The fibration method over real function fields

Let R(C)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb R(C)$$\end{document} be the function field of a smooth, irreducible projective curve over R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb R$$\end{document}. Let X be a smooth, projective, geometrically irreducible variety equipped with a dominant morphism f onto a smooth projective rational variety with a smooth generic fibre over R(C)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb R(C)$$\end{document}. Assume that the cohomological obstruction introduced by Colliot-Thélène is the only one to the local-global principle for rational points for the smooth fibres of f over R(C)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb R(C)$$\end{document}-valued points. Then we show that the same holds for X, too, by adopting the fibration method similarly to Harpaz–Wittenberg.


Introduction
Let C be a smooth, geometrically irreducible projective curve over R. Let R(C) denote the function field of C, and for every x ∈ C(R) let R(C) x be the completion of R(C) with respect to the valuation furnished by x. Now let V be a class of geometrically irreducible projective varieties over R(C). We say that V satisfies the local-global principle for rational points if for every X in V the following holds: is the residue map associated to the discrete valuation ring O X ,x .
Next we need some basic facts about the Galois cohomology of function fields of real algebraic curves, and some form of a residue theorem for them.

Definition 2.2
Let C be a smooth, geometrically irreducible projective curve over R.
Let R(C) denote the function field of C, as above, and for every x ∈ C(R) let R(C) x be the completion of R(C) with respect to the valuation furnished by x. Then we have a residue map as the residue field R(x) of R(C) x is R. Note that as a graded algebra: where t is the generator of the group H 1 et (R, Z/2) of order two. In particular we have a canonical isomorphism H ì et (R, Z/2) ∼ = Z/2. So the residue map is a homomorphism: By slight abuse of notation let ∂ x denote also the composition of the pull-back and this residue map.
The residue theorem, also called the reciprocity law, is the following

Proposition 2.3 Let V ⊆ C(R) be a connected component, let i be at least 2, and let h ∈ H ì et (R(C), Z/2). Then
x∈V ∂ x (h) = 0.

Remark 2.4
It is easy to see that all but finitely many terms of the sum above are zero, so the left hand side is well-defined. For a proof of this fact, and the proposition, see for example Proposition 3.7 of [3] on pages 157-158.
For the sake of simple notation let A C denote the direct product x∈C(R) R(C) x . It is an algebra over R(C). Now let X be a smooth, irreducible projective variety defined over R(C). Clearly Now we are ready to define Colliot-Thélène's obstruction. is the pull-back with respect to the map M x . Note that all but finitely many terms of the sum above are zero, so the left hand side is well-defined, and the image of X (R(C)) in X (A C ) under the diagonal embedding is in X (A C ) CT by Proposition 2.3 above.
One may justify the usage of the more mysterious group H i nr (R(C)(X )/R(C), Z/2) instead of H ì et (X , Z/2) as follows: for proper varieties having a smooth rational point is a birationally invariant property, so the obstruction should also have birational invariance. This holds for the former group, but not the latter. Now that we made explicit what we mean by the CT obstruction, we can make the following bold conjecture, which is motivated by theorems of Witt, Scheiderer and Ducros mentioned in the introduction, and in analogy with Colliot-Thélène's celebrated conjecture (see [4]) saying that the Brauer-Manin obstruction is the only one to the local-global principle for rational points for smooth projective rationally connected varieties over number fields: Conjecture 2. 6 The C T obstruction is the only one to the local-global principle for rational points for smooth projective rationally connected varieties over R(C).
All known classes of varieties for which CT obstruction is the only one to the local-global principle for rational points are rationally connected. Our main result Theorem 1.5 is a contribution to this conjecture.
We finish this section with a lemma which will be used in the proof of Theorem 5.3. Let f : X → Y be a morphism between smooth, projective, irreducible varieties over R(C). The morphism f induces a map X (R(C) x ) → Y (R(C) x ) for every x ∈ C(R), which in turn induces a map X (A C ) → Y (A C ), which we will denote also by f by slight abuse of notation.
for every x ∈ C(R) by naturality, so The claim is now clear.

The topological reinterpretation of the obstruction due to Ducros
Notation 3.1 By resolution of singularities there is an integral, smooth, projective variety X equipped with a projective dominant morphism p : X → C over R whose generic fibre is X → Spec(R(C)). As usual we will call X a model of X over C. For every closed point x of C let O x be the valuation ring of R(C) x , let X x denote the fibre of p over x, let X sm ⊆ X be the smooth locus of p, and let X x,sm = X sm ∩ X x be the smooth locus of X x .
let m x be the special fibre of the section Spec(O x ) → X × C Spec(O x ) associated to M x for every x ∈ C(R). Since X is regular, the point m x lies in X x,sm (R). Whenever convenient, we will denote the map x → m x on C(R) by σ (M).
The topological reformulation of the CT obstruction due to Ducros is the following Theorem 3.6 (Ducros) The following are equivalent: Proof This is Théorème 3.5 of [7] on page 83.

