Proof. Clearly, K is a Jacobson–Tate ring in the sense of [Reference Lourenço13, Def. 3.1]; hence, X is a Jacobson adic space by [Reference Lourenço13, Prop. 3.3(1)]. Thus, (ii), (iii), and the density of $\operatorname {\mathrm {JG}}(X) \subset X$ follow from [Reference Lourenço13, Prop. 3.3(2) and (3)]. The claim $\operatorname {\mathrm {JG}}(U) = \operatorname {\mathrm {JG}}(X) \mathbin {\cap } U$ reduces easily to the case that X is affinoid and U is a rational open subset. Then the claim follows from (ii) and [Reference Lourenço13, Prop. 3.3(2)]. The corresponding statement for Z follows easily by reducing to the affinoid case and then using (ii) and [16, Lem. 00G9].

It remains to prove (iv). Clearly, if $\operatorname {\mathrm {JG}}(X)$ is connected, then so is X. Conversely, assume that $\operatorname {\mathrm {JG}}(X) = U_1' \mathbin {\sqcup } U_2'$ for some open subsets $U_1', U_2' \subset \operatorname {\mathrm {JG}}(X)$ . For $i = 1, 2$ , let $U_i$ be the union of all open subsets $U \subset X$ with $\operatorname {\mathrm {JG}}(U) \subset U_i'$ . Clearly, $U_1$ and $U_2$ are disjoint open subsets of X, so we only have to check that they cover X in order to finish the proof. Let $x \in X$ be given, and let $U = \operatorname {\mathrm {Spa}}(A, A^+)$ be an affinoid open neighborhood of x. Assume that U is not contained in $U_1$ or $U_2$ . Then $U_1'$ and $U_2'$ produce a disconnection of $\mathrm {MaxSpec}\, A$ , and since A is a Jacobson ring (by [Reference Lourenço13, Prop. 3.3(3)]), this produces a disconnection of $\mathrm {Spec}\, A$ and hence of U. (This also proves the second part of the claim.) Then one part of this disconnection is contained in $U_1$ , and the other part is contained in $U_2$ , so that x lies in $U_1$ or $U_2$ .

Proof. Let $U = \operatorname {\mathrm {Spa}}(A, A^+) \mathrel {\hookrightarrow } X$ be an open subspace. Then A is an excellent ring by [Reference Kiehl12, Th. 3.3]. Thus, by [16, Lems. 0FIZ, 0C23, 07QV, and 07NZ], the ring A is normal (resp. reduced) if and only if for all maximal ideals $\mathfrak {m} \subset A$ , the completed localization $\hat A_{\mathfrak {m}}$ is normal (resp. reduced). Now, the claim follows easily from Lemma 2.2.

Proof. As in the proof of Lemma 2.4, A is an excellent ring and in particular a Nagata ring (cf. [16, Lem. 07QV]). We can, moreover, replace A by $A_{\mathrm {red}}$ and thus assume that A is reduced. We want to apply [Reference Conrad3, Th. 1.2.2], for which we need to verify the following three properties:

1. (a) B is flat over A: this is [Reference Huber7, Lem. 1.7.6].

2. (b) The ring ${\tilde A} \otimes _A B$ is normal: to see this, let $\tilde X := \operatorname {\mathrm {Spa}}({\tilde A}, {\tilde A}^+)$ and $\tilde U := \operatorname {\mathrm {Spa}}({\tilde A} \otimes _A B, B^{\prime +})$ , where ${\tilde A}^+$ is the integral closure of $A^+$ in ${\tilde A}$ and $B^{\prime +}$ is the integral closure of $B^+$ inside ${\tilde A} \otimes _A B$ . (Note that $A \to {\tilde A}$ is finite because A is Nagata [see [16, Lem. 035S]].) Then $\tilde U = U \times _X \tilde X$ is an open subset of $\tilde X$ . However, $\tilde X$ is normal because ${\tilde A}$ is normal, hence $\tilde U$ and therefore also ${\tilde A} \otimes _A B$ must be normal by Lemma 2.4.

