Abstract In this work we compare the semialgebraic subsets that are images of regulous maps with those that are images of regular maps. Recall that a map f:Rn→Rm is regulous if it is a rational map that admits a continuous extension to Rn . In case the set of (real) poles of f is empty we say that it is regular map. We prove that if S⊂Rm is the image of a regulous map f:Rn→Rm , there exists a dense semialgebraic subset T⊂S and a regular map g:Rn→Rm such that g(Rn)=T . In case dim(S)=n , we may assume that the difference S⧹T has codimension ≥2 in S . If we restrict our scope to regulous maps from the plane the result is neat: if f:R2→Rmis a regulous map, there exists a regular map g:R2→Rmsuch that Im(f)=Im(g). In addition, we provide in Appendix A a regulous and a regular map f,g:R2→R2whose common image is the open quadrant Q≔{𝚡>0,𝚢>0}. These maps are much simpler than the best-known polynomial maps R2→R2that have the open quadrant as their image. 1. Introduction A map f≔(f1,…,fm):Rn→Rm is regulous if its components are regulous functions, that is, fi=gihi is a rational function that admits a continuous extension to Rn and gi,hi∈R[𝚡] are relatively prime polynomials. By [15, 3.5] {hi=0}⊂{gi=0} and the codimension of {hi=0} is ≥2 . Consequently, the set of indeterminacy of f is an algebraic set of codimension ≥2 . In case {hi=0} is empty for each i=1,…,m , we say that f is a regular map whereas f is polynomial if we may choose h1=⋯=hm=1 . Modern Real Algebraic Geometry was born with Tarskiʼs article [21], where it is proved that the image of a semialgebraic set under a polynomial map is a semialgebraic set. A subset S⊂Rn is semialgebraic when it has a description by a finite boolean combination of polynomial equalities and inequalities, which we will call a semialgebraic description. We are interested in studying what might be called the ‘inverse problem’ to Tarski’s result, that is, to represent semialgebraic sets as images of nice semialgebraic sets (in particular, Euclidean spaces, spheres, smooth algebraic sets, etc.) under semialgebraic maps that enjoy a kind of identity principle, that is, semialgebraic maps that are determined by its restriction to a small neighborhood of a point (for instance, polynomial, regular, regulous, Nash, etc.). In the 1990, Oberwolfach reelle algebraische Geometrie week [17] Gamboa proposed: Problem 1.1 To characterize the (semialgebraic) subsets of Rmthat are either polynomial or regular images of Euclidean spaces. During the last decade, the first and fourth authors jointly with Gamboa have attempted to understand better polynomial and regular images of Rn with the following target: To find obstructions to be either polynomial or regular images [4, 5, 7, 12]. To prove (constructively) that large families of semialgebraic sets with piecewise linear boundary (convex polyhedra, their interiors, their complements and the interiors of their complements) are either polynomial or regular images of Euclidean spaces [6, 9–11, 14, 22, 23]. In [2], appears a complete solution to Problem 1.1 for the one-dimensional case, but the rigidity of polynomial and regular maps makes really difficult to approach Problem 1.1 in its full generality. Taking into account the flexibility of Nash maps, Gamboa and Shiota discussed in 1990 the possibility of approaching the following variant of Problem 1.1 (see [17]). Problem 1.2 To characterize the (semialgebraic) subsets of Rnthat are Nash images of Euclidean spaces. A Nash function on an open semialgebraic set U⊂Rn is a semialgebraic smooth function on U . In [3], the first author solves Problem 1.2 and provides a full characterization of the semialgebraic subsets of Rm that are images under a Nash map on some Euclidean space. A natural alternative approach is to consider regulous maps, which are closer than Nash maps to regular maps, but seem to be less rigid than the latter [19, 20]. We propose to devise the following variant of Problem 1.1. Problem 1.3 To characterize the (semialgebraic) subsets of Rnthat are regulous images of Euclidean spaces. By [15, 3.11], a map f:Rn→Rm is regulous if and only if there exists a composite of a finite sequence of blowings-up with non-singular centers ϕ:Z→Rn such that the composition f◦ϕ:Z→Rm is a regular map. As Z is a non-singular algebraic set, we deduce that f(Rn)=(f◦ϕ)(Z) is a pure dimensional semialgebraic subset of Rn . In addition, as a regulous map on R is regular [15, 3.6], we deduce that f(Rn) is connected by regular images of R . We refer the reader to [15] for a foundational work concerning the ring of regulous functions on an algebraic set. Our experimental approach to the problem of characterizing the regulous images of Euclidean spaces suggests that regular images and regulous images of Euclidean spaces coincide. Although we have not been able to prove this result in its full generality, our main result in this direction is the following. Theorem 1.4 Let f:Rn→Rmbe a regulous map and let S≔f(Rn)be its image. Then, there exists a regular map g:Rn→Rmwhose image T≔g(Rn)⊂Sis dense in S . In addition, if dim(S)=n , we may assume that the difference S⧹Thas codimension ≥2 . The proof of the previous result is reduced to showing the following one, which has interest by its own and concerns the representation of the complement of an algebraic subset of Rn of codimension ≥2 as a polynomial image of Rn . Theorem 1.5 Let X⊂Rnbe an algebraic set of codimension ≥2 . Then the constructible set Rn⧹Xis a polynomial image of Rn . Assume Theorem 1.5 is proved and let us see how the proof of Theorem 1.4 arises. Proof of Theorem 1.4 As we have commented above, the set of indeterminacy X of f has codimension ≥2 . By Theorem 1.5, there exists a polynomial map h:Rn→Rn such that h(Rn)=Rn⧹X . Consider the composition g≔f◦h:Rn→Rm . We have S⧹f(X)⊂T≔g(Rn)=f(Rn⧹X)⊂S. As f is a continuous map and Rn⧹X is dense in Rn , we deduce that T is dense in S . In addition, if dim(S)=n , we have S⧹T⊂f(X) , so dim(S⧹T)≤dim(f(X)) . By [1, Theorem 2.8.8], the following inequalities dim(f(X))≤dim(X)≤n−2=dim(S)−2 hold, so dim(S⧹T)≤dim(S)−2 , as required.