Let $S$ be a scheme. Let $f : U \to X$ be an étale morphism of algebraic spaces over $S$. The functor
induces an equivalence of $U_{spaces, {\acute{e}tale}}$ with the localization $X_{spaces, {\acute{e}tale}}/U$, see Properties of Spaces, Section 66.27. Hence there exist functors
which are left adjoint to
see Modules on Sites, Section 18.19. Warning: This functor, a priori, has nothing to do with cohomology with compact supports! We dubbed this functor “extension by zero” in the reference above. Note that the two versions of $f_!$ agree as $f^* = f^{-1}$ for sheaves of $\mathcal{O}_ X$-modules.
As we are going to use this construction below let us recall some of its properties. Given an abelian sheaf $\mathcal{G}$ on $U_{\acute{e}tale}$ the sheaf $f_!$ is the sheafification of the presheaf
see Modules on Sites, Lemma 18.19.2. Moreover, if $\mathcal{G}$ is an $\mathcal{O}_ U$-module, then $f_!\mathcal{G}$ is the sheafification of the exact same presheaf of abelian groups which is endowed with an $\mathcal{O}_ X$-module structure in an obvious way (see loc. cit.). Let $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ be a geometric point. Then there is a canonical identification
where the sum is over all $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ such that $f \circ \overline{u} = \overline{x}$, see Modules on Sites, Lemma 18.38.1 and Properties of Spaces, Lemma 66.19.13. In the following we are going to study the sheaf $f_!\underline{\mathbf{Z}}$. Here $\underline{\mathbf{Z}}$ denotes the constant sheaf on $X_{\acute{e}tale}$ or $U_{\acute{e}tale}$.
Lemma 69.6.1. Let $S$ be a scheme. Let $f_ i : U_ i \to X$ be étale morphisms of algebraic spaces over $S$. Then there are isomorphisms where $f_{12} : U_1 \times _ X U_2 \to X$ is the structure morphism and
\[ f_{1, !}\underline{\mathbf{Z}} \otimes _{\mathbf{Z}} f_{2, !}\underline{\mathbf{Z}} \longrightarrow f_{12, !}\underline{\mathbf{Z}} \]
\[ (f_1 \amalg f_2)_! \underline{\mathbf{Z}} \longrightarrow f_{1, !}\underline{\mathbf{Z}} \oplus f_{2, !}\underline{\mathbf{Z}} \]
Proof. Once we have defined the map it will be an isomorphism by our description of stalks above. To define the map it suffices to work on the level of presheaves. Thus we have to define a map
We map the element $1_{\varphi _1} \otimes 1_{\varphi _2}$ to the element $1_{\varphi _1 \times \varphi _2}$ with obvious notation. We omit the proof of the second equality. $\square$
Another important feature is the trace map
The trace map is adjoint to the map $\mathbf{Z} \to f^{-1}\underline{\mathbf{Z}}$ (which is an isomorphism). If $\overline{x}$ is above, then $\text{Tr}_ f$ on stalks at $\overline{x}$ is the map
which sums the given integers. This is true because it is adjoint to the map $1 : \mathbf{Z} \to f^{-1}\underline{\mathbf{Z}}$. In particular, if $f$ is surjective as well as étale then $\text{Tr}_ f$ is surjective.
Assume that $f : U \to X$ is a surjective étale morphism of algebraic spaces. Consider the Koszul complex associated to the trace map we discussed above
Here the exterior powers are over the sheaf of rings $\underline{\mathbf{Z}}$. The maps are defined by the rule
where $e_1, \ldots , e_ n$ are local sections of $f_!\underline{\mathbf{Z}}$. Let $\overline{x}$ be a geometric point of $X$ and set $M_{\overline{x}} = (f_!\underline{\mathbf{Z}})_{\overline{x}} = \bigoplus _{\overline{u}} \mathbf{Z}$. Then the stalk of the complex above at $\overline{x}$ is the complex
which is exact because $M_{\overline{x}} \to \mathbf{Z}$ is surjective, see More on Algebra, Lemma 15.28.5. Hence if we let $K^\bullet = K^\bullet (f)$ be the complex with $K^ i = \wedge ^{i + 1}f_!\underline{\mathbf{Z}}$, then we obtain a quasi-isomorphism
We use the complex $K^\bullet $ to define what we call the alternating Čech complex associated to $f : U \to X$.