3. (c) For every minimal prime $\mathfrak p \subset A$ , $B/\mathfrak p B$ is reduced. This follows by a similar argument as in (b): letting $X' = \operatorname {\mathrm {Spa}}(A/\mathfrak p, A^{\prime +})$ and $U' = \operatorname {\mathrm {Spa}}(B/\mathfrak p B, B^{\prime +})$ (where $A^{\prime +}$ and $B^{\prime +}$ are the integral closures of $A^+$ and $B^+$ ), we have $U' = U \times _X X'$ , which is an open subset of $X'$ , so that the reducedness of $X'$ implies the reducedness of $U'$ by Lemma 2.4.

This finishes the proof.

Proof. It follows directly from the definition that $\tilde X$ is a normal adic space. The finiteness of the map $\tilde X \to X$ follows from [16, Lem. 035S] using that all coordinate rings are Nagata as in previous proofs. It remains to show that the map $\tilde X \to X$ is surjective. It is enough to show that $\operatorname {\mathrm {JG}}(\tilde X) \to \operatorname {\mathrm {JG}}(X)$ is surjective because $\tilde X \to X$ is a closed map and $\operatorname {\mathrm {JG}}(X) \subset X$ is dense. This reduces to the affinoid case and then to the analogous result for (affine) Noetherian schemes, which is [16, Lem. 035Q(2)].

Proof. In all cases, we can assume that $X = \operatorname {\mathrm {Spa}}(A, A^+)$ is affinoid.

We start with (i). By [Reference Huber7, Lem. 1.8.6], we have $\dim X = \dim A$ , where $\dim A$ denotes the Krull dimension of X. On the one hand, we have $\dim A = \sup_{\mathfrak {m} \in \mathrm {MaxSpec}\, A} \dim A_{\mathfrak {m}}$ , and on the other hand, $\dim A_{\mathfrak {m}} = \dim \hat A_{\mathfrak {m}} = \dim \hat {\mathcal {O}}_{X,x}$ for all $\mathfrak {m}$ (see [16, Lem. 07NV]), where $x \in \operatorname {\mathrm {JG}}(X)$ is the point of X associated with $\mathfrak {m}$ . This proves (i).

To prove (ii), let $x \in \operatorname {\mathrm {JG}}(X)$ be given and let $\mathfrak {m} = \operatorname {\mathrm {supp}}(x) \in \mathrm {MaxSpec}\, A$ be the corresponding maximal ideal in A. Let $I = \{ \mathfrak p_1, \dots , \mathfrak p_n \}$ be the collection of minimal prime ideals in A, and let $I_{\mathfrak {m}} \subset I$ be the subset of minimal prime ideals that lie inside $\mathfrak {m}$ . For every $\mathfrak p \in I$ , let $Z_{\mathfrak p} = \operatorname {\mathrm {Spa}}(A/\mathfrak p, A^{\mathfrak p+})$ (where $A^{\mathfrak p+}$ is the integral closure of $A^+$ in $A/\mathfrak p$ ), which is a closed adic subspace of X (cf. [Reference Huber7, (1.4.1)]). Let $U := X \setminus \bigcup _{\mathfrak p \in I \setminus I_{\mathfrak {m}}} Z_{\mathfrak p}$ .

Now, choose any $u \in \operatorname {\mathrm {JG}}(U)$ . We claim that $\dim _X u \le \dim _X x$ . To see this, let $\mathfrak n \subset A$ be the maximal ideal corresponding to u, so that $\dim _X u = \dim A_{\mathfrak n}$ (using [16, Lem. 07NV]). By the definition of U, $\mathfrak n = \operatorname {\mathrm {supp}} u$ cannot contain any $\mathfrak p \in I \setminus I_{\mathfrak {m}}$ , so that every maximal chain of prime ideals $\mathfrak p_i = \mathfrak q_0 \subsetneq \mathfrak q_1 \subsetneq \dots \subsetneq \mathfrak n$ has to start with some $\mathfrak p_i \in I_{\mathfrak {m}}$ . We may thus replace A by $A/\mathfrak p_i$ and assume that A is an integral domain. Then [Reference Conrad3, Lem. 2.1.5] implies that A is equidimensional and we are done.

Part (iii) follows easily from (i) and (ii).

Proof. If $X = \operatorname {\mathrm {Spa}}(A, A^+)$ is affinoid (so that A is normal), then A is an integral domain, so it is equidimensional by [Reference Conrad3, Lem. 2.1.5] and hence X is pure-dimensional by Lemma 2.8(i). To handle the general case, for every $d \in \mathbb {Z}_{\ge 0}$ , let $U_d \subset X$ be the union of all open subspaces which are of pure dimension d. From the affinoid case, one deduces that the $U_d$ ’s form a disjoint open partition of X. As X is connected, all but one of the $U_d$ ’s must be empty.