□ As commented above we feel that the following question has a positive answer. Question 1.6 Let f:Rn→Rm be a regulous map. Is there a regular map g:Rn→Rm such that Im(f)=Im(g)? As there are much more regulous maps than regular maps and they are less rigid than regular maps, we hope that a positive answer to the previous question will ease the solution to Problem 1.1 in the regular case. If we restrict our target to regulous maps from the plane, we devise the following neat answer to Question 1.6. Theorem 1.7 Let f:R2→Rmbe a regulous map. Then there exists a regular map g:R2→Rmsuch that Im(f)=Im(g) . The proof of the previous result is reduced to showing the following lemma, which has interest by its own and concerns an ‘alternative’ resolution of the indeterminacy of a locally bounded rational function on R2 to make it regular, adapted to our situation, in which one does not care on the cardinality of the fibers (that are however generically finite). Lemma 1.8 Let f:R2→Rbe a locally bounded rational function on R2 . Then there exists a generically finite surjective regular map ϕ:R2→R2such that f◦ϕis a regular function on R2 . Assume Lemma 1.8 and let us see how the proof of Theorem 1.7 arises. Proof Let f≔(f1,…,fm):R2→Rm be a regulous map. By Lemma 1.8, one can find a surjective regular map ϕ1:R2→R2 such that f1◦ϕ1 is regular. Consider the regulous function g2≔f2◦ϕ1 . Again by Lemma 1.8, one can find a surjective regular map ϕ2:R2→R2 such that g2◦ϕ2 is regular. Clearly, the composition f1◦ϕ1◦ϕ2 is also regular. We proceed inductively with f3,…,fm and obtain surjective regular maps ϕk:R2→R2 such that fk◦ϕ1◦⋯◦ϕk is a regular function for k=1,…,m . At the end, we have a surjective regular map ϕ≔ϕ1◦⋯◦ϕm:R2→R2 such that g≔f◦ϕ is a regular map on R2 . As ϕ is surjective, we have g(R2)=(f◦ϕ)(R2)=f(R2) , as required.□ Structure of the article: The article is organized as follows. In Section 2, we prove Theorem 1.5, while in Section 3, we prove an improved version of Lemma 1.8. Our argument fundamentally relies on the properties of the so-called double oriented blowings-up. These tools allow us to reformulate Lemma 1.8 using regular maps satisfying the arc-lifting property (see Theorem 3.15), which is commonly used in the analytic setting to study singularities [16]. Finally, in Appendix A, we propose a regulous and a regular map f,g:R2→R2 whose common image is the open quadrant Q≔{𝚡>0,𝚢>0} . These maps are much simpler than the best-known polynomial maps R2→R2 that have Q as their image [4, 8, 13]. In addition, the verification that the images of f,g are Q is quite straightforward and does not require the enormous effort needed for the known polynomial maps used in [4, 8, 13] to represent Q . Recall that the systematic study of the polynomial and regular images began in 2005 with the first solution to the open quadrant problem [4]. 2. Complement of an algebraic set of codimension at least 2 The purpose of this section is to prove Theorem 1.5. The proof is conducted in several technical steps and some of them have interest by their own. 2.1. Finite projections of complex algebraic sets Let X⊂Rn be a (real) algebraic set and let IR(X) be the ideal of those polynomials f∈R[𝚡] such that f(x)=0 for all x∈X . The zero-set X˜⊂Cn of IR(X)C[𝚡] is a complex algebraic set that is called the complexification of X . The ideal IC(X˜) of those polynomials F∈C[𝚡] such that F(z)=0 for all z∈X˜ coincides with IR(X)C[𝚡] . Consequently, X˜ is the smallest complex algebraic subset of Cn that contains X and C[𝚡]/IC(X˜)≅(R[𝚡]/IR(X))⊗C. In particular, the real dimension of X coincides with the complex dimension of X˜ . A key result in this section is the following result concerning projections of complex algebraic sets [18, Theorem 2.2.8]. The zero set in Cn of an ideal a⊂C[𝚡] will be denoted Z(a) . Theorem 2.1 (Projection Theorem) Let a⊂C[𝚡]be a non-zero ideal. Suppose that acontains a polynomial Fthat is regular with respect to 𝚡n , that is, F≔𝚡nm+am−1(𝚡′)𝚡nm−1+⋯+a0(𝚡′)for some polynomials ai∈C[𝚡′]≔C[𝚡1,…,𝚡n−1] . Let Π:Cn→Cn−1be the projection (x1,…,xn)↦(x1,...,xn−1)onto the first n−1coordinates of Cn . Define a′≔a∩C[𝚡′]and let X′≔Z(a′)be its zero set. Then Π(X)=X′is a complex algebraic subset of Cn−1 . 2.2. Recovering an algebraic set from its projections Denote π:Rn→Rn−1,(x1,…,xn)→(x1,…,xn−1) the projection onto the first n−1 coordinates of Rn . Pick v⃗≔(v1,…,vn)∈Rn⧹{𝚡n=0} and consider the isomorphism ϕv⃗:Rn→Rn,(x1,…,xn)↦(x1−v1vnxn,…,xn−1−vn−1vnxn,1vnxn) (2.1) that keeps fixed the hyperplane H0≔{𝚡n=0} and maps the vector v⃗ to the vector 𝚎⃗n≔(0,…,0,1) . Its inverse map is ϕv⃗−1:Rn→Rn,(y1,…,yn)↦(y1+v1yn,…,yn−1+vn−1yn,vnyn). Consider the projection πv⃗≔π◦ϕv⃗:Rn→Rn−1×{0}≡Rn−1,(x1,…,xn)↦(x1−v1vnxn,…,xn−1−vn−1vnxn,0)≡(x1−v1vnxn,…,xn−1−vn−1vnxn) (2.2) of Rn onto Rn−1 (identified with {𝚡n=0} ) in the direction of v⃗ . We have πv⃗(y1+v1vnλyn,…,yn−1+vn−1vnλyn,λyn)=(y1,…,yn−1) for each λ∈R (in particular, for λ=vn ). Let X⊂Rn be a real algebraic set of dimension d≤n−2 , let X˜ be its complexification and let a≔IC(X˜)=IR(X)C[𝚡] . Let F∈IR(X)⧹{0} and write F≔F0+F1+⋯+Fm as the sum of its homogeneous components. If Fm(v⃗)≠0 , then F(y1+v1yn,…,yn−1+vn−1yn,vnyn) is regular with respect to yn . We extend πv⃗ to Πv⃗:Cn→Cn−1 and observe that if Fm(v⃗)≠0 , then Πv⃗(X˜) is by Theorem 2.1 an algebraic subset of Cn−1 of dimension ≤d . Consequently, the Zariski closure πv⃗(X)¯zar of πv⃗(X) is contained in Πv⃗(X˜)∩Rn−1 , which is a real algebraic subset of Rn−1 of dimension ≤d . 2.3. Just a small vertical skip Let p∈Rn⧹X and consider the algebraic cone C⊂Rn with vertex p and basis X . As X has real dimension d≤n−2 , the real algebraic set C has (real) dimension d+1≤n−1 . Let C˜⊂Cn be the complexification of C , which is a complex algebraic set of (complex) dimension d+1 . As X⊂C , the inclusion X˜⊂C˜ holds. In addition, as C is a cone, then C˜ is also a cone. Denote C⃗≔{v⃗∈Rn:v⃗=px→,x∈C} and C˜⃗≔{w⃗∈Cn:w⃗=pz→,z∈C˜} . Pick a vector v⃗∈Rn⧹(C⃗∪{Fm=0}) . We claim: Lemma 2.2 We have πv⃗(p)∉πv⃗(X)¯zar. Proof As πv⃗(X)¯zar⊂Πv⃗(X˜)∩Rn−1 , it is enough to check that πv⃗(p)∉Πv⃗(X˜) . If πv⃗(p)∈Πv⃗(X˜) , there exists z∈X˜ such that Πv⃗(z)=πv⃗(p) . Thus, there exists λ∈R and μ∈C such that p+λv⃗=πv⃗(p)=Πv⃗(z)=z+μv⃗. Thus, (λ−μ)v⃗=pz→∈C˜⃗ . As p∉X , it holds p≠z , so λ≠μ and v⃗∈C˜⃗∩Rn=C⃗ , which is a contradiction. Thus, πv⃗(p)∉πv⃗(X)¯zar , as claimed.□ Lemma 2.3 Let X⊂Rnbe a real algebraic set of dimension d≤n−2and let Ω⊂Rn⧹{𝚡n=0}be a non-empty open set. Then, there exist r≤n+1vectors v⃗1,…,v⃗r∈Ωsuch that X=⋂i=1rπi−1(πj(X)¯zar),where πi≔πv⃗ifor i=1,…,r . Proof Given an algebraic set Z⊂Rn define e(Z)≔dim(Z⧹X) , that is, the dimension of the constructible set Z⧹X . If Z1,…,Zk are the irreducible components of Z , then e(Z) is either equal to −1 if Zj⊂X for 1≤j≤k or the maximum of the dimensions of the irreducible components of Z that are not contained in X otherwise. Pick a vector v⃗1∈Ω and denote π1≔πv⃗1 . Let Y11,...,Y1s be the irreducible components of the algebraic set Y1≔π1−1(π1(X)¯zar) , which has dimension ≤d+1 . If e(Y1)=−1 , we have X=Y1 and we are done. Assume e(Y1)≥0 . For each Y1j not contained in X , pick a point pj∈Y1j⧹X . By Lemma 2.2, there exists a vector v⃗2∈Ω such that each pj∉Y2≔π2−1(π2(X)¯zar) , where π2≔πv⃗2 . Let T be an irreducible component of Y1∩Y2 . We claim: dim(T)<e(Y1) . As T is not contained in X but T⊂Y1 , there exists an irreducible component Y1j of Y1 not contained in X such that T⊂Y1j . As pj∉Y2 , we have dim(T)≤dim(Y1j∩Y2)<dim(Y1j)≤e(Y1) , as claimed. Consequently, e(Y1∩Y2) is strictly smaller than e(Y1) . If e(Y1∩Y2)≥0 , we pick a point qj∈Y12j⧹X for each irreducible component Y12j of Y1∩Y2 not contained in X (and indexed j=1,…,ℓ ). By Lemma 2.2, there exists a vector v⃗3∈Ω such that each qj∉Y3≔π3−1(π3(X)¯zar) , where π3≔πv⃗3 . Again, this implies that e(Y1∩Y2∩Y3)<e(Y1∩Y2) . We repeat the process r≤d+3≤n+1 times to find vector v⃗1,…,vr⃗∈Ω such that e(Y1∩⋯∩Yr)=−1 where Yi≔πi−1(πi(X)¯zar) and πi≔πv⃗i . Consequently, X=⋂j=1rYj , as required.