Continuity of subanalytic maps

In the Archimedean setting, Osgood’s theorem asserts that if Ω is an open subset of Cⁿ and f : Ω → Cⁿ an injective holomorphic map, then f(Ω) is open and f induces a biholomorphic map between Ω and f(Ω). See [Nar71, th.5 p.86] or [Ran86, Th. 2.14].

3.1. Dimension of subanalytic sets 143 Proposition 3.1.33. Let K be a complete algebraically closed non-Archimedean field.

Let f : X → Y be a morphism of K-analytic manifolds of the same pure dimension d such that f is injective. Then f(X) is open in Y and the restriction X −−→^f|X f(X) is a homeomorphism.

This proposition is false ifKis not algebraically closed. In characteristicpfor instance, the Frobenius map x→ x^p is an analytic endomorphism of K, which is precisely open if and only ifK is perfect.

Proof. Step 1. It is equivalent to prove thatf is open because then, the mapf_|X : (K^◦)ⁿ→ im(f) is bijective, continuous and open, hence a homeomorphism.

Step 2. To prove this we can work locally, hence restrict to an injective analytic map f : (K^◦)ⁿ→(K^◦)ⁿ.

Step 3. For allx ∈(K^◦)ⁿ, and y=f(x) we must show that for all neighbourhoodV ofx, there existsW ⊂V another open neighbourhood of x such thatf(W) is open (this would show thatf is open).

Since closed balls aroundxform a basis of neighbourhood ofx, it is sufficient to prove that if f : (K^◦)ⁿ → (K^◦)ⁿ is an injective analytic map and x ∈ (K^◦)ⁿ, there exists an open neighbourhoodV ofx such thatf(V) is open.

Step 4. By definition of an analytic map, reducing⁴ the source and the target of f again, we can assume thatf : (K^◦)ⁿ→(K^◦)ⁿ is given by n analytic functions. Hence, f is the restriction to the set of rigid points (here theK-points) of a morphism ofk-affinoid spacesf_Berko:Bⁿ_K →Bⁿ_K.

Step 5. Since the restriction of f_Berko to the set of rigid points is f, hence injective, and since x ∈(Bⁿ_K)_rig by assumption, dim_x(fBerko) = 0. Hence according to [Duc07, Th 3.2], there exists an analytic neighbourhoodV ofx inBⁿ_K such thatf_Berko factorizes as

V −→^α W −→^β T −→^γ BⁿK

whereα is finite, β is the immersion of an analytic domain of T and γ is étale.

Step 6. If z := β(α(x)), since it is a rigid point, γ is a local homeomorphism at z.

Hence if we replaceBⁿ by a small enough ball containingz, we can assume that T =Bⁿ. Hencef_Berko factorizes now as

V −→^α W −→^β Bⁿ withα finite andβ the immersion of an analytic domain.

Step 7. Now if we replace (in the target) Bⁿ_K by a ball small enough that contains y and is included inW, we can even assume thatf_Berko factorizes as

V −→^α Bⁿ_K withα finite.

Step 8. Now remember thatαis still injective at the level of the rigid points, and that V_rig is a neighbourhood of x in (K^◦)ⁿ. Since α is finite, α(V) is a Zariski-closed subset of Bⁿ_K. On the other hand, according to proposition 3.1.22 (ii), since dim(V) = n and sinceαrig is injective, dim(α(V)) = n. Ifα(V) was a proper Zariski-closed subset of Bⁿ_K, we would have dim(α(V)) < n according to lemma 3.1.27 which is impossible. Hence α(V) =Bⁿ_K which is a neighbourhood of y.

4. If we were onC, we would not need to do that.

Proposition 3.1.34. Let f : X → Y be a subanalytic map between subanalytic sets X⊂ (K^alg)^◦nandY ⊂ (K^alg)^◦m. Then there exists a decompositionX=X1∪. . .∪X_N in subanalytic embedded K^alg-analytic manifolds such that for all i f_|Xi is continuous.

Proof. We prove this by induction on dim(X).

If dim(X) = 0,X is finite (proposition 3.1.16 (1)) and the result is clear.

If d = dim(X) > 0, we take a decomposition X = X1 ∪. . .∪XN in subanalytic embeddedK^alg-analytic manifolds (theorem 3.1.30). We can then replacef by the various f_|Xi :X_i →Y. Since dim(X_i)≤dim(X) we can replaceXbyX_iand hence assume thatX is a subanalytic embeddedK^alg-analytic manifold (if it happens that dim(X_i) <dim(X) we can already apply the induction hypothesis). We then have the following commutative diagram:

G= Graph(f)

p1

xx ∼

p2

&&

X f //Y

where G is a subanalytic set of ((K^alg)^◦)^n+m. We apply again the smooth stratification theorem toGand obtain a decompositionG=G1∪. . .∪G_N where eachGiis a subanalytic embedded K^alg-analytic manifold. Since p₁ : G → X is a subanalytic map which is bijective, thanks to proposition 3.1.22, dim(G) = dim(X). Hence, dim(G_i)≤dim(X) for alli. Let us set Xi :=p1(Gi).