Proof. The claim $\dim Y \le \dim X$ follows immediately from the analogous statement for schemes (see [16, Lem. 0ECG]) and [Reference Huber7, Lem. 1.8.6(ii)]. Similarly, the second claim reduces to the statement that if $A \to B$ is a finite morphism of rings such that $\mathrm {Spec}\, B \to \mathrm {Spec}\, A$ is surjective and $\mathfrak {m} \subset A$ is a maximal ideal, then $\dim A_{\mathfrak {m}} = \dim (A_{\mathfrak {m}} \otimes _A B)$ (see, e.g., [16, Lem. 0ECG]).

Proof. It is enough to prove the following: let $Z \subset X$ be a closed adic subspace and suppose that Z contains some open subspace $U \subset X$ ; then $Z = X$ . We can assume that both Z and U are connected. By Corollary 2.9, we know that X is pure-dimensional, say $\dim X =: d$ . Now, let

$$ \begin{align*} Z^{\prime}_1 := \{ x \in \operatorname{\mathrm{JG}}(Z) \mathrel{|} \dim_Z x = d \}, \qquad Z^{\prime}_2 := \{ x \in \operatorname{\mathrm{JG}}(Z) \mathrel{|} \dim_Z x < d \}. \end{align*} $$

We claim that if $V \subset X$ is any affinoid connected open subset with $V \mathbin {\cap } Z^{\prime }_1 \ne \varnothing $ , then $V \subset Z$ . Indeed, let $V = \operatorname {\mathrm {Spa}}(B, B^+)$ , so that $Z \mathbin {\cap } V = \operatorname {\mathrm {Spa}}(B/J, B^{\prime +})$ for some ideal $J \subset B$ (where $B^{\prime +}$ is the integral closure of $B^+$ in $B/J$ ). But then, $\mathrm {Spec}\, B$ is normal and connected (cf. Lemmas 2.2(iv) and 2.4), hence irreducible, and there is some maximal ideal $\mathfrak {m} \subset B/J$ (the one corresponding to x) such that $\dim B = d = \dim (B/J)_{\mathfrak {m}}$ . This implies that $\dim B/J = \dim B$ and hence $J = (0)$ , as desired.

From the previous paragraph, we deduce that $Z^{\prime }_1$ is open in $\operatorname {\mathrm {JG}}(Z)$ and by Lemma 2.8(ii) the same is true for $Z^{\prime }_2$ . However, $\operatorname {\mathrm {JG}}(Z)$ is connected by Lemma 2.2(iv) so that $Z^{\prime }_2 = \varnothing $ because $U \subset Z$ . Hence, $\operatorname {\mathrm {JG}}(Z) = Z^{\prime }_1$ and now the previous paragraph shows that Z is open in X. But then, Z is open and closed in X and thus $Z = X$ .

Proof. We can assume that $X = \operatorname {\mathrm {Spa}}(A, A^+)$ is affinoid; then $Y = \operatorname {\mathrm {Spa}}(B, B^+)$ is also affinoid, f is given by a finite map $f^*\colon A \to B$ , and $B^+$ is the integral closure of $A^+$ in B. Let $J := \ker (f^*)$ , and let $Z = \operatorname {\mathrm {Spa}}(A/J, A^{\prime +})$ , where $A^{\prime +}$ is the integral closure of $A^+$ in $A/J$ . We claim that $\operatorname {\mathrm {im}}(f) = Z$ . Clearly, $\operatorname {\mathrm {im}}(f) \subset Z$ and $\operatorname {\mathrm {im}}(f)$ is closed (because f is proper, e.g., by [Reference Huber7, Lem. 1.4.5(ii)]), so it is enough to show that $\operatorname {\mathrm {im}}(f)$ contains all $x \in \operatorname {\mathrm {JG}}(Z)$ . Using Lemma 2.2, this boils down to the fact that $f^*$ induces a surjective map $\mathrm {MaxSpec}\, B \mathrel {\twoheadrightarrow } \mathrm {MaxSpec}\,(A/J)$ , which is clear.