Definition 69.6.2. Let $S$ be a scheme. Let $f : U \to X$ be a surjective étale morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be an object of $\textit{Ab}(X_{\acute{e}tale})$. The alternating Čech complex1 $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F})$ associated to $\mathcal{F}$ and $f$ is the complex with Hom groups computed in $\textit{Ab}(X_{\acute{e}tale})$.
\[ \mathop{\mathrm{Hom}}\nolimits (K^0, \mathcal{F}) \to \mathop{\mathrm{Hom}}\nolimits (K^1, \mathcal{F}) \to \mathop{\mathrm{Hom}}\nolimits (K^2, \mathcal{F}) \to \ldots \]
The reader may verify that if $U = \coprod U_ i$ and $f|_{U_ i} : U_ i \to X$ is the open immersion of a subspace, then $\check{\mathcal{C}}_{alt}^\bullet (f, \mathcal{F})$ agrees with the complex introduced in Cohomology, Section 20.23 for the Zariski covering $X = \bigcup U_ i$ and the restriction of $\mathcal{F}$ to the Zariski site of $X$. What is more important however, is to relate the cohomology of the alternating Čech complex to the cohomology.
Lemma 69.6.3. Let $S$ be a scheme. Let $f : U \to X$ be a surjective étale morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be an object of $\textit{Ab}(X_{\acute{e}tale})$. There exists a canonical map in $D(\textit{Ab})$. Moreover, there is a spectral sequence with $E_1$-page converging to $H^{p + q}(X, \mathcal{F})$ where $K^ p = \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$.
\[ \check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \longrightarrow R\Gamma (X, \mathcal{F}) \]
\[ E_1^{p, q} = \mathop{\mathrm{Ext}}\nolimits _{\textit{Ab}(X_{\acute{e}tale})}^ q(K^ p, \mathcal{F}) \]
Proof. Recall that we have the quasi-isomorphism $K^\bullet \to \underline{\mathbf{Z}}[0]$, see (69.6.1.1). Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $ in $\textit{Ab}(X_{\acute{e}tale})$. Consider the double complex $\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet )$ with terms $\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{I}^ q)$. The differential $d_1^{p, q} : A^{p, q} \to A^{p + 1, q}$ is the one coming from the differential $K^{p + 1} \to K^ p$ and the differential $d_2^{p, q} : A^{p, q} \to A^{p, q + 1}$ is the one coming from the differential $\mathcal{I}^ q \to \mathcal{I}^{q + 1}$. Denote $\text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))$ the associated total complex, see Homology, Section 12.18. We will use the two spectral sequences $({}'E_ r, {}'d_ r)$ and $({}''E_ r, {}''d_ r)$ associated to this double complex, see Homology, Section 12.25.
Because $K^\bullet $ is a resolution of $\underline{\mathbf{Z}}$ we see that the complexes
are acyclic in positive degrees and have $H^0$ equal to $\Gamma (X, \mathcal{I}^ q)$. Hence by Homology, Lemma 12.25.4 the natural map
is a quasi-isomorphism of complexes of abelian groups. In particular we conclude that $H^ n(\text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))) = H^ n(X, \mathcal{F})$.
The map $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \to R\Gamma (X, \mathcal{F})$ of the lemma is the composition of $\check{\mathcal{C}}^\bullet _{alt}(f, \mathcal{F}) \to \text{Tot}(\mathop{\mathrm{Hom}}\nolimits (K^\bullet , \mathcal{I}^\bullet ))$ with the inverse of the displayed quasi-isomorphism.
