The Stacks project

Proof. By Lemmas 59.73.2 and 59.51.4 we can find a map $\mathcal{G} \to \mathcal{F}$ with $\mathcal{G}$ a constructible abelian sheaf and $\xi $ coming from an element $\zeta $ of $H^ q_{\acute{e}tale}(Z, \mathcal{G}|_ Z)$. Suppose we can find an injective map $\mathcal{G} \to \mathcal{G}'$ of torsion abelian sheaves on $X_{\acute{e}tale}$ such that the image of $\zeta $ in $H^ q_{\acute{e}tale}(Z, \mathcal{G}'|_ Z)$ is zero. Then we can take $\mathcal{F}'$ to be the pushout

and we conclude the result of the lemma holds. (Observe that restriction to $Z$ is exact, so commutes with finite limits and colimits and moreover it commutes with arbitrary colimits as a left adjoint to pushforward.) Thus we may assume $\mathcal{F}$ is constructible.

Assume $\mathcal{F}$ is constructible. By Lemma 59.74.4 it suffices to prove the result when $\mathcal{F}$ is of the form $f_*\underline{M}$ where $M$ is a finite abelian group and $f : Y \to X$ is a finite morphism of finite presentation (such sheaves are still constructible by Lemma 59.73.9 but we won't need this). Since formation of $f_*$ commutes with any base change (Lemma 59.55.3) we see that the restriction of $f_*\underline{M}$ to $Z$ is equal to the pushforward of $\underline{M}$ via $Y \times _ X Z \to Z$. By the Leray spectral sequence (Proposition 59.54.2) and vanishing of higher direct images (Proposition 59.55.2), we find

By Lemma 59.80.9 we can find a finite surjective morphism $Y' \to Y$ of finite presentation such that $\xi $ maps to zero in $H^ q(Y' \times _ X Z, \underline{M})$. Denoting $f' : Y' \to X$ the composition $Y' \to Y \to X$ we claim the map

is injective which finishes the proof by what was said above. To see the desired injectivity we can look at stalks. Namely, if $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ is a geometric point, then

by Proposition 59.55.2 and similarly for the other sheaf. Since $Y' \to Y$ is surjective and finite we see that the induced map on geometric points lifting $\overline{x}$ is surjective too and we conclude. $\square$

Proof. Let $\mathcal{F}$ be a sheaf on $X_{\acute{e}tale}$. There is a canonical (base change) map

of sheaves on $Z_{Zar}$. We will show this map is injective by looking at stalks. The stalk on the left hand side at $z \in Z$ is the stalk of $\mathcal{F}|_{X_{Zar}}$ at $z$. The stalk on the right hand side is the colimit over all elementary étale neighbourhoods $(U, u) \to (X, z)$ such that $U \times _ X Z \to Z$ has a section over a neighbourhood of $z$. As étale morphisms are open, the image of $U \to X$ is an open neighbourhood $U_0$ of $z$ in $X$. The map $\mathcal{F}(U_0) \to \mathcal{F}(U)$ is injective by the sheaf condition for $\mathcal{F}$ with respect to the étale covering $U \to U_0$. Taking the colimit over all $U$ and $U_0$ we obtain injectivity on stalks.

It follows from this and assumption (1) that the map $\Gamma (X, \mathcal{F}) \to \Gamma (Z, i^{-1}_{small}\mathcal{F})$ is injective. By (2) the same thing is true on all $X'$ finite over $X$.

Before we prove the surjectivity, let us introduce some notation. For any object $U$ of $X_{\acute{e}tale}$ and $s \in \mathcal{F}(U)$ we denote $i^{-1}s \in (i_{small}^{-1}\mathcal{F})(U \times _ X Z)$ the pullback (more precisely this is the image of $s$ under the map of Sites, Lemma 7.5.3 combined with the map from the presheaf pullback to the sheaf pullback). This construction is compatible with pullback by morphisms in $X_{\acute{e}tale}$ and by (finite) morphisms $X' \to X$ in a manner which we leave to the reader to clarify.