Proof. Let $\rho \colon \tilde X \to X$ be the normalization of X with connected components $(\tilde X_i)_{i\in I}$ . Each $\tilde X_i$ is irreducible by Lemma 2.12 and maps surjectively onto $X_i$ . It follows easily that $X_i$ is irreducible. (Any $Z_1 \mathbin {\cup } Z_2 \subset X_i$ can be lifted to $(Z_1 \times _{X_i} \tilde X_i) \mathbin {\cup } (Z_2 \times _{X_i} \tilde X_i) = \tilde X$ .) The local finiteness of the cover follows from the fact that for every quasi-compact subset $U \subset X$ , the preimage $\rho ^{-1}(U) \subset \tilde X$ is still quasi-compact and can therefore only meet finitely many connected components.

To prove the second part of the claim, let $i \in I$ be given and let $U := \tilde X_i$ and $V := \tilde X \setminus \tilde X_i$ . It is enough to show that $\rho ^{-1}(\rho (V))$ does not contain U. In fact, we claim that the closed adic subspace $\rho ^{-1}(\rho (V)) \mathbin {\cap } U \subset U$ has strictly lower dimension than U at all points. This can be checked locally, so we can assume that X and $\tilde X = U \mathbin {\sqcup } V$ are affinoid. Then the claim follows easily from the analogous statement for schemes.

Proof. The equivalence of (ii) and (iii) is Lemma 2.12. As in the proof of Lemma 2.15, if $\tilde X$ is irreducible, then so is X. Conversely, assume that $\tilde X$ is reducible, that is, disconnected. Then X has more than one irreducible component and is therefore reducible by Lemma 2.15.

Proof. Let $\rho \colon \tilde X \to X$ be the normalization of X. Then $\tilde X$ is connected by Corollary 2.16, hence pure-dimensional by Corollary 2.9. However, by Lemma 2.10, for every $x \in \operatorname {\mathrm {JG}}(X)$ , there is some $\tilde x \in \operatorname {\mathrm {JG}}(\tilde X)$ with $\dim _X x = \dim _{\tilde X} \tilde x = \dim \tilde X$ .

Proof. Since all of the discussed properties are local (see Lemma 2.4 for normality and reducedness), we can use [Reference Huber7, Lem. 2.2.8] to reduce the claim to $X = \operatorname {\mathrm {Spa}}(A, A^+)$ and $Y = \operatorname {\mathrm {Spa}}(B, B^+)$ affinoid with $A \to B$ finite étale. Then (i) and (iii) follow directly from the analogous property for schemes (see, e.g., [16, Lems. 033B, 033C, and 039S]). Moreover, (ii) follows from (i) and [Reference Conrad3, Th. 1.2.2].

Proof. The argument is similar to [Reference Huber7, Lem. 1.7.9]. As in the reference, we define

$$ \begin{align*} X' &:= \{ v \in \operatorname{\mathrm{Spv}} A \mathrel{|} v(a) \le 1 \text{ for } a \in A^+ \text{ and } v(a) < 1 \text{ for } a \in A^{\circ\circ} \},\\ Y' &:= \{ v \in \operatorname{\mathrm{Spv}} B \mathrel{|} v(b) \le 1 \text{ for } b \in B^+ \text{ and } v(b) < 1 \text{ for } b \in B^{\circ\circ} \}. \end{align*} $$

Moreover, by [Reference Huber5, Prop. 2.6(iii)], there are retractions $r_X\colon \operatorname {\mathrm {Spv}} A \to \operatorname {\mathrm {Spv}}(A, A^{\circ \circ } A)$ and $r_Y\colon \operatorname {\mathrm {Spv}} B \to \operatorname {\mathrm {Spv}}(B, B^{\circ \circ } B)$ . From [Reference Huber5, Th. 3.1], we deduce that $X' \mathbin {\cap } \operatorname {\mathrm {Spv}}(A, A^{\circ \circ } A) = X$ and $Y' \mathbin {\cap } \operatorname {\mathrm {Spv}}(B, B^{\circ \circ } B) = Y$ . It follows easily that, by restricting $r_X$ and $r_Y$ , we obtain retractions $s_X\colon X' \to X$ and $s_Y\colon Y' \to Y$ . Let $f'\colon Y' \to X'$ be the restriction of the natural map $\operatorname {\mathrm {Spv}}(f^*)\colon \operatorname {\mathrm {Spv}} B \to \operatorname {\mathrm {Spv}} A$ . We obtain the following commutative diagram:

