Uniform Quotient Mappings of the Plane

by

W. B. Johnson^{1}^{2}, J. Lindenstrauss^{2} ^{3}, D. Preiss and G. Schechtman^{2} ^{3}

Abstract

It is shown that if f is a mapping of the plane onto itself that is uniformly continuous with modulus of continuity Ω(r) which is o(\sqrt{r}) as r → 0 and f is also co-uniformly continuous then f = P ∘ h where h is a homeomorphism of the plane and P is a complex polynomial. The same conclusion holds also under other assumptions on the moduli of uniform and co-uniform continuity. However, we also present an example showing that this does not hold for all uniform quotient mappings: There is a mapping of the plane onto itself whose moduli of uniform and co-uniform continuity are both of power type but it maps an interval to zero. We also discuss uniform quotient mappings of the plane onto the line.

Subject classification: 54E15, 57N05. Keywords: Lipschitz quotients, Uniform quotient, Co-Lipschitz, Co-uniform continuity.

1 Supported in part by NSF DMS-9623260

2 Supported in part by the U.S.-Israel Binational Science Foundation

3 Participant, Workshop in Linear Analysis and Probability, Texas A&M University 1. Introduction

Let X and Y be metric spaces. As is well known a mapping f : X → Y is said to be uniformly continuous if there is a continuous increasing function Ω(r), r ≥ 0 with Ω(0) = 0 so that d(f(u), f(v)) ≤ Ω(d(u, v)) for all u and v, or in other words, f(B_{r}(x)) ⊂ B_{Ω(r)}(f(x)) for all x ∈ X and r > 0. (B_{r}(x) denotes the open ball with radius r and center x in the appropriate space.) The mapping f is called co-uniformly continuous if there is a continuous increasing function ω(r), r > 0 with ω(r) > 0 for r > 0 so that B_{ω(r)}(f(x)) ⊂ f(B_{r}(x)). The continuity and monotonicity assumptions are made here for convenience and, if not assumed, can be achieved by changing the original functions Ω(r) and ω(r). The only necessary requirement is that the limit of Ω(r), as r → 0, is 0.

A surjective mapping f is said to be a uniform quotient mapping if it is uniformly continuous and co-uniformly continuous. In other words f from X onto Y is a uniform quotient mapping if and only if f ×f : X ×X → Y ×Y maps the uniform neighborhoods of the diagonal in X×X onto the uniform neighborhoods of the diagonal in Y ×Y. Note that if f : X → Y is uniformly continuous and co-uniformly continuous then f (which of course is open) maps X to a closed set; so the image of X is both closed and open. Consequently, if Y is connected, f is automatically surjective. Note also that if f is continuous and open and K is a compact subset of X, then for each r > 0 there is ω(r) > 0 s.t. B_{ω(r)}(f(x)) ⊂ f(B_{r}(x)) is satisfied for x in K. In particular, a continuous open mapping on a compact space is couniformly continuous. Finally, if f is uniformly continuous and co-uniformly continuous, then for all Z ⊂ Y the restriction of f to f^{−1}(Z), when considered as a mapping into Z, is also uniformly continuous and co-uniformly continuous; moreover, the image of every component of f^{−1}(Z) is a component of Z provided that the balls of X are connected and Z ⊂ f(X) is open. A discussion of the notion of co-uniform continuity and uniform quotient mappings (in the context of general uniform spaces) can be found in [Jam]. For normed spaces the moduli always satisfy Ω(r) ≥ Cr and ω(r) ≤ cr for suitable C and c. If Ω(r) ≤ Cr (more precisely: If Ω can be chosen to satisfy Ω(r) ≤ Cr for some 0 < C < ∞ and all r > 0) we say that f is Lipschitz. Similarly, if ω(r) ≥ cr we say that f is co-Lipschitz. A surjective mapping which is Lipschitz and co-Lipschitz is called a Lipschitz quotient mapping.