Proposition 3.7 Let σ be a weakly continuous section of p(R).
Then there is a continuous semi-algebraic section σ of p(R) such that for every x ∈ C(R) the points σ (x) and σ (x) are in the same connected component of X x,sm (R).

Remark 3.8
Note that Theorem 1.3 is an immediate corollary of this proposition.

Proof of Proposition 3.7
This is essentially Proposition 4.1 of [7] on page 85, but there it is stated in a weaker form. However the proof actually shows the stronger form above. We will give an even stronger version incorporating interpolation, see Proposition 3.19 below, using essentially the same methods. I(x) n x ⊂ O B . By slight abuse of notation we will let the symbol S denote this closed subscheme, too. When B is a curve this construction furnishes a bijective correspondence between the set of zero-dimensional closed subschemes of B whose closed points are all real and the set of effective zero cycles on B which are supported on B(R). In this case we will identify these two sets in all that follow.
We say that two maps f , g ∈ C k p (M, N ) are k-equivalent at p if f ( p) = g( p) and for every pair of maps γ ∈ C k 0 (R, M) and  Similarly an interpolation condition φ : S → A × B S is the same data as a morphism S → D of schemes over R, while C k -sections of f (R) can be identified with C k -maps A(R) → D(R). Therefore we will freely apply the concepts of Definitions 3.10 and 3.12 to such functions in all that follows. Proposition 3.14 Let σ be a weakly continuous section of p(R) and let φ : S → X × C S be an interpolation condition of order ≤ k weakly compatible with σ . Then there is a C k -section σ of p(R) such that for every x ∈ C(R) the points σ (x) and σ (x) are in the same connected component of X x,sm (R) and σ is compatible with φ.
Proof Note that the fibre of p(R) over x is a Nash manifold for all but finitely many x ∈ C(R). So by the Nash version of the stratification theorem (see Theorem A of [6] on page 349) there is a finite subset P of C(R), and for every semi-algebraic connected component U of C(R) − P a Nash manifold F U such that p(R) −1 (U ) is Nash-isomorphic to F U × U and the restriction of p(R) onto p(R) −1 (U ) is, modulo the given isomorphism, is the projection onto the second coordinate. By the nature of our construction for every point P in C(R) − P the fibre X P is smooth. By adding finitely many points to the set P, if it is necessary, we may assume that P has at least two points in each semi-algebraic connected component of C(R). Similarly we may assume that P contains every closed point of S without loss of generality.
Set S = (k + 1) P∈P P and let the same symbol denote the unique closed subscheme defined by this zero cycle. Since there is an interpolation condition φ : S → X × C S which subsumes φ, we may assume without loss of generality that the zero cycle defined by S is indeed (k + 1) P∈P P. Write S as a coproduct: where S P is a closed subscheme of S supported on P for each P ∈ P (possibly empty). For every such P let φ P : S P → X × C S P be the interpolation condition which is the pull-back of φ with respect to the closed imbedding S P → S. Let P, Q be a pair of consecutive points of P. Since σ (P) and σ (Q) lie in the smooth locus of p, by the implicit function theorem there are two points P , Q in the open interval ]P Q[ such that P lies before Q , and p(R) has a C ∞ -section σ P (resp. σ Q ) defined over some open neighbourhood of On the other hand, because of the way the set P was constructed, the restriction of such that σ P,Q and σ P are k-equivalent at P , and similarly σ P,Q and σ Q are k-equivalent at Q . Therefore the concatenation of σ P , σ P,Q and σ Q (restricted to Now let P, Q, R be three consecutive points of P (where P = R is allowed). Since both σ P,Q and σ Q,R have extensions to an open neighbourhood of their definitions which are compatible with φ Q , we get that their concatenation is C k at Q. We get that the concatenation of the different sections σ P,Q for all couples P, Q of consecutive points of P is a C k -section σ of p(R) defined over all of C(R) such that for each point x ∈ C(R) the point σ (x) lies in the same connected component of X x,sm (R) as σ (x), and σ is compatible with φ.
We will need a variant of the claim above with Nash sections, since for technical reasons it will be more convenient to work with the latter in the next section. In order to do so we will show two interpolation lemmas first.
Proof Let π j : A m R → A 1 R be the projection onto the j-th coordinate, where j = 1, 2, . . . , m. It will be sufficient to show the claim for φ j = π j • φ and g j = π j • g for each j. In other words we may assume that m = 1 without loss of generality. Since V is affine there is a regular map ψ : V → A 1 R compatible with φ. By replacing g with g − ψ we may assume without loss of generality that φ is the zero map. In this form the claim is a mild variant of Lemma 12.5.5 of [2] on page 321. We include the proof for the reader's convenience. Let h 1 , h 2 , . . . , h n be the generators of the defining ideal of Y . Since V is non-singular, we can represent the germ of g at a point x ∈ V in the form g x = λ 1,x h 1 + · · · + λ n,x h n , where the λ i,x are the germs of C ∞ -functions at x. Using a partition of unity and the compactness of V (R), this allows us to represent g globally as g = λ 1 h 1 + · · · + λ n h n , where λ i ∈ C ∞ (V (R)). Then it suffices to apply Nachbin's version of the Stone-Weierstrass theorem to the functions λ i (see [13]).