By [Reference Huber and Knebusch8, Cor. 2.1.7(iii)], the assumptions on A, B, and $f^*\colon A \to B$ are enough to guarantee that $\operatorname {\mathrm {Spv}}(f^*)\colon \operatorname {\mathrm {Spv}} B \to \operatorname {\mathrm {Spv}} A$ is open. Note also that $B^+$ is the integral closure of $A^+$ in B and $B^{\circ \circ } = A^{\circ \circ } B^+$ by finiteness of f (see [Reference Huber7, (1.4.2)]). This implies that

$$ \begin{align*} Y' &= \{ v \in \operatorname{\mathrm{Spv}} B \mathrel{|} v(b) \le 1 \text{ for } b \in A^+ \text{ and } v(b) < 1 \text{ for } b \in A^{\circ\circ} \}, \end{align*} $$

and hence $f^{\prime -1}(X') = Y'$ . Together with the openness of $\operatorname {\mathrm {Spv}}(f^*)$ , we deduce that $f'$ is open. But then, the above diagram immediately implies that f is open as well.

Proof. Since the claim is local on X, we can assume that $X = \operatorname {\mathrm {Spa}}(A, A^+)$ is affinoid and irreducible. Letting $(Y_i)_{i\in I}$ be the irreducible components of Y, it is enough to show that each $Y_i \to X$ is open; we can thus assume that Y is irreducible as well. Since f is finite and X affinoid, Y must be affinoid as well, say $Y = \operatorname {\mathrm {Spa}}(B, B^+)$ . We can furthermore assume that A and B are reduced, because passing to the reduction does not change the topology. Then A and B are integral domains of dimension d, the induced map $f^*\colon A \to B$ is finite, and A is normal. In order to apply Lemma 2.19, it only remains to verify that $f^*\colon A \to B$ is injective. However, this is clear, as otherwise $f^*$ factors over $A/I$ for some ideal $0 \ne I \subset A$ , but $A/I$ has lower dimension than A and the dimension of B is at most the one of $A/I$ ; contradiction!

Proof. The claim is local on X, so we can assume that $X = \operatorname {\mathrm {Spa}}(A, A^+)$ and $U = \operatorname {\mathrm {Spa}}(A, A^{\prime +})$ for some $A^{\prime +} \supset A^+$ . Then, clearly, $\mathcal {O}_X(X) = A = \mathcal {O}_U(U)$ , and using the same reasoning for every rational open subset of X shows $j_*\mathcal {O}_U = \mathcal {O}_X$ . On the other hand, by [Reference Huber4, Satz 3.3.18] (or [Reference Kedlaya and Liu10, Th. 2.3.3]), the categories $\mathrm {Coh}(\mathcal {O}_U)$ and $\mathrm {Coh}(\mathcal {O}_X)$ are both equivalent to the category of finite A-modules, and using $j_*\mathcal {O}_U = \mathcal {O}_X$ , one checks easily that $j_*$ provides the required equivalence.

Proof. The claim is local on X, so we can assume that $X = \operatorname {\mathrm {Spa}}(A, A^+)$ is affinoid. Let us first consider the case $A^+ = A^{\circ }$ . Then X is quasi-separated and of finite type over K, and hence so is Y. By [Reference Huber6, Prop. 4.5(iv)], both X and Y are induced from rigid-analytic varieties and the same is true for all rational subsets. Thus, the claim follows from the analogous claim in rigid-analytic geometry (see [Reference Kiehl11, Th. 3.3]).

Now, let $A^+$ be general, and let $X' := \operatorname {\mathrm {Spa}}(A, A^{\circ })$ . Then we have an open immersion $j_1\colon X' \to X$ . Let $Y' := Y \times _X X'$ , so that we also have an open immersion $j_2\colon Y' \to Y$ . Letting $f'\colon Y' \to X'$ denote the restriction of f, we get the diagram

Both $j_1$ and $j_2$ satisfy the hypothesis of Lemma 3.1, as one checks easily using the explicit construction of fiber products of adic spaces. Thus, the claim follows from Lemma 3.1 and the fact that it holds for $f'$ by the first paragraph of the proof.