In a recent paper [BJLPS] we dealt with these notions for general Banach spaces X and Y. Here we are interested mainly in the case where X = Y is the plane. (As a matter of notation we shall consider the plane both as IR^{2} and as the complex plane ^{|}C. When we consider it as IR^{2} we use ∥ · ∥ to denote the Euclidean norm while when we consider it as |C we use | · | for that purpose.)

Non trivial examples of Lipschitz quotient mappings from the plane to itself are f_{n}(re^{iθ}) = re^{inθ}, n = 1, 2, . . .. Our main aim is to show that these examples are in a sense typical for general uniform quotient mappings of the plane. We prove, under some conditions on Ω and ω, that any uniform quotient mapping f of the plane is of the form f = P ∘ h where h is a homeomorphism of the plane and P a polynomial. In the examples above f_{n} = P_{n} ∘ h_{n} where h_{n}(re^{iθ}) = r^{1/n}e^{iθ} and P_{n}(z) = z^{n}. Conversely, we show that for any given P there is a homeomorphism h of the plane so that P ∘ h is even a Lipschitz quotient mapping.

We prove the theorem mentioned above in case Ω and ω satisfy at least one of the following three conditions.

1. Ω is arbitrary and ω ≥ cr, i.e., f is uniformly continuous and co-Lipschitz.

2. ω is arbitrary and Ω(r)/\sqrt{r} → 0 as r → 0.

3. There are c, C, p, q with q < 1 + p so that ω(r) ≥ cr^{q} and Ω(r) ≤ Cr^{p}, for 0 < r < 1. The proofs of parts 1 and 2 of this theorem constitute most of Section 2 while the

proof of part 3 is contained in Section 4. The main arguments of the proofs presented here involve checking that under each of the assumptions above f^{−1}(y) is a discrete set for every y. Once this is done the representation theorem (Theorem 2.8) is proved using a result of Stoilow [Sto] which gives a topological characterization of analytic functions.

As a matter of fact we show that for every uniform quotient mapping of the plane there is a number N so that the set f^{−1}(y) has at most N connected components for every y. The assumption 1,2, or 3 is then used to prove that every such component is a singleton.

In Section 3 we present an example showing that some restrictions on the moduli are required. More precisely, there is a uniform quotient mapping f of the plane onto itself with moduli of power type which maps an interval to zero. Of course such a mapping cannot be a superposition of a homeomorphism and a polynomial. As a corollary to this example we also get (in Remark 3.3) a relatively simple construction of an example of a continuous open monotone mapping of the plane onto itself which is not an homeomorphism. Such an example was first given by Anderson [And]. His construction is much more complicated.

Theorem 2.8 mentioned above applies only to mappings defined on the entire plane. However, under the assumption 3 above we also prove in Section 4 a local result. For every uniform quotient mapping f from a domain in the plane into the plane satisfying 3, f^{−1}(y) is discrete for every y in the range. Example 4.1 shows that some restriction on the relation between p and q is needed; it fails for p = 1, q = 3. The same example shows that assumption 2 cannot guarantee that f^{−1}(y) is discrete if f is only assumed to be defined on a domain in the plane.

Section 5 deals with Lipschitz and uniform quotient mappings from IR^{2} to IR . The analysis here is much simpler. We show in particular that for uniform quotient mapping f from from IR^{2} to IR , IR^{2} \ f^{−1}(y) has a bounded number of components for y ranging over IR . If f is a Lipschitz quotient then also f^{−1}(y) has a bounded number of components.

The methods of proof in this paper are particular to the plane. One can ask many natural questions concerning uniform quotient mappings from IR^{n} to IR^{m}, n ≥ max(m, 3). This area of research is wide open. Some comments on these questions as well as results in the infinite dimensional situation are presented in [BJLPS].

2. Global results

We begin with a restatement of Proposition 4.3 of [BJLPS].

Proposition 2.1. Let f : IR^{2} → IR^{2} be a continuous and co-Lipschitz mapping. Then for every y ∈ IR^{2} the set f^{−1}(y) is discrete.