Definition 3.16
Let V be a nonsingular variety over R, and let W ⊂ A m (R) = R m be a Nash manifold. An interpolation condition φ : S → W for some subscheme S ⊂ V of the type considered above is an interpolation condition φ : Since W is compact, each derivative of ρ is bounded in some fixed -neighbourhood of W , and hence sequence ρ • g n approximates g in the C ∞ -topology.

Remark 3.18
Note that for every pair of conjugate point P, P ∈ C(C) − C(R) the complement C = C − P − P is an affine curve, and C (R) = C(R), so this set is compact. Therefore we may apply Lemmas 3.15 and 3.17 to C . In particular the conclusion of Lemma 3.17 holds for C, too.

Proposition 3.19
Let σ be a weakly continuous section of p(R) and let φ : S → X × C S be an interpolation condition weakly compatible with σ . Then there is a Nash section σ of p(R) such that for every x ∈ C(R) the points σ (x) and σ (x) are in the same connected component of X x,sm (R) and σ is compatible with φ.
Proof We may assume that φ is an interpolation condition of order ≤ k, where k is a positive integer. By Lemma 3.14 there is a C k -section σ of p(R) such that for every x ∈ C(R) the points σ (x) and σ (x) are in the same connected component of X x,sm (R) and σ is compatible with φ. By the usual approximation theorems in theory of smooth manifolds the section σ can be arbitrarily well approximated by C ∞ -maps s : Note that if such an s is sufficiently close to σ in the C 1 -topology then p(R) • s is a diffeomorphism and its inverse is very close to the identity map of C(R). Therefore For any smooth section σ : C(R) → X sm (R) sufficiently close to σ in the C 0 -topology and for every x ∈ C(R) the points σ (x) and σ (x) are in the same connected component of X x,sm (R), and hence the same holds for σ (x) and σ (x). The claim now follows at once from Lemma 3.17, as we explained in Remark 3.18.
We finish this section with a convenient condition for weak continuity.

Definition 3.20
We say that a set-theoretical section σ of the map p(R) is mildly in the latter case the image of σ is closed, and its intersection with Therefore the terminology is justified.

Proposition 3.21 Let σ be a semi-algebraic section of p(R) which is mildly continuous at every x ∈ C(R). Then σ is weakly continuous.
Proof Let x and y be two arbitrary different R-valued points of C lying in the same connected component of C(R). We will need the following Proof We may assume without loss of generality that V lies in p(R) −1 (]x y[); otherwise we only need to reverse the roles of x and y. It is also enough to prove the claim for the fibre above x; the proof for the fibre above y is similar. Since V is closed, the intersection V ∩ X x,sm (R) is closed in X x,sm (R). Therefore it will be enough to show that it is also open in X x,sm (R). Let n be the relative dimension of X over C and let z ∈ V ∩ X x,sm (R) be arbitrary. By the implicit function theorem there is a small connected open neighbourhood Let J ⊂ I be the set of points in I lying to the right of x. By shrinking I , if it is necessary, we may assume that J = I ∩]x y[. By assumption Since σ is semi-algebraic, it is continuous at all but finitely many points of C(R). Therefore there is a finite sequence of points z 1 , z 2 , . . . , z n ∈ [x y] such that σ is continuous at every z ∈]x y[ not on this list. We may even assume that z 1 = x, z n = y, and z i lies to the left of z j for every pair of indices i < j. For every i = 1, 2, . . . , n let E i denote the connected component of X z i ,sm (R) containing σ (z i ) and for every i = 1, 2, . . . , n − 1 let J i denote the closure of the image of ]z i z i+1 [ with respect to σ .
Since the restriction of σ onto ]z i z i+1 [ is continuous for every index i < n, the image of ]z i z i+1 [ with respect to σ is connected, and hence its closure J i is connected, too. Since J i is also compact, its image under p(R) is the closure is non-empty. By our assumptions the latter intersection lies in E i , hence J i ∩ E i is non-empty, too. A similar argument shows that J i ∩ E i+1 is also non-empty. Therefore the set by Lemma 3.22 the intersection V ∩ X x,sm (R) contains E z 1 , and hence σ (x). We may argue similarly to deduce that V ∩ X y,sm (R) contains σ (y). In other words V touches σ (x) on the right and σ (y) on the left. Since x and y are arbitrary, we get that σ is weakly continuous.

