Lemma 62.3.17. Let $f : X \to Y$ be a morphism of schemes which is separated and locally quasi-finite. Then

1. for $\mathcal{F}$ in $\textit{Ab}(X_{\acute{e}tale})$ and a geometric point $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ we have

$(f_!\mathcal{F})_{\overline{y}} = \bigoplus \nolimits _{f(\overline{x}) = \overline{y}} \mathcal{F}_{\overline{x}}$

functorially in $\mathcal{F}$, and

2. the functor $f_!$ is exact.

Proof. The functor $f_!$ is left exact by construction. Right exactness may be checked on stalks (Étale Cohomology, Theorem 59.29.10). Thus it suffices to prove part (1).

Let $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ be a geometric point. The scheme $X_{\overline{y}}$ has a discrete underlying topological space (Morphisms, Lemma 29.20.8) and all the residue fields at the points are equal to $k$ (as finite extensions of $k$). Hence $\{ \overline{x} : \mathop{\mathrm{Spec}}(k) \to X : f(\overline{x}) = \overline{y}\}$ is equal to the set of points of $X_{\overline{y}}$. Thus the computation of the stalk follows from the more general Lemma 62.3.11. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).