□ Lemma 2.4 Let X⊂Rnbe an algebraic set of codimension ≥1 . Then there exists a polynomial diffeomorphism ϕ:Rn→Rnsuch that ϕ(X)⊂{−𝚡n−1<𝚡n<𝚡n−1} . Proof Let X^ be the Zariski closure of X in the projective space RPn . As X^ has codimension ≥1 , we may assume that (0:⋯:0:1:0)∉X^ . The sets {𝚡02+⋯+𝚡n−22+𝚡n2≤ε2𝚡n−12} for ε>0 constitute a basis of neighborhoods of (0:⋯:0:1:0) in RPn . Let 0<ε<1 be such that X^∩{𝚡02+⋯+𝚡n−22+𝚡n2≤ε2𝚡n−12}=∅ . As ε2𝚡n−12≥−ε𝚡n−1𝚡0−14𝚡02 (because ε2𝚡n−12+ε𝚡n−1𝚡0+14𝚡02=(ε𝚡n−1+12𝚡0)2 ), we have {𝚡02+⋯+𝚡n−22+𝚡n2≤−ε𝚡n−1𝚡0−14𝚡02}⊂{𝚡02+⋯+𝚡n−22+𝚡n2≤ε2𝚡n−12}. Consequently, X⊂{54+𝚡12+⋯+𝚡n−22+𝚡n2>−ε𝚡n−1} . Consider the polynomial diffeomorphism ϕ:Rn→Rn,x≔(x1,…,xn)↦(x1,…,xn−2,xn−1+2ε(54+x12+⋯+xn−22+xn2),xn) whose inverse ϕ−1:Rn→Rn,x≔(x1,…,xn)↦(x1,…,xn−2,xn−1−2ε(54+x12+⋯+xn−22+xn2),xn) is also polynomial. We have ϕ(X)⊂{𝚡n−1≥1ε(54+𝚡12+⋯+𝚡n−22+𝚡n2)}. As ε<1 , {𝚡n−1≥1ε(54+𝚡12+⋯+𝚡n−22+𝚡n2)}⊂{−𝚡n−1<𝚡n<𝚡n−1}, as required.□ 2.4. Proof of Theorem 1.5 We will represent Rn⧹X as the image of the composition of finitely many polynomial maps fj:Rn→Rn whose images contain constructible sets Rn⧹Yj such that X=⋂jYj . The proof is conducted in several steps: Step 1. Initial preparation. By Lemma 2.4, we assume X⊂Int(K) where K≔{−𝚡n−1≤𝚡n≤𝚡n−1} . Denote the projection onto the first n−1 coordinates by π:Rn→Rn−1,(x1,…,xn)↦(x1,…,xn−1). For each vector v⃗≔(v1,…,vn)∈Rn⧹{𝚡n=0} consider the isomorphism ϕv⃗ that keeps fixed the plane H0≔{𝚡n=0} and maps the vector v⃗ to the vector 𝚎⃗n (see (2.1)) and let πv⃗≔π◦ϕv⃗ be the projection of Rn onto Rn−1 (identified with {𝚡n=0} ) in the direction of v⃗ (see (2.2)). We have ϕv⃗(K)={𝚡n−1−(vn−vn−1)𝚡n≥0,𝚡n−1+(vn+vn−1)𝚡n≥0}. If λ≔vn−vn−1>0 and μ≔vn+vn−1>0 , then πv⃗(K⧹{𝚡n−1=0,𝚡n=0})=π(ϕv⃗(K)⧹{𝚡n−1=0,𝚡n=0})={𝚡n−1>0}⊂Rn−1. Consider the open set Ω≔{𝚡n−𝚡n−1>0,𝚡n+𝚡n−1>0}⊂Rn⧹{𝚡n=0} . By Lemma 2.3, there exist vectors v⃗1,…,v⃗r∈Ω such that X=⋂j=1rπj−1(πj(X)¯zar), where πj≔πv⃗j for j=1,…,r . In particular, each set πj−1(πj(X)¯zar) is an algebraic subset of Rn that contains X . For each j=1,…,r denote ϕj≔ϕv⃗j . Step 2. We claim: There exists a polynomial map f1:Rn→Rnsuch that S1≔f1(Rn⧹Int(K))=Rn⧹(Int(K)∩π1−1(π1(X)¯zar))⊂Rn⧹X. (2.3) Define K1≔ϕ1(K) and X1≔ϕ1(X) . Write K1≔{𝚡n−1−λ1𝚡n≥0,𝚡n−1+μ1𝚡n≥0} for some real numbers λ1,μ1>0 and consider the polynomial map f1′:Rn→Rn,x≔(x1,…,xn)↦(x1,…,xn−1,xn(1−H1(xn−1,xn)2G12(x′))), where H1≔(𝚡n−1−λ1𝚡n)(𝚡n−1+μ1𝚡n)∈R[𝚡n−1,𝚡n] and G1∈R[𝚡′]≔R[𝚡1,…,𝚡n−1] is a polynomial equation of π(X1)¯zar . We claim: f1′(Rn⧹Int(K1))=Rn⧹(Int(K1)∩(π−1(π(X1)¯zar))). Pick a point x′≔(x1,…,xn−1)∈Rn−1 and let ℓx′≔{(x′,t):t∈R} be the line through (x′,0) parallel to 𝚎⃗n . Consider the polynomial Qx′(𝚡n)≔𝚡n(1−H1(x′,𝚡n)2G12(x′))∈R[𝚡n] . We distinguish two cases: • If x′∉π(X1)¯zar , then Qx′(𝚡n) is a polynomial of degree five and negative leading coefficient. Let ax′≤bx′ be real numbers such that ℓx′⧹Int(K1)={x′}×((−∞,ax′]∪[bx′,+∞)). As limxn→±∞Qx′(xn)=∓∞ , Qx′(ax′)=ax′ and Qx′(bx′)=bx′ , we have [ax′,+∞)⊂Qx′((−∞,ax′]),(−∞,bx′]⊂Qx′([bx′,+∞)). Thus, R=(−∞,bx′]∪[ax′,+∞)⊂Qx′(ℓx′⧹Int(K1))⊂R , so f1′(ℓx′⧹Int(K1))=ℓx′ . • If x′∈π(X1)¯zar , then Qx′(𝚡n)=𝚡n and f1′(ℓx′⧹Int(K1))=ℓx′⧹Int(K1) . Putting all together, we deduce f1′(Rn⧹Int(K1))=⋃x′∈Rn−1f1′(ℓx′⧹Int(K1))=⋃x′∉π(X1)¯zarℓx′∪⋃x′∈π(X1)¯zar(ℓx′⧹Int(K1))=Rn⧹(Int(K1)∩(π−1(π(X1)¯zar))). As X1⊂Int(K1)∩π−1(π(X1)¯zar) , we have f1′(Rn⧹Int(K1))=Rn⧹(Int(K1)∩π−1(π(X1)¯zar))⊂Rn⧹X1. (2.4) Recall that π1=π◦ϕ1 and define f1≔ϕ1−1◦f1′◦ϕ1 . If we apply ϕ1−1 to (2.4), we deduce that f1 satisfy condition (2.3) above. Step 3. We claim: There exists a polynomial map f2:Rn→Rnsuch that Rn⧹(Int(K)∩π1−1(π1(X)¯zar)∩(π2−1(π2(X)¯zar)))⊂S2≔f2(S1)⊂Rn⧹X. (2.5) Define K2≔ϕ2(K) and X2≔ϕ2(X) . Write K2≔{𝚡n−1−λ2𝚡n≥0,𝚡n−1+μ2𝚡n≥0} for some real numbers λ2,μ2>0 and consider f2′:Rn→Rn,x≔(x1,…,xn)↦(x1,…,xn−1,xn(1−H22(xn−1,xn)G22(x′))), where H2≔(𝚡n−1−λ2𝚡n)(𝚡n−1+μ2𝚡n)∈R[𝚡n−1,𝚡n] and G2∈R[𝚡′] is a polynomial equation of π(X2)¯zar . Let us check: Rn⧹(Int(K2)∩π−1(π(X2)¯zar)∩ϕ2(π1−1(π1(X)¯zar)))⊂f2′(ϕ2(S1))⊂Rn⧹X2. (2.6) Proceeding as in Step 1 for the vector v⃗1 , we have f2′(Rn⧹Int(K2))=Rn⧹(Int(K2)∩π−1(π(X2)¯zar)). By (2.3) Rn⧹(Int(K)∩π1−1(π1(X)¯zar))=S1 , so Rn⧹(Int(K2)∩ϕ2(π1−1(π1(X)¯zar)))=ϕ2(S1). As π−1(π(X2)¯zar)={G2=0} , we have f2′∣π−1(π(X2)¯zar)=idπ−1(π(X2)¯zar) . Thus, f2′∣X2=idX2 and π−1(π(X2)¯zar)⧹(Int(K2)∩ϕ2(π1−1(π1(X)¯zar)))⊂f2′(ϕ2(S1)) . Consequently, Rn⧹(Int(K2)∩π−1(π(X2)¯zar)∩ϕ2(π1−1(π1(X)¯zar)))=(Rn⧹(Int(K2)∩π−1(π(X2)¯zar)))∪(π−1(π(X2)¯zar)⧹(Int(K2)∩ϕ2(π1−1(π1(X)¯zar))))⊂f2′(ϕ2(S1)). As S1⊂Rn⧹X , we have ϕ2(S1)⊂Rn⧹X2 , so f2′(ϕ2(S1))⊂f2′(Rn⧹X2) . Let us check: f2′−1(X2)=X2 . Once this is proved, f2′(Rn⧹X2)⊂Rn⧹X2 and the remaining part of Equation (2.6) holds. Pick y≔(y′,yn)∈Rn such that f2′(y)∈X2 . Then y′=π(f2′(y))∈π(X2) , so G2(y′)=0 and y=f2′(y)∈X2 . Thus, f2′−1(X2)⊂X2 . As f2′∣X2=idX2 , we deduce f2′−1(X2)=X2 . Recall that π2=π◦ϕ2 and define f2≔ϕ2−1◦f2′◦ϕ2 . If we apply ϕ2−1 to (2.6), we deduce that f2 satisfy condition (2.5) above. Step 4. We proceed similarly with the remaining vectors v⃗j for j=1,…,r to obtain polynomial maps fj:Rn→Rn such that Rn⧹(Int(K)∩⋂i=1jπi−1(πi(X)¯zar))⊂Sj≔fj(Sj−1)⊂Rn⧹X, where S0≔Rn⧹Int(K) . If we write Xj≔ϕj(X) and Kj≔ϕj(K)={𝚡n−1−λj𝚡n≥0,𝚡n−1+μj𝚡n≥0} for some real numbers λj,μj>0 , the sought polynomial map fj is the composition fj≔ϕj−1◦fj′◦ϕj , where fj′:Rn→Rn,x≔(x1,…,xn)↦(x1,…,xn−1,xn(1−Hj2(xn−1,xn)Gj2(x′))), Hj≔(𝚡n−1−λj𝚡n)(𝚡n−1+μj𝚡n)∈R[𝚡n−1,𝚡n] and Gj∈R[𝚡′] is a polynomial equation of π(Xj)¯zar . Step 5. Conclusion. For j=r , we have Rn⧹X=Rn⧹(Int(K)∩⋂j=1rπj−1(πj(X)¯zar))⊂Sr⊂Rn⧹X. Thus, Rn⧹X=Sr=(fr◦⋯◦f1)(Rn⧹Int(K)) . By [14, Theorem 1.3], the semialgebraic set Rn⧹Int(K) is a polynomial image of Rn , so Rn⧹X is a polynomial image of Rn , as required. □ 3. Resolution of the indeterminacy using double oriented blowings-up In this section, we prove Lemma 1.8, that is we can make regular each regulous function (or even each locally bounded rational function) on R2after composing it with a suitable generically finite surjective regular map ϕ:R2→R2 . It is known in general that a regulous function can be