## 59.82 Affine analog of proper base change

In this section we discuss a result by Ofer Gabber, see . This was also proved by Roland Huber, see . We have already done some of the work needed for Gabber's proof in Section 59.80.

Lemma 59.82.1. Let $X$ be an affine scheme. Let $\mathcal{F}$ be a torsion abelian sheaf on $X_{\acute{e}tale}$. Let $Z \subset X$ be a closed subscheme. Let $\xi \in H^ q_{\acute{e}tale}(Z, \mathcal{F}|_ Z)$ for some $q > 0$. Then there exists an injective map $\mathcal{F} \to \mathcal{F}'$ of torsion abelian sheaves on $X_{\acute{e}tale}$ such that the image of $\xi$ in $H^ q_{\acute{e}tale}(Z, \mathcal{F}'|_ Z)$ is zero.

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

$\mathcal{F}' = \mathcal{G}' \amalg _{\mathcal{G}} \mathcal{F}$

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

$H^ q_{\acute{e}tale}(Z, f_*\underline{M}|_ Z) = H^ q_{\acute{e}tale}(Y \times _ X Z, \underline{M}).$

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

$f_*\underline{M} \longrightarrow f'_*\underline{M}$

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

$(f_*\underline{M})_{\overline{x}} = \bigoplus \nolimits _{f(\overline{y}) = \overline{x}} M$

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. $\square$

The lemma above will take care of higher cohomology groups in Gabber's result. The following lemma will be used to deal with global sections.

Lemma 59.82.2. Let $X$ be a quasi-compact and quasi-separated scheme. Let $i : Z \to X$ be a closed immersion. Assume that

1. for any sheaf $\mathcal{F}$ on $X_{Zar}$ the map $\Gamma (X, \mathcal{F}) \to \Gamma (Z, i^{-1}\mathcal{F})$ is bijective, and

2. for any finite morphism $X' \to X$ assumption (1) holds for $Z \times _ X X' \to X'$.

Then for any sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $\Gamma (X, \mathcal{F}) = \Gamma (Z, i^{-1}_{small}\mathcal{F})$.

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

$i^{-1}(\mathcal{F}|_{X_{Zar}}) \longrightarrow (i_{small}^{-1}\mathcal{F})|_{Z_{Zar}}$

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 $X$.

Let $s \in \Gamma (Z, i^{-1}_{small}\mathcal{F})$. By construction of $i^{-1}_{small}\mathcal{F}$ there exists an étale covering $\{ V_ j \to Z\}$, 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 $s|_{V_ j}$ is the pullback of $s_ j$. 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 37.44.7 we can find a finite surjective morphism $X' \to X$ such that $X' \to X$ Zariski locally factors through $\coprod U_ j \to X$. It follows that $s|_{Z'}$ Zariski locally comes from a section of $\mathcal{F}|_{X'}$. In other words, $s|_{Z'}$ comes from $t' \in \Gamma (X', \mathcal{F}|_{X'})$ by assumption (2). By injectivity we conclude that the two pullbacks of $t'$ to $X' \times _ X X'$ are the same (after all this is true for the pullbacks of $s$ to $Z' \times _ Z Z'$). Hence we conclude $t'$ comes from a section of $\mathcal{F}$ over $X$ by Remark 59.55.6. $\square$

Lemma 59.82.3. Let $Z \subset X$ be a closed subset of a topological space $X$. Assume

1. $X$ is a spectral space (Topology, Definition 5.23.1), and

2. for $x \in X$ the intersection $Z \cap \overline{\{ x\} }$ is connected (in particular nonempty).

If $Z = Z_1 \amalg Z_2$ with $Z_ i$ closed in $Z$, then there exists a decomposition $X = X_1 \amalg X_2$ with $X_ i$ closed in $X$ and $Z_ i = Z \cap X_ i$.

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. $\square$

Lemma 59.82.4. Let $Z \subset X$ be a closed subset of a topological space $X$. Assume

1. $X$ is a spectral space (Topology, Definition 5.23.1), and

2. for $x \in X$ the intersection $Z \cap \overline{\{ x\} }$ is connected (in particular nonempty).

Then for any sheaf $\mathcal{F}$ on $X$ we have $\Gamma (X, \mathcal{F}) = \Gamma (Z, \mathcal{F}|_ Z)$.