We repeat the proof from [BJLPS]. We first state the following simple lemma concerning the lifting of Lipschitz curves:

Lemma 2.2. Suppose that f : IR^{n} → X is continuous and co-Lipschitz with constant one, f(x) = y, and ξ : [0, ∞) → X is a curve with Lipschitz constant one, and ξ(0) = y. Then there is a curve ϕ : [0, ∞) → IR^{n} with Lipschitz constant one such that ϕ(0) = x and f(ϕ(t)) = ξ(t) for t ≥ 0.

Proof. For m = 1, 2, … define ϕ_{m}(0) = x, and, by induction, assuming that f(ϕ_{m}(\frac{k}{m})) = ξ(\frac{k}{m}), choose ϕ_{m}(\frac{k+1}{m}) such that ∥ϕ_{m}(\frac{k+1}{m}) − ϕ_{m}(\frac{k}{m})∥ ≤ \frac{1}{m} and f(ϕ_{m}(\frac{k+1}{m})) = ξ(\frac{k+1}{m}). Extend ϕ_{m}(t) to a Lipschitz curve ϕ_{m} : [0, ∞) → IR^{n} having Lipschitz constant one. The limit ϕ of any convergent subsequence of ϕ_{m} has the desired properties.

Proof of Proposition 2.1. Without loss of generality, assume that B_{r}(f(x)) ⊂ f(B_{r}(x)) for every x in IR^{2} and r > 0, y = 0, and f(0) = 0. Let u_{k} = e^{kπi/3} and S = {tu_{k} : t ≥ 0, k = 0, 2, 4}. Let also 0 < δ < 1 be such that ∥x∥, ∥y∥ ≤ 2 and ∥x − y∥ < δ imply that ∥f(x) − f(y)∥ < 1/2.

For each x ∈ B_{1}(0) ∩ f^{−1}(0) and k = 1, 3, 5, use Lemma 2.2 to choose ϕ_{k,x} : [0, ∞) → IR^{2} having Lipschitz constant one such that ϕ_{k,x}(0) = x and f(ϕ_{k,x}(t)) = tu_{k} for t ≥ 0. Let D_{k,x} be the component of IR^{2} \ f^{−1}(S) containing ϕ_{k,x}(0, ∞). Noting that B_{δ}(ϕ_{k,x}(1)) ⊂ D_{k,x} ∩ B_{3}(0), a comparison of areas shows that the set of all such D_{k,x} has at most 9δ^{−2} elements. Suppose now that B_{1}(0) ∩ f^{−1}(0) has more than N = [(9δ^{−2})^{3}] elements, then it contains elements x ≠ y such that {D_{1,x}, D_{3,x}, D_{5,x}} = {D_{1,y}, D_{3,y}, D_{5,y}}. Hence D_{k,x} = D_{k,y} for k = 1, 3, 5, since the (connected) image of D_{k} := D_{k,x} contains u_{k} and so can contain no other u_{j}, and we infer that there are simple curves ψ_{k} : [0, 1] → IR^{2} such that ψ_{k}(0) = x, ψ_{k}(1) = y and ψ_{k}(t) ∈ D_{k} for 0 < t < 1. For each pair k, l = 1, 3, 5 of different indices, let G_{k,l} be the interior of the Jordan curve (ψ_{k} − ψ_{l}) (difference in the sense of oriented curves). If j ≠ k, l, we note that G_{k,l} ∩ D_{j} = ∅ since otherwise D_{j} would be bounded. In particular, G_{1,3} ∩ ∂G_{3,5} = ∅, so either G_{1,3} ⊂ G_{3,5} or G_{1,3} ∩ G_{3,5} = ∅. In the former case we would get a contradiction from ψ_{1}(0, 1) ⊂ G_{3,5}, since ψ_{1}(0, 1) ⊂ D_{1}. In the latter case ∂(G_{1,5}) = ∂(\overline{G_{1,3} ∪ G_{3,5}}). This intuitively clear fact follows for example from the theorem about θ curves (see e.g. [Kur, ch. 10, § 61, II, Theorem 2]) or from Schoenflies’ extension theorem. It follows that G_{1,5} ⊃ G_{1,3} and we get a contradiction from ψ_{3}(0, 1) ⊂ G_{1,5}.