The Stone-Weierstrass approximation theorem with interpolation
Definition 4.1 Let π : C × P 1 (R) = C(R) × P 1 (R) → C(R) denote the projection onto the first factor. Let R ⊆ C(R) × P 1 (R) be a semi-algebraic subset. We say that R is admissible if it is the union of an open semi-algebraic set and finitely many points. The kissing points of an admissible semi-algebraic set R as above are all points of R which are not in the interior of R. We say that R does not have topological obstruction if there is a Nash section s : C(R) → C × P 1 (R) whose image lies in R.
The key result we need is an analogue of Conjecture 9.1 in [11] which we will formulate next. It is essentially a refined version of the classical the Stone-Weierstrass approximation theorem with interpolation conditions. Theorem 4.2 Let R ⊆ C(R) × P 1 (R) be an admissible semi-algebraic subset, and for some closed subscheme S ⊂ C let φ : S → (P 1 ×C)× C S = P 1 ×S be an interpolation condition such that there is a Nash section s : C(R) → C × P 1 (R) compatible with φ and whose image lies in R. Then there is a regular section f : C → C × P 1 R of the first projection compatible with φ such that f (C(R)) lies in R.
In particular we get that if R ⊆ C(R)×P 1 (R) is an admissible semi-algebraic subset which does not have topological obstruction then there is a morphism f : C → P 1 of schemes over R such that f (C(R)) lies in R. We are going to prove the theorem above via a sequence of lemmas.

Lemma 4.3
For some closed subscheme Z ⊂ C let h : Z → P 1 R be an interpolation condition. Then there is a morphism f : C → P 1 R of schemes over R such that f is compatible with h and f has no poles on C(R) outside of Z .
Proof First assume that h is actually a map h : Z → A 1 R . By our usual abuse of notation let Z also denote the effective divisor defining this closed subscheme and let d denote its degree. Choose an effective real divisor D on C which is supported outside of C(R) and whose degree is bigger than 2g + d where g is the genus of C. Then by the Riemann-Roch theorem we have dim H 1 (C, O C (D)) = 0, so the pull-back induces a surjection: Therefore there is a real rational function f on C compatible with h whose polar divisor is a sub-divisor of D.
Now consider the general case. We may assume without loss of generality that Z is non-empty. Since Z is a finite scheme over R there is an interpolation condition h : be the unique map such that h 1 | Z is h 1 , and h 1 | 2O 2 is f 2 | 2O 2 . By the above there is a rational function f 1 on C compatible with h 1 whose polar divisor is supported outside of C(R). Since Z is non-empty, the functions f 1 , f 2 are not both identically zero, and hence there is a non- The composition of f and the projection

Proof Let D ⊂ A(R) be a semi-algebraic connected component, and let D 1 ⊂ D be the collection of all points x ∈ D such that x has a semi-algebraic open neighbourhood
U ⊂ D such that the semi-algebraic set U ∩ f −1 (Z ) has dimension strictly less than D. Clearly D 1 is open (in the usual semi-algebraic topology). Set D 2 = f −1 (Z )− D 1 ; since f −1 (Z ) is closed, the set D 2 is closed, too. It will be sufficient to show that D 2 is also open. Clearly we only need to verify the latter Zariski-locally, that is, we may assume without loss of generality that A is affine.
Now let x ∈ D 2 be arbitrary; then there are a Zariski-open affine neighbourhood V ⊂ B of f (x) and a finite set of regular maps g 1 , g 2 , . . . , g n from V to A 1 R such that Z ∩ V is the common zero locus of g 1 , g 2 , . . . , g n . Let U ⊂ f −1 (V (R)) be a connected semi-algebraic open neighbourhood of x in D. Then g i (R) • f : U → R is a Nash function for each index i, so by Proposition 8.1.10 of [2] on page 166 its zero set Z i ⊂ U has either dimension strictly less than U , or this set Z i is equal to U .
Since Z i contains f −1 (Z )(R) ∩ U the former is not possible, so Z i = U for every index i. This implies that f −1 (Z )(R) ∩ U also equal to U .