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

$\Gamma (X, \mathcal{F}) = \Gamma (Z, \mathcal{F}|_ Z) \cap \Gamma (X, \mathcal{G})$

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

$\Gamma (X, (Z_ i \to X)_*\underline{S_ i}) = \Gamma (Z_ i, \underline{S_ i}) \quad \text{and}\quad \Gamma (X, (Z_ i \to X)_*\underline{S_ i}|_ Z) = \Gamma (Z \cap Z_ i, \underline{S_ i})$

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. $\square$

Example 59.82.5. Lemma 59.82.4 is false if $X$ is not spectral. Here is an example: Let $Y$ be a $T_1$ topological space, and $y \in Y$ a non-open point. Let $X = Y \amalg \{ x \}$, endowed with the topology whose closed sets are $\emptyset$, $\{ y\}$, and all $F \amalg \{ x \}$, where $F$ is a closed subset of $Y$. Then $Z = \{ x, y\}$ is a closed subset of $X$, which satisfies assumption (2) of Lemma 59.82.4. But $X$ is connected, while $Z$ is not. The conclusion of the lemma thus fails for the constant sheaf with value $\{ 0, 1\}$ on $X$.

Lemma 59.82.6. Let $(A, I)$ be a henselian pair. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/I)$. For any sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $\Gamma (X, \mathcal{F}) = \Gamma (Z, \mathcal{F}|_ Z)$.

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. $\square$

Finally, we can state and prove Gabber's theorem.

Theorem 59.82.7 (Gabber). Let $(A, I)$ be a henselian pair. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/I)$. For any torsion abelian sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $H^ q_{\acute{e}tale}(X, \mathcal{F}) = H^ q_{\acute{e}tale}(Z, \mathcal{F}|_ Z)$.

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

$0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{Q} \to 0$

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

$\xymatrix{ H^{q - 1}_{\acute{e}tale}(X, \mathcal{F}') \ar[d] \ar[r] & H^{q - 1}_{\acute{e}tale}(X, \mathcal{Q}) \ar[d] \ar[r] & H^ q_{\acute{e}tale}(X, \mathcal{F}) \ar[d] \ar[r] & H^ q_{\acute{e}tale}(X, \mathcal{F}') \ar[d] \\ H^{q - 1}_{\acute{e}tale}(Z, \mathcal{F}'|_ Z) \ar[r] & H^{q - 1}_{\acute{e}tale}(Z, \mathcal{Q}|_ Z) \ar[r] & H^ q_{\acute{e}tale}(Z, \mathcal{F}|_ Z) \ar[r] & H^ q_{\acute{e}tale}(Z, \mathcal{F}'|_ Z) }$

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. $\square$

Lemma 59.82.8. Let $X$ be a scheme with affine diagonal which can be covered by $n + 1$ affine opens. Let $Z \subset X$ be a closed subscheme. Let $\mathcal{A}$ be a torsion sheaf of rings on $X_{\acute{e}tale}$ and let $\mathcal{I}$ be an injective sheaf of $\mathcal{A}$-modules on $X_{\acute{e}tale}$. Then $H^ q_{\acute{e}tale}(Z, \mathcal{I}|_ Z) = 0$ for $q > n$.

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

$H^ q_{\acute{e}tale}(Z, \mathcal{I}|_ Z) = H^ q_{\acute{e}tale}(X^ h, \mathcal{I}|_{X^ h})$

By Theorem 59.51.3 we have

$H^ q_{\acute{e}tale}(X^ h, \mathcal{I}|_{X^ h}) = \mathop{\mathrm{colim}}\nolimits _{A \to B} H^ q_{\acute{e}tale}(\mathop{\mathrm{Spec}}(B), \mathcal{I}|_{\mathop{\mathrm{Spec}}(B)})$

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

$H^{q - 1}_{\acute{e}tale}(U \cap V \cap Z, \mathcal{I}|_ Z) \to H^ q_{\acute{e}tale}(Z, \mathcal{I}|_ Z) \to H^ q_{\acute{e}tale}(U \cap Z, \mathcal{I}|_ Z) \oplus H^ q_{\acute{e}tale}(V \cap Z, \mathcal{I}|_ Z)$

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

Comment #1923 by Guignard on

Lemma $50.73.8$ is false ; the collection $(s_x)_x$ of germs constructed in the proof doesn't necessarily satisfy the continuity condition (*). Here is a counterexample : let $Y$ be a $T_1$ topological space, and $y \in Y$ a non-open point. Let $X = Y \sqcup \{ x \}$, endowed with the topology whose closed sets are $\varnothing,\{y\}$, and all $F \sqcup \{ x \}$, where $F$ is a closed subset of $Y$. Then $Z = \{ x,y\}$ is a closed subset of $X$, which satisfies the assumptions of Lemma $50.73.8$. But $X$ is connected, while $Z$ isn't. The conclusion of the lemma thus fails for the constant sheaf $\mathcal{F} = \{ 0,1 \}_X$.