Finally, consider the spectral sequence $({}'E_ r, {}'d_ r)$. We have
This proves the lemma. $\square$
It follows from the lemma that it is important to understand the ext groups $\mathop{\mathrm{Ext}}\nolimits _{\textit{Ab}(X_{\acute{e}tale})}(K^ p, \mathcal{F})$, i.e., the right derived functors of $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F})$.
Lemma 69.6.4. Let $S$ be a scheme. Let $f : U \to X$ be a surjective, étale, and separated morphism of algebraic spaces over $S$. For $p \geq 0$ set where the fibre product has $p + 1$ factors. There is a free action of $S_{p + 1}$ on $W_ p$ over $X$ and functorially in $\mathcal{F}$ where $K^ p = \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$.
\[ W_ p = U \times _ X \ldots \times _ X U \setminus \text{all diagonals} \]
\[ \mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F}) = S_{p + 1}\text{-anti-invariant elements of } \mathcal{F}(W_ p) \]
Proof. Because $U \to X$ is separated the diagonal $U \to U \times _ X U$ is a closed immersion. Since $U \to X$ is étale the diagonal $U \to U \times _ X U$ is an open immersion, see Morphisms of Spaces, Lemmas 67.39.10 and 67.38.9. Hence $W_ p$ is an open and closed subspace of $U^{p + 1} = U \times _ X \ldots \times _ X U$. The action of $S_{p + 1}$ on $W_ p$ is free as we've thrown out the fixed points of the action. By Lemma 69.6.1 we see that
where $f^{p + 1} : U^{p + 1} \to X$ is the structure morphism. Looking at stalks over a geometric point $\overline{x}$ of $X$ we see that
is the quotient whose kernel is generated by all tensors $1_{\overline{u}_0} \otimes \ldots \otimes 1_{\overline{u}_ p}$ where $\overline{u}_ i = \overline{u}_ j$ for some $i \not= j$. Thus the quotient map
factors through $(W_ p \to X)_!\underline{\mathbf{Z}}$, i.e., we get
This already proves that $\mathop{\mathrm{Hom}}\nolimits (K^ p, \mathcal{F})$ is (functorially) a subgroup of
To identify it with the $S_{p + 1}$-anti-invariants we have to prove that the surjection $(W_ p \to X)_!\underline{\mathbf{Z}} \to \wedge ^{p + 1}f_!\underline{\mathbf{Z}}$ is the maximal $S_{p + 1}$-anti-invariant quotient. In other words, we have to show that $\wedge ^{p + 1}f_!\underline{\mathbf{Z}}$ is the quotient of $(W_ p \to X)_!\underline{\mathbf{Z}}$ by the subsheaf generated by the local sections $s - \text{sign}(\sigma )\sigma (s)$ where $s$ is a local section of $(W_ p \to X)_!\underline{\mathbf{Z}}$. This can be checked on the stalks, where it is clear. $\square$
Lemma 69.6.5. Let $S$ be a scheme. Let $W$ be an algebraic space over $S$. Let $G$ be a finite group acting freely on $W$. Let $U = W/G$, see Properties of Spaces, Lemma 66.34.1. Let $\chi : G \to \{ +1, -1\} $ be a character. Then there exists a rank 1 locally free sheaf of $\mathbf{Z}$-modules $\underline{\mathbf{Z}}(\chi )$ on $U_{\acute{e}tale}$ such that for every abelian sheaf $\mathcal{F}$ on $U_{\acute{e}tale}$ we have
\[ H^0(W, \mathcal{F}|_ W)^\chi = H^0(U, \mathcal{F} \otimes _{\mathbf{Z}} \underline{\mathbf{Z}}(\chi )) \]