Lemma 4.5 In the proof of Theorem 4.2 we may assume that R ⊆ C(R) × A 1 (R) without loss of generality.
Proof By Lemma 4.4 for every connected component D ⊂ C(R) either s is constant on D, or s takes every value only finitely many times on D. Therefore for all but finitely many x ∈ P 1 (R) the function s takes x only finitely many times on C(R), and hence after applying an automorphism of P 1 over R, if this is necessary, we may assume that the set T of points t ∈ C(R) where s(t) = ∞ is finite without loss of generality. We may even assume that φ does not take ∞ as a value. Since Nash maps are analytic, there are an effective zero divisor Z with support on C(R) and an interpolation condition h : Z → P 1 R compatible with s such that for every Nash map r : C(R) → P 1 (R) compatible with h the limit lim x→t (s(x) − r (x)) (4.5.1) exists and finite for every t ∈ T . We may even assume that h subsumes φ without loss of generality by choosing a Z such that Z − S is effective. By Lemma 4.3 there is a morphism r : C → P 1 R of schemes over R such that r is compatible with h and r has no poles on C(R) outside of Z .
By our assumptions both s and r take values in R on C(R) − T , so their difference s = s − r is a Nash function C(R) − T → R. Since the limit (4.5.1) exists and finite for every t ∈ T , this map s extends uniquely to a continuous map C(R) → R which we will also denote by s by slight abuse of notation. The latter is also Nash and it is compatible with a unique interpolation condition φ : Z → A 1 R . Let R ⊆ C(R)×A 1 (R) be the union of the image of s and the set The set R is admissible, and it contains the image of s. So if we assume that the claim of the theorem holds for R then there is a map f : C → P 1 R of schemes over R compatible with φ such that f (C(R)) lies in R. The rational function f +r : C P 1 R extends to a map f : C → P 1 R which is compatible with φ and f (C(R)) lies in R.

Proposition 4.6 Let R ⊆ C × A 1 (R) be an admissible semi-algebraic subset, and let s : C(R) → C × A 1 (R) be a Nash section whose image lies in R. Then there are an open neighbourhood U of s in the
Consider the real analytic function The zero set of f is s(V ), hence the zero set of the restriction f | A is just Z . By the Łojasiewicz inequality (in the form of Corollaire to Théorème 1 of section 18 in [12]) applied to our f , G, A, E, and Z , there are d, N > 0 such that Choose an integer M ≥ N /2, and setd = min √ d, 1 . Then Indeed, the projection π is a contraction, and dist(t, T ) ≤ 1 since U was imbedded into (0, 1). So (4.2) implies (4.3) for (t, a) ∈ E. On the other hand if (t, a) / ∈ E then |a| > S + 1, hence a − s 2 (t) > 1, and (4.3) follows again.
With the divisor Y = MT let ψ : Y → C × A 1 R be the (unique) interpolation condition (of order M) compatible with s. Suppose now that a section r is compatible with ψ, and its C M+1 -distance from s is less than, say, 1. If D denotes the maximum of s Combining (4.3) and (4.4) we obtain that, after possibly shrinking K , r (t) = t, r 2 (t) / ∈ A for t ∈ K \ T , hence r (K ) ⊂ R.
On the other hand Q = s C(R) \ K • is a compact set in R • . If the C 0 -distance of r from s is smaller than dist(Q, A), then r C(R) \ K ⊂ R as well.
Proof of Theorem 4.2 By Lemma 4.5 we may assume that R ⊆ C(R) × A 1 (R) without loss of generality. By Proposition 4.6 there is an open neighbourhood U of s in the C ∞ -topology and an interpolation condition ψ : Y → A 1 R compatible with s with the following property: for every C ∞ -section r : C(R) → C × A 1 (R) which lies in U and compatible with ψ, the image of r lies in R. Let T be a zero-dimensional closed subscheme of C whose closed points are all real and which contains both S and Y as a closed subscheme. Let φ : T → A 1 R be the unique interpolation condition compatible with s. Since s is compatible both with φ and ψ, the interpolation condition φ subsumes both φ and ψ. By Lemma 3.15 there is a regular section r : C → C × A 1 compatible with φ such that r (R) lies in U . Since φ subsumes φ, the section r is compatible with φ. Since φ subsumes ψ, the section r is compatible with ψ, too. Therefore the image of r (R) lies in R.