Proof. We first note that whether f has connected fibers or not can be checked on the rank-1 points of X: if $x \in X$ is any point, let $x_0 \in X$ be the maximal generalization of X. Then there is a natural open immersion $Y_{x_0} \mathrel {\hookrightarrow } Y_x$ and each point of $Y_x$ has a generalization in $Y_{x_0}$ . Thus, every nontrivial disjoint open cover of $Y_x$ produces a nontrivial disjoint open cover of $Y_{x_0}$ ; therefore, if $Y_{x_0}$ is connected, then so is $Y_x$ . We immediately deduce that (ii) implies (iii) (cf. [Reference Huber7, Prop. 1.7.5]).

We next show that (iii) implies (ii): given $x'\colon \operatorname {\mathrm {Spa}}(k, \mathcal {O}_k) \to X$ with image $x \in X$ , it is enough to show that there is an étale map $\alpha \colon U \to X$ with a point $u \in U$ such that $x' = \alpha \mathbin {\circ } u$ . To see this, we can assume that $X = \operatorname {\mathrm {Spa}}(A, A^+)$ is affinoid. Then $\hat k(x)$ is the completion of the quotient field $\kappa (\mathfrak p) = \mathrm {Quot}(A/\mathfrak p)$ , where $\mathfrak p = \operatorname {\mathrm {supp}} x$ (see [Reference Kedlaya and Liu9, Lem. 2.4.17(a)]). By Krasner’s lemma (cf. [Reference Bosch, Güntzer and Remmert2, Prop. 3.4.2/5]), there is a finite separable extension $\kappa '$ of $\kappa (\mathfrak p)$ such that $k = \kappa ' \otimes _{\kappa (\mathfrak p)} k(x)$ . By the proof of [16, Lem. 00UD], after possibly replacing A by the localization $A_f$ for some $f \in A$ (which corresponds to an open subspace in the adic world), there is a finiteo étale A-algebra $A'$ and a prime ideal $\mathfrak p' \subset A'$ lying above $\mathfrak{p}$ such that $\kappa (\mathfrak p') = \kappa '$ . Letting $U = \operatorname {\mathrm {Spa}}(A', A^{\prime +})$ , where $A^{\prime +}$ is the integral closure of $A^+$ in $A'$ , one sees easily that there is a canonical injection $u\colon \operatorname {\mathrm {Spa}}(k, \mathcal {O}_k) \mathrel {\hookrightarrow } U$ whose image point is the rank-1 valuation with support $\mathfrak p'$ .

It is clear that (i) implies (ii): if $Y_{x'}$ is disconnected for any $x'$ as in (ii), then $Y_{\overline x'} = Y_{x'} \times _{x'} \overline x'$ is also disconnected, where $\overline x'\colon \operatorname {\mathrm {Spa}}(\overline k, \mathcal {O}_{\overline k}) \to X$ is the obvious map which factors over $x'$ . It remains to show that (ii) implies (i). We can then assume that $X = \operatorname {\mathrm {Spa}}(k, \mathcal {O}_k)$ for some non-archimedean field k. One also easily reduces to the case that $Y = \operatorname {\mathrm {Spa}}(B, B^+)$ is affinoid. By (ii), for every finite separable extension $k'$ of k, the space $Y \times _X \operatorname {\mathrm {Spa}}(k', \mathcal {O}_{k'}) = \operatorname {\mathrm {Spa}}(B \otimes _k k', B^{\prime +})$ is connected (where $B^{\prime +}$ is the integral closure of $B^+$ in $B \otimes _k k'$ ). However, this means that $\mathrm {Spec}\,(B \otimes _k k')$ is connected (cf. Lemma 2.2(iv)), which overall implies that $\mathrm {Spec}\, B$ is geometrically connected over k (see [16, Lem. 0389]). Thus, given any algebraically closed non-archimedean field $\overline k$ over k, $\mathrm {Spec}\,(B \otimes _k \overline k)$ is connected. Then $Y \times _X \operatorname {\mathrm {Spa}}(\overline k, \mathcal {O}_{\overline k}) = \operatorname {\mathrm {Spa}}(B \otimes _k \overline k, B^{\prime +})$ is connected by Lemma 2.2(iv).