Proof. The quotient morphism $q : W \to U$ is a $G$-torsor, i.e., there exists a surjective étale morphism $U' \to U$ such that $W \times _ U U' = \coprod _{g \in G} U'$ as spaces with $G$-action over $U'$. (Namely, $U' = W$ works.) Hence $q_*\underline{\mathbf{Z}}$ is a finite locally free $\mathbf{Z}$-module with an action of $G$. For any geometric point $\overline{u}$ of $U$, then we get $G$-equivariant isomorphisms
where the second $=$ uses a geometric point $\overline{w}_0$ lying over $\overline{u}$ and maps the summand corresponding to $g \in G$ to the summand corresponding to $g(\overline{w}_0)$. We have
because $q_*\mathcal{F}|_ W = \mathcal{F} \otimes _\mathbf {Z} q_*\underline{\mathbf{Z}}$ as one can check by restricting to $U'$. Let
be the subsheaf of sections that transform according to $\chi $. For any geometric point $\overline{u}$ of $U$ we have
It follows that $\underline{\mathbf{Z}}(\chi )$ is locally free of rank 1 (more precisely, this should be checked after restricting to $U'$). Note that for any $\mathbf{Z}$-module $M$ the $\chi $-semi-invariants of $M[G]$ are the elements of the form $m \cdot \sum \nolimits _ g \chi (g) g$. Thus we see that for any abelian sheaf $\mathcal{F}$ on $U$ we have
because we have equality at all stalks. The result of the lemma follows by taking global sections. $\square$
Now we can put everything together and obtain the following pleasing result.
Lemma 69.6.6. Let $S$ be a scheme. Let $f : U \to X$ be a surjective, étale, and separated morphism of algebraic spaces over $S$. For $p \geq 0$ set (with $p + 1$ factors) as in Lemma 69.6.4. Let $\chi _ p : S_{p + 1} \to \{ +1, -1\} $ be the sign character. Let $U_ p = W_ p/S_{p + 1}$ and $\underline{\mathbf{Z}}(\chi _ p)$ be as in Lemma 69.6.5. Then the spectral sequence of Lemma 69.6.3 has $E_1$-page and converges to $H^{p + q}(X, \mathcal{F})$.
\[ W_ p = U \times _ X \ldots \times _ X U \setminus \text{all diagonals} \]
\[ E_1^{p, q} = H^ q(U_ p, \mathcal{F}|_{U_ p} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p)) \]
Proof. Note that since the action of $S_{p + 1}$ on $W_ p$ is over $X$ we do obtain a morphism $U_ p \to X$. Since $W_ p \to X$ is étale and since $W_ p \to U_ p$ is surjective étale, it follows that also $U_ p \to X$ is étale, see Morphisms of Spaces, Lemma 67.39.2. Therefore an injective object of $\textit{Ab}(X_{\acute{e}tale})$ restricts to an injective object of $\textit{Ab}(U_{p, {\acute{e}tale}})$, see Cohomology on Sites, Lemma 21.7.1. Moreover, the functor $\mathcal{G} \mapsto \mathcal{G} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p))$ is an auto-equivalence of $\textit{Ab}(U_ p)$, whence transforms injective objects into injective objects and is exact (because $\underline{\mathbf{Z}}(\chi _ p)$ is an invertible $\underline{\mathbf{Z}}$-module). Thus given an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $ in $\textit{Ab}(X_{\acute{e}tale})$ the complex
computes $H^*(U_ p, \mathcal{F}|_{U_ p} \otimes _\mathbf {Z} \underline{\mathbf{Z}}(\chi _ p))$. On the other hand, by Lemma 69.6.5 it is equal to the complex of $S_{p + 1}$-anti-invariants in
which by Lemma 69.6.4 is equal to the complex
which computes $\mathop{\mathrm{Ext}}\nolimits ^*_{\textit{Ab}(X_{\acute{e}tale})}(K^ p, \mathcal{F})$. Putting everything together we win. $\square$
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