The main theorem and some easy reductions
Definition 5.1 Note that for every x ∈ C(R) the discrete valuation of R(C) x induces a topology on the projective space P n (R(C) x ), and hence on the R(C) x -valued points of any quasi-projective variety defined over R(C) x . Moreover this topology is canonical in the sense that it does not depend on the choice of the embedding into a projective space. We will call this the x-adic topology. Now let X be again a smooth, irreducible projective variety defined over R(C). We will equip the direct product with the direct product of the x-adic topologies.

Remark 5.2
Let X be as above, and let X be a model of X over C. It is possible to give a simple description of a basis for the topology on X (A C ) defined above in terms of X as follows. By slight abuse of notation let X (O x ) denote the set of sections for every x ∈ C(R). As we already noted we have a bijection X (R(C) x ) ∼ = X (O x ) for every x ∈ C(R) by the valuative criterion of properness, so we have a bijection too. For every interpolation condition φ : S → X × C S let be the subset of all those sections whose pull-back under the closed immersion is φ. These sets form a basis for the topology of X (A C ) under the map in (5.2.1).
For every morphism f : X → Y of varieties over R(C) and for every c ∈ Y (R(C)) let X c denote the fibre of f above c. Our main result Theorem 1.5 follows from the following

Remark 5.4
Note that Theorem 1.5 is trivially true when C(R) is empty. Indeed in this case there is no CT obstruction both for the smooth fibres of f over rational points and for X itself. By assumption all such fibres will have rational points, so X has rational points, too.
Proof We start the proof with two easy reduction steps. For the sake of simple notation set n-times . Proposition 5. 5 We may assume that Y = Y n for some n without loss of generality.
Proof Using resolutions of singularities it follows from the assumption that there is a diagram of birational morphisms between smooth projective varieties over R(C) (for some n). Let X → X × Y Y be a desingularisation of the pull-back X × Y Y of X via φ which is isomorphic over the nonsingular part, and let f : X → Y be the composition of this desingularisation and the base change of f with respect to φ. Let V ⊆ Y be a non-empty Zariski-open subset such that the restrictions of both φ and ψ onto V are isomorphisms onto their images, and f is smooth over φ(V ). Then we have a commutative diagram: In particular ρ is birational. Let Z ⊂ X be the complement of f −1 (φ(V )), and let Z ⊂ X be the complement of f −1 (V ). Now let M be an element of X (A C ) CT , as in the claim above. We may assume without loss of generality that its given open neighbourhood U is of the form The set I is finite. Now let X be a model of X over C and for every x ∈ C(R) let T x ⊆ X (R(C) x ) be the x-adic open neighbourhood of M x which under the bijection X (R(C) x ) ∼ = X (O x ) corresponds to those sections Spec(O x ) → X × C Spec(O x ) whose special fibre is the same as the special fibre of the section corresponding to M x . Since Z is a proper Zariski-closed subscheme of X , the set Z (R(C) x ) is nowhere dense in X (R(C) x ) with respect to the x-adic topology, so there is a non-empty W x ⊆ U x ∩ T x , open with respect to the x-adic topology, such that W x and Z (R(C) x ) have empty intersection, for every x ∈ C(R). By the above for every Therefore we may assume, without loss of generality, that M x / ∈ Z (R(C) x ) for every x ∈ C(R), the set I is non-empty, and U x ∩ Z (R(C) x ) = ∅ for every x ∈ I . Since ρ it is an open neighbourhood of M. Since the map ψ • f : X → Y n satisfies the conditions of theorem, we get that there is a point c ∈ Y n (R(C)) such that X c is smooth, and an N ∈ X c (A C ) CT such that N ∈ U . Let x be now an element of I . Then c lies in the image of U x with respect to ψ • f , so it must lie in ψ(V (R(C) x )). Therefore there is a unique c ∈ V (R(C)) such that c = ψ(c). Set c = φ(c) and N = ρ( N ). Clearly c ∈ φ(V (R(C))) and hence X c is smooth. Moreover ρ maps X c isomorphically onto X c , so N ∈ X c (A C ) CT . Finally N ∈ U since ρ maps U into U .
Lemma 5. 6 We may assume that Y = P 1 R(C) without loss of generality.
Proof We may immediately reduce the case when Y = Y n to the case when Y = P 1

R(C)
via an easy induction on n, so the claim follows from the proposition above.