made regular after composition with a finite sequence of algebraic blowings-up, but of course the source space is no longer the plane. We prove in this section that we can achieve the desired generically finite surjective regular map via a finite composition of double oriented blowings-up. Let us begin with some examples. Examples 3.1 (1) Consider the regulous function f:R2→R given by f(x,y)≔x3x2+y2 . The polynomial map φ:R2→R2 defined by φ(u,v)≔(u(u2+v2),v(u2+v2)) is surjective, and (f◦φ)(u,v)=u3 is polynomial. Note that φ is a polynomial homeomorphism. (2) Consider the rational function on R2 given by g(x,y)≔xx2+y2 defined and continuous on R2⧹{(0,0)} . The polynomial map ϕ:R2→R2 defined by ϕ(u,v)≔(v3(uv+1),v(uv−1)) is surjective. Actually, if ϕ(u,v)=(a,b)≠(0,0) , then v≠0 and uv=1+b/v . Thus, it is enough to prove that there exists v∈R⧹{0} satisfying 2v3+bv2−a=0 . This is true if a≠0 because the previous equation in v has odd degree whereas, in case a=0 , we take v=−b2 . In addition, the composition (g◦ϕ)(u,v)=v(uv+1)v4(uv+1)2+(uv−1)2 is regular. However, the surjective regular map ϕ is not proper because ϕ−1((0,0))={v=0} . Remark 3.2 The previous surjective regular map ϕ:R2→R2 has the form (u,v)↦(Pk(u,v)Q(u,v),P(u,v)R(u,v)), where k≥1 is an odd integer and P,Q,R∈R[𝚞,𝚟] are polynomials such that {PQ=0}∩{R=0}=∅ . The latter condition enables us to soften g by making that the denominator has empty zero-set after increasing numerator’s multiplicity. Another (more complicate!) possible surjective regular map such that g◦ϕ is regular could be ϕ:R2→R2,(u,v)↦((u+v)k(uv2(u+v)−1)),(u+v)(uv2(u+v)+1). A natural question that arise at this point is the following. Question 3.3 Which rational functions f≔PQ∈R(𝚞,𝚟) can be softened by means of a surjective regular map ϕ:R2→R2? Remark 3.4 A necessary condition is Z(Q)⊂Z(P), but it is not sufficient as one can check considering the rational function f(x,y)≔x2+y2x4+y4 . If we compute the multiplicity of f◦ϕ at any preimage of the origin under a surjective regular function ϕ≔(ϕ1,ϕ2):R2→R2 , we observe that it is always negative (because the multiplicity of the denominator doubles the one of the numerator), so f◦ϕ cannot be regular. In the following, we will pay a special attention to locally bounded rational functions, that is, rational functions on R2 that are locally bounded in a neighborhood of its indeterminacy points. As it happens with regulous functions, a locally bounded rational function on R2 admits only a finite number of poles. Lemma 3.5 Let f≔PQbe a locally bounded rational function on R2where P,Q∈R[𝚡,𝚢]are relatively prime polynomials. Then the zero-set of Qconsists of finitely many points. Proof As f=PQ is locally bounded and the polynomials P,Q∈R[𝚡,𝚢] are relatively prime, one has {Q=0}⊂{P=0} . Let Q1 be an irreducible factor of Q . Using the criterion [1, Theorem 4.5.1] for principal real ideals of R[𝚡,𝚢] , we deduce that the ideal (Q1)R[𝚡,𝚢] is not real. Otherwise, the ideal J({Q1=0}) of all polynomials of R[𝚡,𝚢] vanishing identically on {Q1=0} is (Q1)R[𝚡,𝚢] , so Q1 divides P against the coprimality of P and Q . Consequently, the zero-set of Q1 has by [1, Theorem 4.5.1] dimension ≤0 , that is, it is a finite set. Applying this to all the irreducible factors of Q , we conclude that {Q=0} is a finite set, so f has only finitely many poles, as required.□ As we have already mentioned, regulous functions can be made regular after composition with a finite sequence of blowings-up along smooth algebraic centers [15]. Actually those rational functions on Rn that become regular after such compositions are exactly the locally bounded rational functions. Theorem 3.6 Let fbe a rational function on Rn . Then fis locally bounded if and only if there exists a finite sequence σof blowings-up along non-singular centers such that f◦σis regular. Sketch of proof Assume first that such a sequence exists. Then the preimage under σ of a compact Euclidean neighborhood of an indeterminacy point of f is compact by properness of σ . Thus, the continuous function f◦σ is bounded on such preimage. As a consequence f is locally bounded at the corresponding indeterminacy point. The converse implication can be proved as [15, Theorem 3.11]. Let us provide an idea on how the proof works. By means of a finite sequence σ:M→Rn of blowings-up along non-singular centers, it is possible to make normal crossings the numerator and the denominator of f . If we choose a suitable local system of coordinates at a point p∈M , the rational function f◦σ is equal in such a neighborhood of p to a unit times a quotient of products of the variables (with exponents). As f◦σ is also locally bounded, the denominator must divide the numerator, so f◦σ is regular at p , as required.□ 3.1. Double oriented blowings-up In the following, we restrict our target to some particular type of surjective regular maps: compositions of finitely many double oriented blowings-up. The oriented blowing-up π+ of R2 at the origin corresponds to the passage to polar coordinates and it can be achieved using the analytic map π+:[0,+∞)×S1→C≡R2,(ρ,v)↦ρv. The previous map provides a Nash diffeomorphism between the cylinder C≔(0,+∞)×S1 and the punctured plane P≔R2⧹{(0,0)} whose inverse is (π+∣C)−1:P→S,u↦(∥u∥,u∥u∥) . The inverse image of the origin under π+ is the circle {0}×S1 . We choose a rational parameterization of S1 (consider for instance the stereographic projection from the North pole) and we obtain the map π0:[0,+∞)×R→R2,(ρ,t)↦(ρ2tt2+1,ρt2−1t2+1). The image of this map is R2⧹{(0,t):t>0} . We consider the regular extension π of π0 to R2 , whose image is R2 because π(ρ,0)=(0,−ρ) for each ρ∈R . Definition 3.7 The surjective regular map π:R2→R2,(ρ,t)↦(ρ2tt2+1,ρt2−1t2+1) is called the double oriented blowing-up of R2at the origin. The double oriented blowing-up of R2 at an arbitrary point p∈R2 is the composition τ0p→◦π◦τ−0p→ where τv⃗ denotes the translation τv⃗:R2→R2,x↦x+v⃗ for each v⃗∈R2 . Let Π:C2⧹{t2+1=0}→C2 be the natural rational extension of π to C2⧹{t2+1=0} . Lemma 3.8 The double oriented blowing-up of R2at the origin π:R2→R2satisfies the following properties: πis a surjective regular map. π−1((0,0))={ρ=0}is the t-axis. π∣{t=0}:{t=0}→{x=0},(ρ,0)→(0,−ρ)provides a bijection between the ρ-axis and the y-axis. The determinant of the Jacobian matrix of πat the point (ρ,t)values 2ρt2+1 . Both π∣R2⧹{ρ=0}and Π∣C2⧹{(t2+1)ρ=0}are local diffeomorphisms. Both restrictions π∣R2⧹{ρt=0}:R2⧹{ρt=0}→R2⧹{x=0}and Π∣:C2⧹{ρt(t2+1)=0}→C2⧹{x(x2+y2)=0}are double covers. In fact, π(−ρ,−1t)=π(ρ,t) and Π(−ρ,−1t)=Π(ρ,t) . We can relate the double oriented blowing-up to the classical blowing-up as follows. Denote σ:M→R2 the blowing-up of R2 at the origin. We describe M as the subset of R2×P1 given by the equation M≔{((x,y),[u:v])∈R2×P1:xv=yu}, whereas σ:M→R2,((x,y),[u:v])↦(x,y) . Lemma 3.9 The map ψ:R2→M,(ρ,t)↦((ρ2tt2+1,ρt2−1t2+1),[2t:t2−1])satisfies the following properties: It is a surjective regular map. The restriction ψ∣{t=0}:{t=0}→L≔{((0,−ρ),[0:1]):ρ∈R}⊂Mis bijective. The restriction ψ∣R2⧹{t=0}:R2⧹{t=0}→M⧹Lis a double cover and it holds ψ(−ρ,−1t)=ψ(ρ,t) . π=σ◦ψ . Example 3.10 Consider the locally bounded rational function on R2 given by the formula f(x,y)≔x2x2+y2 . Then (f◦π)(ρ,t)=4t24t2+(t2−1)2 is regular. 