Proof of Lemma 3.5

The claim is local on $X'$ , and it clearly holds if g is an open immersion, so by using [Reference Huber7, Lem. 2.2.8], we can assume that $X = \operatorname {\mathrm {Spa}}(A, A^+)$ and $X' = \operatorname {\mathrm {Spa}}(A', A^{\prime +})$ are affinoid with $A \to A'$ being finite étale. By Proposition 3.2, the sheaf $f_* \mathcal M$ is coherent, and hence equals the coherent sheaf associated with the finite A-module $M_A := \Gamma (Y, \mathcal M)$ . Then $g^* f_* \mathcal M$ is the coherent sheaf on $X'$ , which is associated with the finite $A'$ -module $M_A \otimes _A A'$ (cf. [16, Lem. 01BJ]). Using a similar treatment for $f^{\prime }_* g^{\prime *} \mathcal M$ , we see that the claim boils down to showing

$$ \begin{align*} \Gamma(Y, \mathcal M) \otimes_A A' = \Gamma(Y', g^{\prime *} \mathcal M). \end{align*} $$

Again, by [16, Lem. 01BJ], we have $\Gamma (Y', g^{\prime *} \mathcal M) = \Gamma (Y, \mathcal M) \otimes _{\mathcal {O}_Y(Y)} \mathcal {O}_{Y'}(Y')$ , which reduces the claim to the case $\mathcal M = \mathcal {O}_Y$ . Choose a finite cover $Y = \bigcup _{i = 1}^n U_i$ with all $U_i = \operatorname {\mathrm {Spa}}(B_i, B_i^+)$ affinoid. For every $i = 1, \dots , n$ , let $U_i' := U_i \times _X X' = \operatorname {\mathrm {Spa}}(B_i \otimes _A A', B_i^{\prime +})$ , where $B_i^{\prime +}$ is the integral closure of $B_i^+$ in $B_i \otimes _A A'$ (cf. proof of [Reference Huber7, Prop. 1.2.2]). The sheaf property of $\mathcal {O}_Y$ and $\mathcal {O}_{Y'}$ provides exact sequences

$$ \begin{align*} 0 \to \mathcal{O}_Y(Y) \to \bigoplus_{i=1}^n \mathcal{O}_Y(U_i) \to \bigoplus_{i,j=1}^n \mathcal{O}_Y(U_i \mathbin{\cap} U_j),\\ 0 \to \mathcal{O}_{Y'}(Y') \to \bigoplus_{i=1}^n \mathcal{O}_{Y'}(U_i') \to \bigoplus_{i,j=1}^n \mathcal{O}_{Y'}(U_i' \mathbin{\cap} U_j'). \end{align*} $$

Note that $\mathcal {O}_{Y'}(U_i') = \mathcal {O}_Y(U_i) \otimes _A A'$ by the explicit description of $U_i$ and $U_i'$ . Moreover, since f is separated, all the intersections $U_i \mathbin {\cap } U_j$ are affinoid (by the same argument as in the case of schemes, using that fiber products and closed subspaces of affinoids are affinoid; for the latter, see [Reference Huber4, Prop. 3.6.27]), and hence satisfy the similar relation $\mathcal {O}_{Y'}(U_i' \mathbin {\cap } U_j') = \mathcal {O}_Y(U_i \mathbin {\cap } U_j) \otimes _A A'$ . Therefore, the second of the above exact sequences reads

$$ \begin{align*} 0 \to \mathcal{O}_{Y'}(Y') \to \bigoplus_{i=1}^n \mathcal{O}_Y(U_i) \otimes_A A' \to \bigoplus_{i,j=1}^n \mathcal{O}_Y(U_i \mathbin{\cap} U_j) \otimes_A A'. \end{align*} $$

Comparing this to the first exact sequence and noting that $A'$ is flat over A, we deduce that $\mathcal {O}_{Y'}(Y') = \mathcal {O}_Y(Y) \otimes _A A'$ , as desired.

Proof. In the following argument, we denote by ${\lvert Z \rvert }$ the topological space underlying an adic space Z. We also use the associated Berkovich spectrum of all the spaces involved: for any analytic adic space Z, let ${\lvert Z \rvert }^B$ be the subset of all maximal points of Z. There is a natural continuous projection ${\lvert Z \rvert } \mathrel {\twoheadrightarrow } {\lvert Z \rvert }^B$ by [Reference Huber7, Lem. 8.1.7(ii)], and if Z is taut (e.g., qcqs), then ${\lvert Z \rvert }^B$ is Hausdorff by [Reference Huber7, Lem. 8.1.8(ii)].