The fibration method
Let us begin the main part of the proof of Theorem 5.3. By the above we may assume without loss of generality that Y = P 1 R(C) . By resolution of singularities there is an integral, smooth, projective variety X equipped with a projective dominant morphism f : X → C × P 1 R over R whose generic fibre is f : X → P 1 R(C) . In particular X is a model of X over C with respect to the composition p of f with the projection π : C × P 1 R → C onto the first factor. Since the generic fibre of f is smooth, the same holds for f, too. Therefore there is a closed subscheme Z ⊂ C × P 1 R of positive codimension such that f| f −1 (U ) : Proof By our assumption U is of the form The set I is finite. For every connected component D ⊂ C(R) for all but finitely many x ∈ D the intersection Z ∩ π −1 (x) is a zero dimensional scheme, since Z has positive codimension in C × P 1 R . Since p is generically smooth and has geometrically irreducible fibres, for all but finitely many x ∈ D the fibre X x is smooth and geometrically irreducible. Therefore for all D as above we may choose a point x(D) ∈ D which does not lie in I , the intersection Z(R) ∩ π −1 (x D )(R) is finite, and X x(D) is smooth and geometrically irreducible.
For every D as above let E D ⊂ X x(D) (R) denote the connected component con- Assume that this is not the case for some D ∈ π 0 (C(R)). Then the image of E D under f(R) lies in the finite set Z(R) ∩ π −1 (x(D))(R). But E D is connected, so is its image under the continuous map f(R), which therefore must be a point p D ∈ Z(R) ∩ π −1 (x(D))(R). Note that E D is Zariski-dense in X x (D) . Indeed suppose that this is not the case; as the latter is geometrically irreducible, the Zariskiclosure of E D has dimension strictly less than the dimension d of X x(D) . Therefore the dimension of the semi-algebraic set E D is also less than d. But X x(D) is smooth, so the dimension of E D is d by the inverse function theorem, which is a contradiction. We get that X x(D) lies in f −1 ( p D ), and hence the fibre of f over any other point in π −1 (x(D)) is empty. But this is a contradiction, so our original assumption on the image of E D with respect to f(R) is false.
For every D as above choose an N x(D) ∈ X (R(C) x(D) ) such that the special fibre of the section Replacing M by M we may assume without loss of generality that σ (M) is Nash, compatible with φ, and f(R) • σ (M) only intersects Z(R) in finitely many points. Recall that a Nash map a : A → B of Nash manifolds over R is Nash trivial if there is a Nash manifold L and a Nash diffeomorphism b : L × B → A (over R) such that a • b : L × B → B is the projection onto the second coordinate.

Proposition 6.2 There is an admissible semi
by open semi-algebraic subsets which are Nash diffeomorphic to an affine space over R, necessarily of dimension 2, by Lemma 3.2 of [9] on page 1217. For every j ∈ J the restriction Let B be the union of the end-points of these open intervals for all j ∈ J , where by end-points we mean accumulation points not in the interval. Since the set of these intervals is finite, the set B is also finite. The complement of B in C(R) is the union of finitely many pair-wise disjoint open intervals; let K denote the set of these open intervals. Since the sets {V j } j∈J cover all but finitely many points of I, for every H ∈ K the pre-image π(R) −1 (H ) ∩ I lies in V j for some j ∈ J . For every such H fix such a V j and let W H ⊂ V j be a semi-algebraic tubular neighbourhood of

Definition 6.3
For every x ∈ C × P 1 (R) and every 1-dimensional subspace L of the tangent space of C × P 1 at x let X L,sm ⊂ X x be the largest open sub-scheme such that for every closed point P of X L,sm the image of the differential of f at P contains L.
Let σ : C(R) → X (R) be a Nash section of p(R) and let R + ⊆ C(R) × P 1 (R) be an admissible semi-algebraic subset which contains the image Im(f(R) • σ ) of f(R) • σ . For every point x on Im(f(R) • σ ) let L(x) denote the tangent line of f(R) • σ at x; it is a 1-dimensional subspace of the tangent space of C × P 1 at x. A butterfly extension of σ on R + is a semi-algebraic section β : Proof We will need the following easy semi-algebraic separation lemma.