Let $t \in \Gamma (Z, i^{-1}_{small}\mathcal{F})$. By construction of $i^{-1}_{small}\mathcal{F}$ there exists an étale covering $\{ V_ j \to Z\} _{j \in J}$, étale morphisms $U_ j \to X$, sections $s_ j \in \mathcal{F}(U_ j)$ and morphisms $V_ j \to U_ j$ over $X$ such that $t|_{V_ j}$ is the pullback of $i^{-1}s_ j$ by $V_ j \to U_ j \times _ X Z$. Since $V_ j \to U_ j \times _ X Z$ is étale, the image is an open subscheme. Since $U_ j \times _ X Z \subset U_ j$ is closed, we may, after replacing $U_ j$ by an open subscheme, assume that $V_ j \to U_ j \times _ X Z$ is surjective. Then $\{ V_ j \to U_ j \times _ X Z\} $ is an étale covering and we conclude that $t|_{U_ j \times _ X Z} = i^{-1}s_ j$. Observe that every nonempty closed subscheme $T \subset X$ meets $Z$ by assumption (1) applied to the sheaf $(T \to X)_*\underline{\mathbf{Z}}$ for example. Thus we see that $\coprod U_ j \to X$ is surjective. By More on Morphisms, Lemma 37.45.7 we can find a finite surjective morphism $X' \to X$ such that $X' \to X$ Zariski locally factors through $\coprod U_ j \to X$. Say $X' = \bigcup _{\alpha \in A} W_\alpha $ is an open covering and $j : A \to J$ is a map, such that there exist morphisms $g_\alpha : W_\alpha \to U_{j(\alpha )}$ over $X$. Denote $\mathcal{F}'$ the pullback of $\mathcal{F}$ to $X'_{\acute{e}tale}$. Denote $s'_\alpha \in \mathcal{F}'(W_\alpha )$ the pullback of $s_{j(\alpha )}$ by $g_\alpha $. Set $Z' = X' \times _ X Z$. Denote $i' : Z' \to X'$ the base change of $i$. Then $(i'_{small})^{-1}\mathcal{F}' = (Z' \to Z)_{small}^{-1}i_{small}^{-1}\mathcal{F}$. Via this identification we may denote $t' \in \Gamma (Z', (i')^{-1}_{small}\mathcal{F}')$ the pullback of $t$ to $Z'$. Then $t'|_{W_\alpha \times _ X Z}$ is equal to $(i')^{-1}s'_\alpha $ by construction. We conclude that $t'$ is a section of the subsheaf

By assumption (2) the section $t'$ comes from a section $s' \in \Gamma (X', \mathcal{F}'|_{X'_{Zar}}) = \Gamma (X', \mathcal{F}')$. By the injectivity proved in the second paragraph, we conclude that the two pullbacks of $s'$ to $X' \times _ X X'$ are the same (after all this is true for the pullbacks of $t'$ to $Z' \times _ Z Z'$). Hence we conclude $s'$ comes from a section of $\mathcal{F}$ over $X$ by Remark 59.55.6. $\square$

Proof. Observe that $Z_ i$ is quasi-compact. Hence the set of points $W_ i$ specializing to $Z_ i$ is closed in the constructible topology by Topology, Lemma 5.24.7. Assumption (2) implies that $X = W_1 \amalg W_2$. Let $x \in \overline{W_1}$. By Topology, Lemma 5.23.6 part (1) there exists a specialization $x_1 \leadsto x$ with $x_1 \in W_1$. Thus $\overline{\{ x\} } \subset \overline{\{ x_1\} }$ and we see that $x \in W_1$. In other words, setting $X_ i = W_ i$ does the job. $\square$

Proof. If $x \leadsto x'$ is a specialization of points, then there is a canonical map $\mathcal{F}_{x'} \to \mathcal{F}_ x$ compatible with sections over opens and functorial in $\mathcal{F}$. Since every point of $X$ specializes to a point of $Z$ it follows that $\Gamma (X, \mathcal{F}) \to \Gamma (Z, \mathcal{F}|_ Z)$ is injective. The difficult part is to show that it is surjective.

Denote $\mathcal{B}$ be the set of all quasi-compact opens of $X$. Write $\mathcal{F}$ as a filtered colimit $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ where each $\mathcal{F}_ i$ is as in Modules, Equation (17.19.2.1). See Modules, Lemma 17.19.2. Then $\mathcal{F}|_ Z = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i|_ Z$ as restriction to $Z$ is a left adjoint (Categories, Lemma 4.24.5 and Sheaves, Lemma 6.21.8). By Sheaves, Lemma 6.29.1 the functors $\Gamma (X, -)$ and $\Gamma (Z, -)$ commute with filtered colimits. Hence we may assume our sheaf $\mathcal{F}$ is as in Modules, Equation (17.19.2.1).

Suppose that we have an embedding $\mathcal{F} \subset \mathcal{G}$. Then we have

where the intersection takes place in $\Gamma (Z, \mathcal{G}|_ Z)$. This follows from the first remark of the proof because we can check whether a global section of $\mathcal{G}$ is in $\mathcal{F}$ by looking at the stalks and because every point of $X$ specializes to a point of $Z$.