Remark. If we assume in addition that f is uniformly continuous, then a more careful analysis of the proof shows that the cardinality of f^{−1}(y) is finite and moreover is bounded, independently of y, by a constant depending only on the co-Lipschitz constant of f and its modulus of uniform continuity. We do not expand on this since we shall present a different proof of it, using Lemma 2.7 below. See the beginning of the proof of Theorem 2.8.

We now come to the main result of this paper which is a version of Proposition 2.1 in which the co-Lipschitz condition is weakened to mere co-uniformity but the continuity assumption is strengthened to uniform continuity with modulus strictly better than \sqrt{r}

Theorem 2.3. Let f : IR^{2} → IR^{2} satisfy B_{ω(r)}(f(z)) ⊂ f(B_{r}(z)) ⊂ B_{Ω(r)}(f(z)), for all r > 0 and z ∈ IR^{2}, where Ω(r), ω(r) : [0, ∞) → [0, ∞) are continuous strictly increasing functions such that Ω(0) = ω(0) = 0. If lim_{r→0} Ω(r)/\sqrt{r} = 0, then the inverse images of points under f are discrete. Moreover, there is a number N depending only on Ω and ω such that the cardinality of f^{−1}(y) is bounded by N for all y ∈ IR^{2}.

For the proof we need a sequence of lemmas. In all of these lemmas (2.4 - 2.7) we assume that f : IR^{2} → IR^{2} satisfies

(2.1)  B_{ω(r)}(f(z)) ⊂ f(B_{r}(z)) ⊂ B_{Ω(r)}(f(z)),

for all r > 0 and z ∈ IR^{2}, where Ω(r), ω(r) : [0, ∞) → [0, ∞) are continuous strictly increasing functions such that Ω(0) = ω(0) = 0. The additional assumption, lim_{r→0} Ω(r)/\sqrt{r} = 0, is not used in these Lemmas.

Lemma 2.4. Let f : IR^{2} → IR^{2} satisfy (2.1). For every r_{0} > 0 there is a constant R_{0} = R_{0}(r_{0}) < ∞ depending only on Ω, ω and r_{0} such that, for every y ∈ IR^{2} and every r ≥ r_{0}, every component of f^{−1}(B_{r}(y)) has diameter at most R_{0}r.

Proof. If not, there are for every k = 1, 2, … functions f_{k} : IR^{2} → IR^{2} satisfying f_{k}(0) = 0, B_{ω(s)}(f_{k}(z)) ⊂ f_{k}(B_{s}(z)) ⊂ B_{Ω(s)}(f_{k}(z)), for all s > 0 and z ∈ IR^{2}, and numbers r_{k} ≥ r_{0}, such that the component C_{k} of f_{k}^{−1}(B_{r_{k}}(0)) containing 0 has diameter at least kr_{k}. Observe (or see [BJLPS, Remark 3.3]) that a uniformly continuous g is Lipschitz for large distances in the sense that ∥g(z_{1}) − g(z_{2})∥ ≤ 2Ω(1)∥z_{1} − z_{2}∥ if ∥z_{1} − z_{2}∥ ≥ 1. Similarly for a couniformly continuous function g, g(B_{s}(z)) ⊃ B_{sω(1)/2}(g(z)) for s ≥ 1. Hence there is a subsequence of f_{k}(kr_{k}z)/kr_{k} converging to a Lipschitz and co-Lipschitz g : IR^{2} → IR^{2}. It follows that g^{−1}(0) contains a connected set of diameter at least 1/2 in contradiction to Proposition 2.1.