Lemma 6.5 Let F and G be two disjoint closed semi-algebraic subsets of X (R). Then there are two disjoint open semi-algebraic subsets A, B ⊂ X (R) such that F ⊂ A and G ⊂ B.
Proof Since X is projective, by Theorem 3.4.4 of [2] on page 72 there is a continuous semi-algebraic embedding ι : X (R) → R m for some positive integer m. Because X is projective, the semi-algebraic set X (R) is compact, so the same holds for its closed subsets F and G. Therefore ι(F) and ι(G) are closed in R m , and by elimination of quantifiers these sets are also semi-algebraic. Then d = dist(F, G) > 0, where dist stands for the Euclidean distance. The sets It has a unique extension s : For every point x on I let L(x) denote the tangent line of f(R)•σ (M) at x. Note that for every x ∈ I we have s(x) = σ (M)(π(R)(x)), and σ (M) is a Nash section, so s(x) lies in X L(x),sm (R). Therefore it will be enough to show that there is an admissible semi-algebraic subset R + ⊆ R containing I such that for every kissing point x of R + on I the intersection of the closure of the image of s with X x (R) lies in the connected component of s(x) in X L(x),sm . Let K denote the set of kissing points of R.
For every x on I let X L(x),bad be the complement of X L(x),sm in X x . For every x ∈ K the subscheme X L(x),bad of X is Zariski-closed, so the semi-algebraic set X L(x),bad (R) is closed in X (R), and does not contain s(x), hence by Lemma 6.5 we may pick two disjoint open semi-algebraic subsets A x , B x ⊂ X (R) such that X L(x),bad (R) ⊂ A x and s(x) ∈ B x . Now let E be any semi-algebraic connected component of X L(x),sm (R) which does not contain s(x), and let E denote its closure in X x (R). Since E is closed in X L(x),sm (R), and s(x) lies in X L(x),sm (R), the set E does not contain s(x), so by Lemma 6.5 we may pick two disjoint open semi-algebraic subsets A E , B E ⊂ X (R) such that E ⊂ A E and s(x) ∈ B E .
For every x ∈ K let W x be the intersection of B x and the B E for all E as above. The set of connected components of X L(x),sm (R) is finite, therefore the set W x is an open semi-algebraic neighbourhood of s(x) such the intersection of its closure with X x (R) lies in the connected component of s(x) in X L(x),sm . Since σ (M) is continuous, for every x as above π(R)(x) has an open connected neighbourhood V x in C(R) such that σ (M) maps V x into W x . We may assume that the sets V x are pair-wise disjoint by shrinking them, if this is necessary. For every x ∈ K let R x be the intersection of π(R) −1 (V x ) with the image of W x ∩ f(R) −1 (R o ) with respect to f. Since Let T be the interior of the complement of the union of V x for all x ∈ K in C(R). Then the set is open, semi-algebraic and contains all but finitely many points of I. Therefore R + = R ++ ∪ I is an admissible subset of R. For every x ∈ K the intersection of the closure of the image of s| R + with X x (R) lies in the connected component of s(x) in X L(x),sm by construction. If x is a kissing point of R + not in K , then x ∈ R o , so s is continuous at x, and hence the intersection of the closure of the image of s| R + with X x (R) is just s(x).
Let X sm ⊆ X be the smooth locus of f . For every x ∈ I let P x denote the formal completion of the Nash section σ (M) around σ (M)(x); it is a section Spec(O x ) → X × C Spec(O x ). By slight abuse of notation let the same symbol P x denote its generic fibre, too. Since σ (M) only intersects Z(R) in finitely many points, we have P x ∈ X sm (R(C) x ). By property (ii) of Proposition 6.1 we have P x ∈ U x for every x ∈ I , so by Lemma 6.6 for every x ∈ I there is an open neighbourhood V x of f (P x ) in the x-adic topology such that for every z ∈ V x the set f −1 (z)(R(C) x ) ∩ U x is non-empty. We may assume that V x = P 1 (R(C) x ) whenever U x = X (R(C) x ). There is an interpolation condition κ : T → P 1 compatible with f(R) • σ (M) such Now let R + ⊆ R be an admissible semi-algebraic subset containing the image I of f(R) • σ (M) and let s : R + → X (R) be a butterfly extension of σ (M). By removing every kissing point of R + which does not lie on I we may even assume that every kissing point of R + lies on I. Let κ : T → P 1 be an interpolation condition subsuming κ : T → P 1 , compatible with f(R) • σ (M) such that for every kissing point z of R + the point π(R)(z) has coefficient at least 2 in T .
By Theorem 4.2 there is a regular map c : C → P 1 compatible with κ such that the graph c of c in C × P 1 (R) lies in R + . Let c ∈ P 1 (R(C)) be the generic point of c. Let X c denote the closed subscheme f −1 ( c ) ⊂ X and let π c : X c → C be the composition of f and π . Since all but finitely many points of c lie in R o + ⊂ U(R), the generic fibre X c of X c (with respect to π c ) is smooth. Let X c,sm ⊆ X c be the smooth locus of π c . By resolution of singularities there is a sequence of blow-ups r : X c → X c such that X c is a model of X c over C and contains X c,sm as an open sub C-scheme, i.e. the restriction r | r −1 (X c,sm ) : r −1 (X c,sm ) → X c,sm is an isomorphism.