If dim(G_i) < dim(X), then dim(X_i) < dim(X) so by induction hypothesis, we can again stratifyXi inX_i,j⁰ ssuch thatf_|Xi,j is continuous.

If dim(Xi) = dim(X), then p1|G_i : Gi → X is an injective subanalytic map. Hence (p1)_|Gi)

Kd^alg : Gi(K^dalg) → X(K^dalg) is an injective analytic map between K^dalg-analytic manifolds of the same dimension. Hence according to proposition 3.1.33, it induces a homeomorphism betweenGi(K^dalg) and Xi(K^dalg). Hence p1|G_i also induces a homeomor-phism betweenGi and Xi. Eventually,f|X_i =p2◦(p1|G_i)⁻¹ is then continuous.

Corollary 3.1.35. Let f : X → Y be a subanalytic map between subanalytic sets X ⊂ (K^alg)^◦n and Y ⊂ (K^alg)^◦m. Let

D:={x∈X f is not continuous at x}.

ThenD is a subanalytic set and

dim(D)<dim(X).

Proof. First D is subanalytic because continuity at x can be expressed by a first order formula.

Then, let us assume that dim(D) = dim(X). According to corollary 3.1.25, D has nonempty interior inX, so replacingX by

◦

D (which is subanalytic), we can assume that D=X, i.e. that f is nowhere continuous.

But according to proposition 3.1.34, there is a decomposition in subanalytic setsX = X1∪. . .∪Xn such that for all i, f|X_i is continuous. There must exist some i such that dim(Xi) = dim(X), so according to corollary 3.1.25 again, Xi has nonempty interior in X, so if x ∈ X^◦i, f is continuous at x which contradicts the fact that f is nowhere continuous.

3.1. Dimension of subanalytic sets 145 Proposition 3.1.36. Let X ⊂ (K^alg)^◦m×Γⁿ be a subanalytic set, and f : X → Γ a subanalytic map. Then there exists a decompositionX=X1∪. . .∪XN in subanalytic sets such that for alli f_|Xi :X_i →Γ is continuous.

Actually, it is even defined as

f_|Xi = ^ai q

|g_i(x)|α^ui

wheregi is a continuous subanalytic mapUi →K^algfor some subanalyticUi ⊂ (K^alg)^◦m, and u_i ∈ Zⁿ, a_i ∈ N^∗. We denote by x = (x₁, . . . , x_m) ∈ (K^alg)^◦m the field variables, and by α= (α₁, . . . , α_n)∈Γⁿ the value group variables.

Proof. Let G:= Graph(f) ⊂ (K^alg)^◦m×Γⁿ×Γ which is subanalytic. We will use the variablex(resp. α,γ) for the variable of (K^alg)^◦m(resp. Γⁿ, Γ). It is enough to consider the case whereGis a finite intersection of subsets defined by inequalities

|F(x)|γ^a./|G(x)|α^u

wherea∈ N,u ∈Zⁿ,./∈ {=, >, <} and F, G: (K^alg)^◦m → K^alg are D-functions. We can then restrict to subanalytic sets where F(x) and G(x) 6= 0, hence their quotient ^F_G will be D-functions (K^alg)^◦m → K^alg. Hence G appears as a finite union of subsets of the form

I_j ={(x, α, γ)∈ (K^alg)^◦m×Γⁿ×Γ |g_j(x)|α^uj < γ^bj <|h_j(x)|α^vj}j = 1. . . M and

{(x, α, γ)∈ (K^alg)^◦m×Γⁿ×Γ |f_i(x)|α^ui =γ^ai} i= 1. . . N plus some additional conditions involving only the variables (x, α).

Since by assumption, for all (x, α)∈X,G_(x,α)={γ ∈Γ(x, α, γ)∈G}is a singleton, we can in fact remove the intervalsI_j. We then set temporarily

Xi={(x, α)∈X f(x, α) = ^ai q

|f_i(x)|α^ui}.

We then have

f_|Xi(x, α) = ^aiq|f_i(x)|α^ui

with f_i : (K^alg)^◦n → K^alg an D-function. Now if we apply proposition 3.1.34 to the mapf_i : (K^alg)^◦n→K^alg we know that we can shrink theX_i⁰sso that the map

Xi → K^alg (x, α) 7→ fi(x) is continuous. As a consequence,

X_i → Γ

(x, α) 7→ |f_i(x)|

is continuous, and then

f|X_i : Xi → Γ (x, α) 7→ ^aip|f_i(x)|α^ui is also continuous.