Comment #1925 by on

@#1923: OK, thanks very much. Seems quite wrong in fact. I will try to fix it with additional assumptions on the space, but I am not hopeful. More tomorrow.

Comment #1926 by on

@#1925. Actually, I tried a bit more and I have come to the conclusion that the lemma should be OK for spectral spaces. Here are some details.

As a sanity check I will first prove that if $Z$ is disconnected, then $X$ is disconnected. Namely, suppose that $Z = Z_1 \amalg Z_2$ with $Z_1$ and $Z_2$ closed and nonempty. Let $W_i \subset X$ be the set of points specializing to $Z_i$. Then $X = W_1 \cup W_2$ by the assumption that every point of $X$ specializes to a point of $Z$. We will use Lemma 5.24.7 which shows that $W_i$ is the intersection of the quasi-compact opens containing $Z_i$ and hence closed in the constructible topology. Then the closure of $W_i$ consists of specializations of points of $W_i$ by Lemma 5.23.6. But the assumption that specializations of points of $x$ always meet $Z$ in connected sets implies that $W_1 \cap \overline{W_2} = \emptyset$ and vice versa. Thus $X = W_1 \amalg W_2$ is a disjoint union decomposition into closed subsets.

Next, as an exercise I proved it when $X$ is a finite sober space. In this case the argument works for the following silly reason: if $x \in X$ is a point then the set $E = \{x' \in X \mid x' \leadsto x\}$ is open. Thus it is clear that the construction of the family $(s_x)$ in the proof of the lemma does satisfy the continuity property because the stalk of $\mathcal{F}$ at $x$ is $\mathcal{F}(E)$.

For the general case the idea is to reduce to the case of a finite sober space. First we note that it is clear that the map $\Gamma(X, \mathcal{F}) \to \Gamma(Z, \mathcal{F}|_Z)$ is injective. Surjectivity is the difficult part.

Suppose that $s \in \Gamma(Z, \mathcal{F}|_Z)$. Then we can find finitely many quasi-compact opens $U_i$ and $s_i \in \mathcal{F}(U_i)$ such that $s|_{Z \cap U_i} = s_i|_{Z \cap U_i}$. For each pair $i, j$ we can choose a quasi-compact open $W_{i, j}$ with $Z \cap U_i \cap U_j \subset W_{i, j} \subset U_i \cap U_j$ with $s_i|_{W_{i, j}} = s_j|_{W_{i, j}}$. Then we see that $s$ comes from a section of where $h_U$ denotes the extension by the empty set of the singleton sheaf on $U$. This sheaf $\mathcal{G}$ is an example of what I would call a constructible sheaf of sets on $X$.

Then there are two ways to finish the proof. Method I: write the sheaf $\mathcal{G}$ as a pullback of a sheaf from a finite sober space (using the structure theorem for spectral spaces being filtered limits of finite sober spaces) and use the result for finite spaces. This doesn't quite work immediately and you have to reformulate the lemma a bit to make it work. Method II: you can embed $\mathcal{G}$ in a finite product of pushforwards $i_*S$ where $i : T \to X$ is the embedding of a constructible closed subset and $S$ indicates the constant sheaf with value $S$ on $T$. Since the pair $(T, Z \cap T)$ satisfies the assumptions of the lemma the result for this pair will then follow from the sanity check performed at the beginning of this comment. Finally, you show that if $\mathcal{G} \subset \mathcal{G}'$ and the image of $s$ in $H^0(Z, \mathcal{G}'|_Z)$ comes from a section of $\mathcal{G}'$ over $X$, then it works for $s$. To see this you can go back to the description in terms of $s_x$ and deduce the continuity from the fact that we know continuity after mapping into $\mathcal{G}'$.

So, I think this saves the thing... I will update the proof and statwment of the lemma soon! Thanks again!

Comment #1927 by on

@#1923: OK, I fixed the statement of the lemma and will update the proof hopefully later today. See here.

Comment #1928 by on

@#1923: New version online now. The fix exactly as predicted.

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