By Modules, Lemma 17.19.4 there is an injection $\mathcal{F} \to \prod (Z_ i \to X)_*\underline{S_ i}$ where the product is finite, $Z_ i \subset X$ is closed, and $S_ i$ is finite. Thus it suffices to prove surjectivity for the sheaves $(Z_ i \to X)_*\underline{S_ i}$. Observe that

Moreover, conditions (1) and (2) are inherited by $Z_ i$; this is clear for (2) and follows from Topology, Lemma 5.23.5 for (1). Thus it suffices to prove the lemma in the case of a (finite) constant sheaf. This case is a restatement of Lemma 59.82.3 which finishes the proof. $\square$

Proof. Recall that the spectrum of any ring is a spectral space, see Algebra, Lemma 10.26.2. By More on Algebra, Lemma 15.11.16 we see that $\overline{\{ x\} } \cap Z$ is connected for every $x \in X$. By Lemma 59.82.4 we see that the statement is true for sheaves on $X_{Zar}$. For any finite morphism $X' \to X$ we have $X' = \mathop{\mathrm{Spec}}(A')$ and $Z \times _ X X' = \mathop{\mathrm{Spec}}(A'/IA')$ with $(A', IA')$ a henselian pair, see More on Algebra, Lemma 15.11.8 and we get the same statement for sheaves on $(X')_{Zar}$. Thus we can apply Lemma 59.82.2 to conclude. $\square$

Proof. The result holds for $q = 0$ by Lemma 59.82.6. Let $q \geq 1$. Suppose the result has been shown in all degrees $< q$. Let $\mathcal{F}$ be a torsion abelian sheaf. Let $\mathcal{F} \to \mathcal{F}'$ be an injective map of torsion abelian sheaves (to be chosen later) with cokernel $\mathcal{Q}$ so that we have the short exact sequence

of torsion abelian sheaves on $X_{\acute{e}tale}$. This gives a map of long exact cohomology sequences over $X$ and $Z$ part of which looks like

Using this commutative diagram of abelian groups with exact rows we will finish the proof.

Injectivity for $\mathcal{F}$. Let $\xi $ be a nonzero element of $H^ q_{\acute{e}tale}(X, \mathcal{F})$. By Lemma 59.82.1 applied with $Z = X$ (!) we can find $\mathcal{F} \subset \mathcal{F}'$ such that $\xi $ maps to zero to the right. Then $\xi $ is the image of an element of $H^{q - 1}_{\acute{e}tale}(X, \mathcal{Q})$ and bijectivity for $q - 1$ implies $\xi $ does not map to zero in $H^ q_{\acute{e}tale}(Z, \mathcal{F}|_ Z)$.

Surjectivity for $\mathcal{F}$. Let $\xi $ be an element of $H^ q_{\acute{e}tale}(Z, \mathcal{F}|_ Z)$. By Lemma 59.82.1 applied with $Z = Z$ we can find $\mathcal{F} \subset \mathcal{F}'$ such that $\xi $ maps to zero to the right. Then $\xi $ is the image of an element of $H^{q - 1}_{\acute{e}tale}(Z, \mathcal{Q}|_ Z)$ and bijectivity for $q - 1$ implies $\xi $ is in the image of the vertical map. $\square$

Proof. We will prove this by induction on $n$. If $n = 0$, then $X$ is affine. Say $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/I)$. Let $A^ h$ be the filtered colimit of étale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism. Then $(A^ h, IA^ h)$ is a henselian pair and $A/I = A^ h/IA^ h$, see More on Algebra, Lemma 15.12.1 and its proof. Set $X^ h = \mathop{\mathrm{Spec}}(A^ h)$. By Theorem 59.82.7 we see that

By Theorem 59.51.3 we have

where the colimit is over the $A$-algebras $B$ as above. Since the morphisms $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ are étale, the restriction $\mathcal{I}|_{\mathop{\mathrm{Spec}}(B)}$ is an injective sheaf of $\mathcal{A}|_{\mathop{\mathrm{Spec}}(B)}$-modules (Cohomology on Sites, Lemma 21.7.1). Thus the cohomology groups on the right are zero and we get the result in this case.

Induction step. We can use Mayer-Vietoris to do the induction step. Namely, suppose that $X = U \cup V$ where $U$ is a union of $n$ affine opens and $V$ is affine. Then, using that the diagonal of $X$ is affine, we see that $U \cap V$ is the union of $n$ affine opens. Mayer-Vietoris gives an exact sequence

and by our induction hypothesis we obtain vanishing for $q > n$ as desired. $\square$