3.2. Multiplicity of an affine complex curve at a point For the sake of completeness, we recall how one can compute the multiplicity at a point of a polynomial equation of a planar curve. In the following, we use the letter ω to denote the order of a power series. Let Q∈C[𝚡,𝚢] be a non-constant polynomial and denote C≔{Q=0}⊂C2 . Let p∈C be a point of C and assume after a translation that p is the origin. Let Γ1,…,Γr be the complex branches of C at the origin and let γi≔(γi,1,γi,2)∈C{𝚜}2 be a primitive parameterization of Γi . We mean with primitive parameterization of Γi a couple of convergent power series γi,1,γi,2∈C{𝚜} such that both series do not belong simultaneously to the ring C{𝚜k} for any k≥2 , Γi is the germ at the origin of the set {(γi,1(s),γi,2(s)):∣s∣<ε} for some ε>0 small enough. To compute the multiplicity mult0(Q) of Q at the origin, write Q as the sum of its homogeneous components Q=Qm+Qm+1+⋯+Qd , where each Qk is either zero or a homogeneous polynomial of degree k and Qm≠0 . Then mult0(Q)=m . It is possible to express mult0(Q) in terms of some invariant associated to the complex branches Γ1,…,Γr of C at the origin. Write Q≔P1e1⋯Pses , where each Pi∈C[𝚡,𝚢] is an irreducible polynomial and ei≥1 . As mult0(Q)=e1mult0(P1)+⋯+esmult0(Ps) , it is enough to express mult0(Pi) in terms of the complex branches of C that correspond to the factor Pi . Thus, we assume in the following that Q is an irreducible polynomial. After a linear change of coordinates, we may assume that Q is a regular series with respect to 𝚢 of order mult0(Q) , that is, ω(Q(0,𝚢))=mult0(Q)=m . By Weierstrass’ Preparation Theorem Q=Q*U , where Q*∈C{𝚡}[𝚢] is a distinguished polynomial of degree m and U∈C{𝚡,𝚢} is a unit, that is, U(0,0)≠0 . Let Q1*,…,Qℓ*∈C{𝚡}[𝚢] be the irreducible factors of Q* in C{𝚡}[𝚢] , which are distinguished polynomials with respect to 𝚢 such that deg𝚢(Qi*)=ω(Qi*) . As Q∈C[𝚡,𝚢] is irreducible, it has no multiple factor in C{𝚡}[𝚢] . Thus, Q*=Q1*⋯Qℓ* and mult0(Q)=mult0(Q*)+mult0(U)=mult0(Q*)=∑j=1ℓmult0(Qj*). Consequently, it is enough to compute mult0(P) for an irreducible distinguished polynomial P∈C{𝚡}[𝚢] with deg𝚢(P)=ω(P) . Let ξ∈C{𝚡*} be a Puiseux root of P . It holds that deg𝚢(P) coincides with the smallest q≥1 such that ξ≔β(𝚡1/q)∈C{𝚡1/q} for some β∈C{𝚜} . In addition, the polynomial P is associated with exactly one complex branch Γ for which any primitive parameterization γ≔(γ1,γ2) satisfies q=min{ω(γ1),ω(γ2)} . Thus, if we define mult(Γ) as the minimum order of the components of a primitive parameterization of Γ , we have mult0(P)=q=mult(Γ) . 3.3. Alternative resolution of indeterminacy We know that regulous functions become regular after a finite composition of blowings-up (and this is even true for locally bounded functions as claimed in Theorem 3.6). However, the source space of the regular function obtained is no longer the same as that of the regulous function we started with. The following result, which is proved below, is an improvement of Lemma 1.8 from which we deduce it. Theorem 3.11 Let fbe a locally bounded rational function on R2 . Then there exists a finite composition ϕ:R2→R2of finitely many double oriented blowings-up and polynomial isomorphisms such that f◦ϕis a regular function on R2 . The strategy to prove Theorem 3.11 is to follow the same process as in classical resolution of indeterminacy [18, Section 5.3], but replacing the usual blowing-up along a point by the double oriented blowing-up. The difficulty is to control the number of indeterminacy points, since the double oriented blowing-up is no longer an isomorphism outside the center and it is generically a double cover. A crucial step consists in separating all complex branches passing through an indeterminacy point with different tangents. Note, however, that after applying the double oriented blowing-up at an indeterminacy point, the preimage of that point consists of several points each of these with less complex branches than the original indeterminacy point, but possibly with more different tangents! To face this issue, we state first an auxiliary result that will be needed constantly in the proof of Theorem 3.11. Lemma 3.12 Let F≔{p1,…,pn}⊂R2be a finite set and let L′⊂R2be a line. Pick i0∈{1,…,n}and q∈L′ . Then there exists a polynomial isomorphism φ:R2→R2that maps Finto a half-line in L′issued from φ(pi0)=q . Moreover, consider an additional line Lthrough pi0 , a collection {H1,…,Hr}of complex lines through pi0different from L , a collection {H1′,…,Hs′}of complex lines through qdifferent from L′ .Then we may assume in addition that the differential of φat pi0maps Lonto L′and the polynomial extension Φ:C2→C2of φto C2does not map any of the lines Hithrough pi0into the finite union ⋃j=1sHj′of lines through q . Proof After an affine change of coordinates, we may assume that L′ is the x-axis and the restriction to F of the projection of R2 onto the x-axis is injective. Write pi≔(ai,bi) and let P∈R[𝚡] be a univariate polynomial such that P(ai)=bi for i≠i0 and P(ai0)≠bi0 . Then the polynomial isomorphism (x,y)↦(x,y−P(x)) maps the points of F except for pi0 into the x-axis. Let R≠L be a line through pi0 non-parallel to the x-axis such that the intersection R∩{y=0}={(λ,0)} leaves all the points pi for i≠i0 in the half-line {y≥λ} . Consider an affine change of coordinates that keeps the x-axis invariant and maps the line R to the y-axis. After this change of coordinates pi0=(0,μi0) for some μi0≠0 and L is not parallel to the y-axis. Let Q∈R[𝚡] be a polynomial such that its graph {y−Q(x)=0} passes through the points pi and whose tangent at the point pi0 is L . The polynomial isomorphism φ:R2→R2,(x,y)↦(x,y−Q(x)) maps the point pi0 to the origin, keeps the points pi for i≠i0 inside the x-axis, and the differential of the polynomial isomorphism φ at pi0 maps L to the x-axis (that is, to the line L′ ). After a translation parallel to the x-axis, we may assume in addition that φ maps pi0 to q . As we have much freedom to choose Q , we may assume that the polynomial extension Φ:C2→C2 of φ to C2 does not map any of the lines Hi through pi0 into the finite union ⋃j=1sHj′ of lines through q , as required.