Lemma 59.70.9. Consider a cartesian diagram of schemes

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

where $f$ is finite, $g$ is étale, and $j$ is an open immersion. Then $f_* \circ j'_! = j_! \circ g_*$ as functors $\textit{Ab}(U_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$.

Proof. Let $\mathcal{F}$ be an object of $\textit{Ab}(U_{\acute{e}tale})$. Let $\overline{y}$ be a geometric point of $Y$ not contained in the open $V$. Then

$(f_*j'_!\mathcal{F})_{\overline{y}} = \bigoplus \nolimits _{\overline{x},\ f(\overline{x}) = \overline{y}} (j'_!\mathcal{F})_{\overline{x}} = 0$

by Proposition 59.55.2 and because the stalk of $j'_!\mathcal{F}$ at $\overline{x} \not\in U$ are zero by Lemma 59.70.4. On the other hand, we have

$j^{-1}f_*j'_!\mathcal{F} = g_*(j')^{-1}j'_!\mathcal{F} = g_*\mathcal{F}$

by Lemmas 59.55.3 and Lemma 59.70.4. Hence by the characterization of $j_!$ in Lemma 59.70.4 we see that $f_*j'_!\mathcal{F} = j_!g_*\mathcal{F}$. We omit the verification that this identification is functorial in $\mathcal{F}$. $\square$

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