Lemma 59.55.5. Let $f : X \to Y$ be a surjective finite morphism of schemes. Set $f_ n : X_ n \to Y$ equal to the $(n + 1)$-fold fibre product of $X$ over $Y$. For $\mathcal{F} \in \textit{Ab}(Y_{\acute{e}tale})$ set $\mathcal{F}_ n = f_{n, *}f_ n^{-1}\mathcal{F}$. There is an exact sequence

$0 \to \mathcal{F} \to \mathcal{F}_0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \ldots$

on $X_{\acute{e}tale}$. Moreover, there is a spectral sequence

$E_1^{p, q} = H^ q_{\acute{e}tale}(X_ p, f_ p^{-1}\mathcal{F})$

converging to $H^{p + q}(Y_{\acute{e}tale}, \mathcal{F})$. This spectral sequence is functorial in $\mathcal{F}$.

Proof. If we prove the first statement of the lemma, then we obtain a spectral sequence with $E_1^{p, q} = H^ q_{\acute{e}tale}(Y, \mathcal{F})$ converging to $H^{p + q}(Y_{\acute{e}tale}, \mathcal{F})$, see Derived Categories, Lemma 13.21.3. On the other hand, since $R^ if_{p, *}f_ p^{-1}\mathcal{F} = 0$ for $i > 0$ (Proposition 59.55.2) we get

$H^ q_{\acute{e}tale}(X_ p, f_ p^{-1}\mathcal{F}) = H^ q_{\acute{e}tale}(Y, f_{p, *}f_ p^{-1} \mathcal{F}) = H^ q_{\acute{e}tale}(Y, \mathcal{F}_ p)$

by Proposition 59.54.2 and we get the spectral sequence of the lemma.

To prove the first statement of the lemma, observe that $X_ n$ forms a simplicial scheme over $Y$, see Simplicial, Example 14.3.5. Observe moreover, that for each of the projections $d_ j : X_{n + 1} \to X_ n$ there is a map $d_ j^{-1} f_ n^{-1}\mathcal{F} \to f_{n + 1}^{-1}\mathcal{F}$. These maps induce maps

$\delta _ j : \mathcal{F}_ n \to \mathcal{F}_{n + 1}$

for $j = 0, \ldots , n + 1$. We use the alternating sum of these maps to define the differentials $\mathcal{F}_ n \to \mathcal{F}_{n + 1}$. Similarly, there is a canonical augmentation $\mathcal{F} \to \mathcal{F}_0$, namely this is just the canonical map $\mathcal{F} \to f_*f^{-1}\mathcal{F}$. To check that this sequence of sheaves is an exact complex it suffices to check on stalks at geometric points (Theorem 59.29.10). Thus we let $\overline{y} : \mathop{\mathrm{Spec}}(k) \to Y$ be a geometric point. Let $E = \{ \overline{x} : \mathop{\mathrm{Spec}}(k) \to X \mid f(\overline{x}) = \overline{y}\}$. Then $E$ is a finite nonempty set and we see that

$(\mathcal{F}_ n)_{\overline{y}} = \bigoplus \nolimits _{e \in E^{n + 1}} \mathcal{F}_{\overline{y}}$

by Proposition 59.55.2 and Lemma 59.36.2. Thus we have to see that given an abelian group $M$ the sequence

$0 \to M \to \bigoplus \nolimits _{e \in E} M \to \bigoplus \nolimits _{e \in E^2} M \to \ldots$

is exact. Here the first map is the diagonal map and the map $\bigoplus _{e \in E^{n + 1}} M \to \bigoplus _{e \in E^{n + 2}} M$ is the alternating sum of the maps induced by the $(n + 2)$ projections $E^{n + 2} \to E^{n + 1}$. This can be shown directly or deduced by applying Simplicial, Lemma 14.26.9 to the map $E \to \{ *\}$. $\square$

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