□ Proof of Theorem 3.11 The set of indeterminacy points of f coincides with the zero-set {Q=0} , and this is a finite set by Lemma 3.5. We may assume in addition that Q is non-negative on R2 . Consider the complex algebraic curve C≔{z∈C2:Q(z)=0} , which is invariant under complex conjugation in C2 . This means that if p∈C is a real point and Γ is a complex branch of C at p , then the conjugated branch Γ¯ is also a complex branch of C at p . If Tp is the tangent line to Γ at p , then its complex conjugated Tp¯ is the tangent line to Γ¯ at p . Thus, Tp=Tp¯ if and only if Tp admits a linear equation with real coefficients. We are going to solve the indeterminacy points of f by applying a finite chain of double oriented blowings-up centered at the indeterminacy points and polynomial isomorphisms of R2 like those provided by Lemma 3.12 (that maps a finite subset of R2 inside a half-line). Denote F≔{p1,…,pn} the set of indeterminacy points of f . For each point p∈F , let multp(Q) be the multiplicity of Q at p . Let pi0∈F be such that M≔multpi0(C)≥multp(C) for each p∈F . By Lemma 3.12, there exists a polynomial isomorphism φ:R2→R2 such that if we substitute f by f◦φ−1 we may assume that the indeterminacy points of f belong to the negative half y-axis {(0,y):y≤0} and pi0 is the origin. In case the germ Cpi0 has only one tangent line at the origin, we know that it has to be a real line. If C has a real tangent line at the origin, we may assume in addition by Lemma 3.12 that this tangent line is the y-axis, whereas the remaining ones are different from 𝚡+𝚒𝚢=0 and 𝚡−𝚒𝚢=0 . Note that the origin is now an indeterminacy point of f with maximal multiplicity and the y-axis is one of its tangents. Consider the composition f◦π , where π denotes the double oriented blowing-up introduced in Definition 3.7. Denote F′≔{pi:i≠i0} and G≔π−1(F′) and let Π:C2⧹{t2+1=0}→C2 be the natural rational extension of π to C2⧹{t2+1=0} . As π∣R2⧹{ρ=0} is a local diffeomorphism and π∣{t=0}:{t=0}→{x=0} is a bijection between the ρ-axis and the y-axis, the sets G and F′ have the same number of points and, if q∈G , then Π−1(C) has at q the same number of complex branches and different tangents as C has at π(q) . Let us analyze now what happens at the origin. Let Γ1,…,Γr be the complex branches of C at the origin. The germ C0 equals the union of branches ⋃i=1rΓi and M≔mult0(Q)=∑i=1reimult(Γi) for some positive integers ei . Fix a complex branch Γi and let Ti be the tangent line to Γi . We have Π−1(Γi)={ρ=0}ai1∪{ρ=0}ai2∪Λi1∪Λi2 , where ai1,ai2∈{ρ=0}⊂C2⧹{t2+1=0} and Λij is a complex branch at aij (as we will see in the following, sometimes there are exactly two complex branches Λi1,Λi2 and sometimes there is only one and we will consider Λi1=Λi2 ). As π∣{t=0}:{t=0}→{x=0},(ρ,0)→(0,−ρ) and π∣R2⧹{ρt=0}:R2⧹{ρt=0}→R2⧹{x=0} is a double cover, one has exactly one point ai≔ai1=ai2 if Ti is the y-axis and two different points ai1,ai2 otherwise. Let us analyze the germs Λij for j=1,2 . Choose a primitive parameterization γi≔(γi1,γi2) of Γi . We may assume that it has the form {γi1≔𝚜ki,γi2≔ci𝚜ℓi+⋯,ifki≤ℓior{γi1≔ci𝚜ki+⋯,γi2≔𝚜ℓi,ifℓi<ki with ki,ℓi≥1 and ci∈C⧹{0} . Recall that mult(Γi)=min{ki,ℓi} . Observe that the tangent line to Γi at the origin is Ti≔{{𝚢=0}ifki<ℓi,{ci𝚡−𝚢=0}ifki=ℓi,{𝚡=0}ifki>ℓi. If we apply the double oriented blowing-up π at the origin, we make x=ρ2tt2+1,y=ρt2−1t2+1, so ρ2=x2+y2 and the order of ρ for each complex branch Λij with respect to the variable 𝚜 is min{ki,ℓi} (recall that 1+ci2≠0 since the tangent line to Γi is different from 𝚡+𝚒𝚢=0 and 𝚡−𝚒𝚢=0 ). We distinguish several situations: If ki<ℓi , then ρ=±𝚜ki+⋯ and t=±1+⋯ . Thus, Λi1 and Λi2 are two complex branches at the points ai1≔(0,1) and ai2≔(0,−1) of multiplicities smaller than or equal to ki=mult(Γi) . If ki=ℓi , then ρ=±𝚜ki(1+ci2)+⋯ . As 1+ci2≠0 , one has t=(±(1+ci2)+ci)+⋯ . Thus, Λi1 and Λi2 are two complex branches at the points ai1≔(0,(1+ci2)+ci) and ai2≔(0,−(1+ci2)+ci) of multiplicities smaller than or equal to ki=mult(Γi) . If ki>ℓi , then ρ=−𝚜ℓi+⋯ and t=ci12𝚜ki−ℓi+⋯ . Thus, there is only one complex branch that we write Λi≔Λi1=Λi2 is a complex branch at the point ai≔ai1=ai2=(0,0) of multiplicity smaller than or equal to ℓi=mult(Γi) . Starting from the locally bounded rational function f=P/Q , we have constructed the rational function f◦π=P′/Q′ for some polynomials P′,Q′∈R[𝚛,𝚝] . As f◦π remains locally bounded, {q∈R2:Q′(q)=0} is a finite set. We have seen above that if two tangent lines Ti and Ti′ are different, so are the points aij and ai′j′ for j,j′∈{1,2} . Thus, multaij(Q′)=∑i′:Ti′=Tiei′mult(Λi′sj)≤∑i′:Ti′=Tiei′mult(Γi). Consequently, if C has more than one tangent line at the origin, multaij(Q′)≤∑i′:Ti′=Tiei′mult(Γi)<∑i=1reimult(Γi)=mult0(Q)=M for each point aij and, although we have increased the cardinality of the zero set of the new denominator Q′ , we have dropped the number of real points on which the denominator has multiplicity M . Next assume that C has only one tangent line at the origin (which is the y-axis). As in the usual desingularization process, the multiplicity will decrease, but after a finite number of steps. Namely, the unique tangent to C at the origin is {𝚡=0} and ki>ℓi (case (iii) above) for i=1,…,r . For each complex branch Λi ( =Λi1=Λi2 ) at the origin, we have the following two possibilities (after reseting the names of the variables and calling 𝚡 the first variable and 𝚢 the second variable): (iii.1) if ki′≔ki−ℓi<ℓi , then mult(Λi)<mult(Γi) and the tangent line to Λi is {𝚢=0} , (iii.1) if ki′≔ki−ℓi≥ℓi , then mult(Λi)=mult(Γi) and we can parameterize the complex branch Λi by a primitive parameterization λi≔(λi1,λi2)∈C{𝚜}2 such that λi1≔di𝚜ki′+⋯ and λi2≔𝚜ℓi . In this case, the tangent line is either {𝚡=0} if ki′>ℓi or {𝚢−di𝚡=0} if ki′=ℓi . If situation (iii.1) arises for one of the complex branches of C at the origin, we have mult0(Q′)<mult0(Q)=M and we have dropped the number of real points p∈{Q=0} such that multp(Q)=M . Otherwise, all the branches Λi are in situation (iii.2) and mult0(Q′)=mult0(Q)=M . The worst situation arises if C′≔{z∈C2:Q′(z)=0} has only one tangent line the origin. We can apply to Q′ the previous procedure at the origin: We construct (using Lemma 3.12) a polynomial isomorphism that maps all the real zeros of Q′ into the half line {(0,y):y≤0} and keeps the origin invariant. If C′ has a real tangent line, we assume that one of such real tangent lines has equation 𝚡=0 . We apply the double oriented blowing-up at the origin and repeat the previous discussion. As the parametrization of the involved complex branches are primitive and Λi∩R2={(0,0)} , if we follow the previous algorithm we realize that the only possibility to get (apparently) stuck in situation (iii.2) for all the complex branches (which is the only case that do not drop the multiplicity) is that ℓi=1 for each i=1,…,r . Assume that such is the case and choose a parameterization γi≔(γi1,γi2) of the complex branch Γi (at the origin), that is, {γi1≔c2𝚜2+⋯+cm𝚜m+ci,m+1*𝚜m+1+⋯,γi2≔𝚜, where c2,…,cm∈R (these coefficients are the same for each i because we are ‘apparently’ stuck in situation (iii.2)), ci,m+1∈C for i=1,…,r and for instance c1,m+1*∈C⧹R (because the branches of C are all complex as C∩R2 is a finite set). Thus, if we apply our algorithm m+1 times at the origin, we achieve from the complex branch Γ1 a complex branch Θ1 that admits a primitive parameterization θ1≔(θ11,θ12)∈C{𝚜}2 such that {θ11≔c1,m+1*𝚜+⋯,θ12≔𝚜. The tangent line to Θ1 is {𝚡−c1,m+1*𝚢=0} whereas the tangent line to its conjugated complex branch Θ¯1 is {𝚡−c¯1,m+1*𝚢=0} , which is different from {𝚡−c1,m+1*𝚢=0} . Hence, it is not possible to get stuck in situation (iii.2) indefinitely. Thus, we are always able to reduce the number of points of multiplicity M with respect to the denominator, using finitely many (polynomial isomorphisms and) double oriented blowings-up. Consequently, after applying suitably the algorithm above finitely many times, we find a regular map ϕ:R2→R2 which is a finite composition of (polynomial isomorphisms and) double oriented blowings-up such that f◦ϕ is a locally bounded rational function whose denominator has empty zero-set, that is, f◦ϕ is a regular function, as required.□ 3.4. Arc-lifting property As we have seen above, any locally bounded rational function on the plane becomes regular after a composition with a generically finite surjective regular map, which can be chosen as a composition of a finite sequence of double oriented blowings-up and polynomial isomorphisms. We have seen also in Example 3.1 that other rational functions may becomes regular after composing with a surjective regular map. We propose here to add a geometric condition to the involved surjective regular maps in order to characterize this property. Definition 3.13 A map ϕ:R2→R2 satisfies the arc-lifting property if for any analytic arc γ:(R,0)→(R2,p) , where p∈R2 , there exists an analytic arc γ˜:(R,0)→(R2,p˜) , where p˜∈R2 , such that ϕ◦γ˜=γ . Arc-lifting property is a natural condition when dealing with blowing-analytic equivalence [16]. In particular, a blowing-up along a non-singular center satisfies the arc-lifting property, but also a blowing-up along an ideal, or even a real modification [16, p. 99]. Proposition 3.14 A double oriented blowing-up satisfies the arc-lifting property. It is possible to use Lemma 3.9 together with the arc-lifting property of the blowing-up along a point to prove the arc-lifting property of a double oriented blowing-up. However, we prefer to produce a direct elementary proof of this fact in order to enlighten the special behavior of a double oriented blowing-up, namely to describe in full details when an arc admits one or two liftings. Proof of Proposition 3.14 Let γ≔(γ1,γ2):(R,0)→(R2,p) be an analytic arc in R2 defined on a neighborhood of the origin in R . If we consider the double oriented blowing-up at the origin of R2 , the statement is obvious if p is not the origin. Assume in the following: p≔(0,0) and consider γ1,γ2 as elements of the ring R{𝚜} of analytic series in one variable with coefficients in R . By definition of the double oriented blowing-up at the origin, we must find analytic series ρ,t∈R{𝚜} such that (ρ2tt2+1,ρt2−1t2+1)=(γ1,γ2), (3.1) hence the tuple γ˜≔(ρ,t) satisfies the required conditions. If γ1=0 , we take t≔0 and ρ≔−γ2 . If γ2=0 , we take t≔1 and ρ≔γ1 . Thus, we assume in the following γ1,γ2≠0 and write γ1≔a𝚜k+⋯andγ2≔b𝚜ℓ+⋯, where a,b∈R⧹{0} and ℓ,k are positive integers. Using Equation (3.1), we deduce ρ=εγ12+γ22andt=δρ+γ2ρ−γ2 (3.2) for some ε,δ∈{−1,+1} , as soon as the previous expressions have sense and provide analytic series. We analyze the following three situations: Assume first that k<ℓ . Then ρ2=γ12+γ22=a2𝚜2k(1+⋯) so that the two possible analytic choices for ρ are given by ρ≔εa𝚜k1+⋯=εa𝚜k(1+⋯), where ε=±1 . Analogously t2=ρ+γ2ρ−γ2=(εa𝚜k+⋯)+(b𝚜ℓ+⋯)(εa𝚜k+⋯)−(b𝚜ℓ+⋯)=1+⋯. Therefore, we obtain two possible analytic solutions t≔δ1+⋯=δ(1+⋯) , where δ=±1 . The analytic arc γ˜=(ρ,t) is a lifting of γ if Equation (3.1) is satisfied, which is the case if and only if εδ=+1 . Thus, we have obtained two analytic liftings for γ . Assume next k=ℓ . Proceeding as in the previous case ρ≔εa2+b2𝚜k+⋯ (with ε=±1 ), whereas t2=εa2+b2+b+⋯εa2+b2−b+⋯=cε+⋯ and cε≔a2+b2+εba2+b2−εb>0 for both choices of ε=±1 , so t≔δcε+⋯ for some δ=±1 . Using again Equation (3.1), there exist two analytic liftings of γ corresponding to the choice εδ=sign(a) . The situation in the remaining case k>ℓ is slightly different. The analytic series ρ≔εγ12+γ22=ε∣b∣𝚜ℓ+⋯ for some ε=±1 , whereas t must satisfy the equation t2=ρ+γ2ρ−γ2=(ε∣b∣+⋯)+(b+⋯)(ε∣b∣+⋯)−(b+⋯). (3.3) The previous equation has an analytic solution t if and only if ε=−sign(b) , that is, ρ≔−b𝚜ℓ+⋯ . We rewrite Equation (3.3) as t2=ρ+γ2ρ−γ2=−sign(b)γ12+γ22+sign(b)γ22−sign(b)γ12+γ22−sign(b)γ22=1+(γ1/γ2)2−11+(γ1/γ2)2+1=a24b2𝚜2(k−ℓ)+⋯. Thus, there exist two possible analytic solutions to (3.3) given by the formula t≔δa2b𝚜k−ℓ+⋯, where δ=±1 . The couple (ρ,t) provides analytic lifting of γ if Equation (3.1) holds and this happens if and only if δ=−1 , that is, t≔−a2b𝚜k−ℓ+⋯ . Consequently, in this case, we have only one analytic lifting of γ . After the analysis made, we conclude that the double oriented blowing-up at the origin satisfies the arc-lifting property, as required.□ The last result allows us to establish that the locally bounded rational functions are exactly those rational functions on R2 that become regular after composition with a surjective regular map satisfying the arc-lifting property (compare this with Theorem 3.6). Theorem 3.15 Let fbe a rational function on R2 . There exists a surjective regular map ϕ:R2→R2satisfying the arc-lifting property and such that f◦ϕis regular if and only if fis locally bounded. Proof By Theorem 3.11 and Proposition 3.14, it is enough to prove that if ϕ:R2→R2 is a surjective regular map satisfying the arc-lifting property and f◦ϕ is a regular function, then f is locally bounded. Otherwise, there exists an analytic arc γ:(R,0)→(R2,p) such that (f◦γ)(s) goes to infinity as s tends to zero. By the arc-lifting property, there exists an analytic arc γ˜:(R,0)→(R2,p˜) such that ϕ◦γ˜=γ . In particular, the analytic arc f◦γ=f◦ϕ◦γ˜ is not bounded at the origin, which is a contradiction.