The claim is local on X, so we can assume that X is qcqs (e.g., affinoid); then Y is also qcqs, so in particular X and Y are taut. Clearly, ${\lvert {Y_x} \rvert }^B = {\lvert {Y_x} \rvert } \mathbin {\cap } {\lvert {Y} \rvert }^B$ and ${\lvert {Y_x} \rvert }^B = {\lvert {V_1} \rvert }^B \mathbin {\sqcup } {\lvert {V_2} \rvert }^B$ with ${\lvert {V_1} \rvert }^B, {\lvert {V_2} \rvert }^B \subset {\lvert {Y_x} \rvert }^B$ open subsets. By the discussion at the beginning of the proof, ${\lvert {Y} \rvert }^B$ , ${\lvert {V_1} \rvert }^B$ , and ${\lvert {V_2} \rvert }^B$ are quasi-compact Hausdorff spaces and in particular T4 spaces. Hence, there exist disjoint open neighborhoods $U_1"$ and $U_2"$ of ${\lvert {V_1} \rvert }^B$ and ${\lvert {V_2} \rvert }^B$ inside ${\lvert {Y} \rvert }^B$ . Let $U_1', U_2' \subset Y$ be their preimages under the projection ${\lvert Y \rvert } \to {\lvert {Y} \rvert }^B$ . Then $U_1'$ and $U_2'$ are disjoint open subsets of Y such that $V_i \subset U_i'$ for $i = 1, 2$ .

Let $Z := {\lvert Y \rvert } \setminus ({\lvert {U_1'} \rvert } \mathbin {\cup } {\lvert {U_2'} \rvert })$ , which is a closed subset of ${\lvert Y \rvert }$ . By the properness of f, $f(Z) \subset {\lvert X \rvert } $ is closed. Now, let $U := {\lvert X \rvert } \setminus f(Z)$ . Then U is an open neighborhood of x and $f^{-1}(U) = U_1 \mathbin {\sqcup } U_2$ for $U_i := f^{-1}(U) \mathbin {\cap } U_i'$ , as desired.

Proof. Let $U \to X$ be any étale morphism, and let $f_U\colon Y_U := Y \times _X U \to U$ be the base change. By Lemma 3.4, it is enough to show that $f_U$ has connected fibers. However, by Lemma 3.5, we have $f_{U*} \mathcal {O}_{Y_U} = \mathcal {O}_U$ , and hence we can replace U by X and reduce to showing that f has connected fibers. Let $x \in X$ be given and assume that $Y_x$ is disconnected. By Lemma 3.7, there is an open neighborhood $V \subset X$ of x such that $f^{-1}(V) = V_1 \mathbin {\sqcup } V_2$ , where both $V_1$ and $V_2$ are nonempty over x. In particular, there are idempotents $e_1, e_2 \in \mathcal {O}_Y(f^{-1}(V)) = (f_*\mathcal {O}_Y)(V) = \mathcal {O}_X(V)$ corresponding to $V_1$ and $V_2$ . These idempotents satisfy $e_1 = 1 - e_2$ and hence correspond to two disjoint subsets of X. On the other hand, both of these subsets must contain x because both $e_1$ and $e_2$ are nonzero in any neighborhood of x (since $V_1$ and $V_2$ are nonempty over x). Contradiction!

Proof. Given any coherent sheaf $\mathcal M$ on X, there is a unique adic space M with a (necessarily finite) map $m\colon M \to X$ such that for any affinoid $U \subset X$ with preimage $V = m^{-1}(U)$ , we have $\mathcal {O}_M(V) = \mathcal M(U)$ . Indeed, if X is affinoid, this is clear, and the general case is easily obtained by glueing. Now, by Proposition 3.2, the sheaf $f_*\mathcal {O}_Y$ is a coherent sheaf of $\mathcal {O}_X$ -modules on X. Applying the gluing construction to $\mathcal M = f_*\mathcal {O}_Y$ , we obtain a finite map $g\colon Z \to X$ with $g_*\mathcal {O}_{Z} = f_*\mathcal {O}_Y$ . From the construction of Z, one sees easily that there is a map $h\colon Y \to Z$ such that $f = g \mathbin {\circ } h$ and $h_*\mathcal {O}_Y = \mathcal {O}_Z$ . Then Proposition 3.8 implies that h has geometrically connected fibers.