Lemma 59.91.5. Let $f : X \to Y$ be a proper morphism of schemes. Let $g : Y' \to Y$ be a morphism of schemes. Set $X' = Y' \times _ Y X$ with projections $f' : X' \to Y'$ and $g' : X' \to X$. Let $\mathcal{F}$ be any sheaf on $X_{\acute{e}tale}$. Then $g^{-1}f_*\mathcal{F} = f'_*(g')^{-1}\mathcal{F}$.
Proof. There is a canonical map $g^{-1}f_*\mathcal{F} \to f'_*(g')^{-1}\mathcal{F}$. Namely, it is adjoint to the map
\[ f_*\mathcal{F} \longrightarrow g_*f'_*(g')^{-1}\mathcal{F} = f_*g'_*(g')^{-1}\mathcal{F} \]
which is $f_*$ applied to the canonical map $\mathcal{F} \to g'_*(g')^{-1}\mathcal{F}$. To check this map is an isomorphism we can compute what happens on stalks. Let $y' : \mathop{\mathrm{Spec}}(k) \to Y'$ be a geometric point with image $y$ in $Y$. By Lemma 59.91.4 the stalks are $\Gamma (X'_{y'}, \mathcal{F}_{y'})$ and $\Gamma (X_ y, \mathcal{F}_ y)$ respectively. Here the sheaves $\mathcal{F}_ y$ and $\mathcal{F}_{y'}$ are the pullbacks of $\mathcal{F}$ by the projections $X_ y \to X$ and $X'_{y'} \to X$. Thus we see that the groups agree by Lemma 59.39.5. We omit the verification that this isomorphism is compatible with our map. $\square$
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