□ Funding The first author was supported by Spanish GRAYAS MTM2014-55565-P, MTM2017-82105-P and Grupos UCM 910444. This article has been mainly written during several research stays of the first author in the Institut de Recherche Mathématiques de Rennes of the Université de Rennes 1. The first author would like to thank the department for the invitation and the very pleasant working conditions. References 1 J. Bochnak , M. Coste and M.-F. Roy , Real algebraic geometry. Ergeb. Math. Vol. 36 , Springer-Verlag , Berlin , 1998 . 2 J. F. Fernando , On the one dimensional polynomial and regular images of Rn , J. Pure Appl. Algebra 218 ( 2014 ), 1745 – 1753 . Google Scholar Crossref Search ADS 3 J. F. Fernando , On Nash images of Euclidean spaces. Preprint RAAG ( 2016 ). arXiv:1503.05706. 4 J. F. 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Ueno , On unbounded convex polyhedra as polynomial images of Rn , Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) XXX ( 2017 ), XXX – XXX . (to appear). 10 J. F. Fernando and C. Ueno , On complements of convex polyhedra as polynomial and regular images of Rn , Int. Math. Res. Not. IMRN 2014 ( 2014 ), 5084 – 5123 . Google Scholar Crossref Search ADS 11 J. F. Fernando and C. Ueno , On the complements of 3-dimensional convex polyhedra as polynomial images of R3 , Int. J. Math . 25 ( 2014 ), 1450071 (18 pages). Google Scholar Crossref Search ADS 12 J. F. Fernando and C. Ueno , On the set of points at infinity of a polynomial image of Rn , Discrete Comput. Geom. 52 ( 2014 ), 583 – 611 . Google Scholar Crossref Search ADS 13 J. F. Fernando and C. Ueno , A short proof for the open quadrant problem , J. Symbolic Comput. 79 ( 2017 ), 57 – 64 . Google Scholar Crossref Search ADS 14 J. F. Fernando and C. Ueno , On complements of convex polyhedra as polynomial images of Rn . Preprint RAAG ( 2015 ). arXiv:1412.5107. 15 G. Fichou , J. Huisman , F. Mangolte and J.-P. Monnier , Fonctions Régulues , J. Reine Angew. Math. 718 ( 2016 ), 103 – 151 . 16 T. Fukui and L. Paunescu , On blow-analytic equivalence. ‘Arc spaces and additive invariants in real algebraic and analytic geometry’, 87–125, Panor. Synthèses, 24, Soc. Math. France, Paris ( 2007 ). 17 J. M. Gamboa , Algebraic images of the real plane, Reelle algebraische Geometrie, June, 10th–16th ( 1990 ), Oberwolfach. 18 T. de Jong and G. Pfister , Local analytic geometry. Basic theory and applications. Advanced Lectures in Mathematics , Friedr. Vieweg & Sohn , Braunschweig , 2000 . 19 W. Kucharz , Rational maps in real algebraic geometry , Adv. Geom. 9 ( 2009 ), 517 – 539 . Google Scholar Crossref Search ADS 20 W. Kucharz , Approximation by continuous rational maps into spheres , J. Eur. Math. Soc. 16 ( 2014 ), 1555 – 1569 . Google Scholar Crossref Search ADS 21 A. Tarski , A decision method for elementary algebra and geometry. Prepared for publication by J.C.C. Mac Kinsey, Berkeley, 1951 . 22 C. Ueno , A note on boundaries of open polynomial images of R2 , Rev. Mat. Iberoam. 24 ( 2008 ), 981 – 988 . Google Scholar Crossref Search ADS 23 C. Ueno , On convex polygons and their complements as images of regular and polynomial maps of R2 , J. Pure Appl. Algebra 216 ( 2012 ), 2436 – 2448 . Google Scholar Crossref Search ADS Appendix A. The open quadrant As announced in the Introduction, we represent the open quadrant Q≔{𝚡>0,𝚢>0} as the image of simple regulous and regular maps f,g:R2→R2 . A.1. A regulous map whose image is the open quadrant Consider the regulous map f≔(f1,f2):R2→R2,(x,y)↦(x2(x2y2x2+y2)+(xy+1)21+(x+y)2,y2(x2y2x2+y2)+(xy+1)21+(x+y)2). We claim: f(R2)=Q . Proof Observe first that f1,f2 are strictly positive on R2 . Thus, f(R2)⊂Q . Let us prove next the converse inclusion. Pick a point (a,b)∈Q . Let us show first: if a=b, then (a,a)∈g(R2) . We have f(t,t)=(t42+(t2+1)21+4t2,t42+(t2+1)21+4t2) and the image of the previous map contains the half-line {(s,s):s≥1} . In addition, f(t,0)=(1t2+1,1t2+1) and its image contains the segment {(s,s):0<s≤1} . Thus, (a,a)∈f(R2) for each a>0 . If a≠b , we may assume a<b . We search x,y∈R such that f(x,y)=(a,b) . Write y=λx where λ∈R . We obtain x2(x2y2x2+y2)+(xy+1)21+(x+y)2=x4λ21+λ2+(x2λ+1)21+x2(1+λ)2=a,y2(x2y2x2+y2)+(xy+1)21+(x+y)2=x4λ41+λ2+(x2λ+1)21+x2(1+λ)2=b. Consequently, b−a=x4λ2(λ2−1)1+λ2↝x=(1+λ2)(b−a)λ2(λ2−1)4 for some λ>1 (recall that a<b ). We deduce φ(λ)≔(1+λ2)(b−a)λ2(λ2−1)λ21+λ2+((1+λ2)(b−a)λ2(λ2−1)λ+1)21+(1+λ)2(1+λ2)(b−a)λ2(λ2−1)=a for some λ>1 . Simplifying we conclude φ(λ)=(b−a)λ2−1+(±(1+λ2)(b−a)(λ2−1)+1)21+(1+λ)2∣λ∣(1+λ2)(b−a)(λ2−1). Observe that limλ→+∞φ(λ)=0 while limλ→1+φ(λ)=+∞ . Thence, there exists λ0>1 such that φ(λ0)=a . If we take x0≔(1+λ02)(b−a)λ02(λ02−1)4andy0≔λ0x0, we have f(x0,y0)=(a,b) , as required.□ Remark A.1 Observe that the set of indeterminacy of f is {(0,0)} and f(0,0)=(1,1) . In addition, f(14−6+273,14−6+273)=(1,1) . Thus, if h:R2→R2 is a polynomial map such that h(R2)=R2⧹{(0,0)} (see Lemma 1.5 and [4, Ex. 1.4(iii)]), then g≔f◦h:R2→R2 is a regular map such that g(R2)=Q . The polynomial map h:R2→R2 proposed in [4, Ex. 1.4(iii)] that has the punctured plane R2⧹{(0,0)} as image is given by the formula (x,y)↦(xy−1,(xy−1)x2−y) . A.2. A regular map whose image is the open quadrant A simpler regular map g:R2→R2 than the one proposed in Remark A.1 that has Q as its image is provided next. Consider the regular map g≔(g1,g2):R2→R2,(x,y)↦(x2(xy−1)2+(xy+1)21+(x+y)2,y2(xy−1)2+(xy+1)21+(x+y)2). We claim: g(R2)=Q . Proof We proceed analogously to the preceding case, but replacing the lines y=λx by hyperbolas y=λ/x . As g1,g2 are strictly positive on R2 , we have g(R2)⊂Q . To prove the converse inclusion, pick a point (a,b)∈Q . Let us show first: if a=b, then (a,a)∈g(R2) . We have g(t,t)=(t2(t2−1)2+(t2+1)21+4t2,t2(t2−1)2+(t2+1)21+4t2). The image of g(t,t) contains the half-line {(s,s):s≥45} (fort=1,wehaveg(1,1)=(45,45)) . In addition, g(t,1/t)=(41+(t+1/t)2,41+(t+1/t)2)=(4t2t2+(t2+1)2,4t2t2+(t2+1)2) and its image contains the segment {(s,s):0<s≤45} (fort=1wehaveg(1,1)=(45,45)) . Consequently, (a,a)∈g(R2) . If a≠b , we may assume a>b . We search x,y∈R such that g(x,y)=(a,b) . Write y=λ/x for some λ∈R . Then a−b=g1(x,λ/x)−g2(x,λ/x)=(λ−1)2(x2−λ2/x2). Consequently, for λ≠1 x4−a−b(λ−1)2x2−λ2=0↝x2=r(λ)≔(a−b(λ−1)2+(a−b)2(λ−1)4+4λ2)/2. We have limλ→+∞r(λ)=+∞ and limλ→1+(λ−1)2r(λ)=a−b . Consider the equation f1(x,λ/x)=a , that is, (λ−1)2x2+(λ+1)2x2x2+(x2+λ)2=a, and replace x2 by r(λ) to obtain φ(λ)≔(λ−1)2r(λ)+(λ+1)2r(λ)r(λ)+(r(λ)+λ)2=a. Observe that limλ→+∞φ(λ)=+∞ while limλ→1+φ(λ)=a−b . As a>a−b , there exists λ0>1 such that φ(λ0)=a . If we take x0≔(a−b(λ0−1)2+(a−b)2(λ0−1)4+4λ02)/2andy0≔λ0/x0, we have g(x0,y0)=(a,b) , as required.□ © The Author(s) 2018. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/open_access/funder_policies/chorus/standard_publication_model)
The Quarterly Journal of Mathematics – Oxford University Press
Published: Dec 1, 2018
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