & = \prod \nolimits _{\overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'} ((g')^{-1}\mathcal{F})_{\overline{x}'} \\ & = \prod \nolimits _{\overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'} \mathcal{F}_{g' \circ \overline{x}'} \\ & = \prod \nolimits _{\overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f \circ \overline{x} = g \circ \overline{y}'} \mathcal{F}_{\overline{x}} \\ & = (f_*\mathcal{F})_{g \circ \overline{y}'} \\ & = (g^{-1}f_*\mathcal{F})_{\overline{y}'} \end{align*}

The first equality by Proposition 59.55.2. The second equality by Lemma 59.36.2. The third equality holds because the diagram is a cartesian square and hence the map

$\{ \overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'\} \longrightarrow \{ \overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f \circ \overline{x} = g \circ \overline{y}'\}$

sending $\overline{x}'$ to $g' \circ \overline{x}'$ is a bijection. The fourth equality by Proposition 59.55.2. The fifth equality by Lemma 59.36.2. $\square$

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