UNIFORMLY RIGID SPACES 1. Introduction Let K be a non ...

28 CHRISTIAN KAPPENof uniformly rigid K-spaces has fibered products as well and that theseare constructed by glueing semi-affinoid fibered products of semi-affinoidsubspaces. It is clear from this description that the urig-functor preservesfibered products.Semi-affinoid subspaces of semi-affinoid spaces can be described in the styleof the Gerritzen-Grauert Theorem [5] 7.3.5/3:Lemma 2.47. Let X be a semi-affinoid K-space, and let U ⊆ X be a semiaffinoidsubspace. Then U admits a leaflike covering (U i ) i∈I such that eachU i is a semi-affinoid subdomain in X.Proof. By Lemma 2.46, U admits an admissible covering (V j ) j∈J by semiaffinoidsubdomains V j of X; by Proposition 2.33, this covering is refinedby a leaflike covering (U i ) i∈I of U. Via pullback, the V j are semi-affinoidsubdomains of U. Let ϕ: I → J denote a refinement map. By Corollary2.26 (i), for each i ∈ I the set U i is a semi-affinoid subdomain in V ϕ(i) and,hence, in X, as desired.□A morphism of uniformly rigid K-spaces is called flat in a point of itsdomain if it induces a flat homomorphisms of stalks in this point, and itis called flat if it is flat in all points. Clearly a morphism of semi-affinoidK-spaces is flat in the sense of locally G-ringed K-spaces if and only if theunderlying homomorphism of rings of global sections is flat.2.4.1. Comparison with rigid geometry. In Section 2.3.1, we have definedthe rigid space X r associated to a semi-affinoid K-space X = sSp A togetherwith a universal K-homomorphism A → Γ(X r , O X r) which inducesa bijection X r → X of physical points and isomorphisms of completedstalks. We will show that this universal homomorphism extends to a morphismcomp X : X r → X of locally G-ringed K-spaces which is final amongall morphisms from rigid K-spaces to X. To do so, we first show that theabove bijection is continuous, that is, that the rigid G-topology T rig is finerthan T urig . We will need the following elementary fact from rigid geometry;the proof is left as an exercise to the reader:Lemma 2.48. Let X be an affinoid K-space, and let U ⊆ X be a subsetadmitting a covering (U i ) i∈I by admissible open subsets U i ⊆ X such thatfor any affinoid K-space Y and any morphism ϕ: Y → X with image inU, the induced covering (ϕ −1 (U i )) i∈I of Y has a refinement which is a finitecovering by affinoid subdomains. Then U ⊆ X is admissible.Proposition 2.49. The rigid G-topology T rig on X is finer than the uniformlyrigid G-topology T urig .Proof. It is clear that T aux admissible subsets and T aux -admissible coveringsare T rig -admissible. Let U ⊆ X be a T urig -admissible subset. To check thatU is T rig -admissible, we may work locally on X r . Let V ′ ⊆ X r be an affinoid