Lemma 2.5. Let f : IR^{2} → IR^{2} satisfy (2.1). For every x ∈ IR^{2} and every unit vector u there is a closed unbounded set Γ_{x,u} such that x ∈ Γ_{x,u}, f(Γ_{x,u}) = {f(x) + tu; t ≥ 0}, and {w ∈ Γ_{x,u}; ∥f(w) − f(x)∥ ≤ τ} is compact and connected for every τ ≥ 0. Proof. For m = 1, 2, … define ϕ_{m}(0) = x, and, by induction, assuming that f(ϕ_{m}(\frac{k}{m})) = f(x) + \frac{k}{m}u, choose ϕ_{m}(\frac{k+1}{m}) such that ∥ϕ_{m}(\frac{k+1}{m}) − ϕ_{m}(\frac{k}{m})∥ ≤ ω^{−1}(\frac{1}{m}) and f(ϕ_{m}(\frac{k+1}{m})) = f(x) + \frac{k+1}{m}u. Extend ϕ_{m}(t) to a Lipschitz curve ϕ_{m} : [0, ∞) → IR^{2} having Lipschitz constant at most mω^{−1}(\frac{1}{m}). Since \frac{k}{m} = ∥f(ϕ_{m}(\frac{k}{m})) − f(x)∥ ≤ Ω(∥ϕ_{m}(\frac{k}{m}) − x∥), ∥ϕ_{m}(t) − x∥ → ∞ as t → ∞.

For any t ≥ 0 of the form \frac{k}{m} choose the largest s_{m}(t) such that f(ϕ_{m}(s_{m}(t))) = f(x) + tu. By Lemma 2.4, if m is large enough, diam(ϕ_{m}[0, s_{m}(t)]) ≤ R_{0} · (1 + \frac{t}{2}), where R_{0} = R_{0}(1). Let m_{j} be chosen so that, for every rational t > 0, s_{m_{j}}(t) is eventually defined and the sequence ϕ_{m_{j}}[0, s_{m_{j}}(t)] of continua converges to a continuum C_{t}. Note that f(C_{t}) = [f(x), f(x) + tu] and that t′ > t implies that C_{t} ⊃ C_{t′} ∩ f^{−1}[f(x), f(x) + tu). In particular, Ω(∥w − x∥) ≥ t for w ∈ C_{t′} \ C_{t}, which shows that Γ_{x,u} = ⋃_{t} C_{t} is closed and unbounded. Clearly x ∈ Γ_{x,u} and f(Γ_{x,u}) = {f(x) + tu; t ≥ 0}. Moreover, {w ∈ Γ_{x,u}; f(w) ∈ [f(x), f(x) + τu]} = ⋂_{t>τ} C_{t}, so it is compact and connected. Lemma 2.6. Let f : IR^{2} → IR^{2} satisfy (2.1). Suppose that a, b belong to different components of f^{−1}(y), r ≥ 4(R_{0} + ∥a − b∥), where R_{0} = R_{0}(1) of Lemma 4, and u is a unit vector. Then

Ω^{−1} (\frac{r}{6R_{0}} ) ≤ dist(\{z ∈ Γ_{a,u}; ∥z − a∥ ≥ r\}, \{z ∈ Γ_{b,u}; ∥z − b∥ ≥ r\})
+ dist({z ∈ Γ_{a,−u}; ∥z − a∥ ≥ r}, {z ∈ Γ_{b,−u}; ∥z − b∥ ≥ r}).

Proof. Note that by Lemma 2.5 the sets whose distances we estimate are always nonempty, and suppose that dist({z ∈ Γ_{a,u}; ∥z − a∥ ≥ r}, {z ∈ Γ_{b,u}; ∥z − b∥ ≥ r}) + dist({z ∈ Γ_{a,−u}; ∥z − a∥ ≥ r\}, \{z ∈ Γ_{b,−u}; ∥z − b∥ ≥ r\}) < Ω^{−1}(\frac{r}{6R_{0}} ).

Case I. Γ_{a,u} ∩Γ_{b,u} ≠ ∅ and Γ_{a,−u} ∩Γ_{b,−u} ≠ ∅. Then for sufficiently large τ, A = {w ∈ Γ_{a,u} ∪ Γ_{b,u}; ∥f(w) − y∥ ≤ τ} and B = {w ∈ Γ_{a,−u} ∪ Γ_{b,−u}; ∥f(w) − y∥ ≤ τ} are continua. But {a, b} ⊂ A ∩ B ⊂ f^{−1}(y), so A ∩ B is not connected since a and b belong to different components of f^{−1}(y). So by [Kur, ch. 10, § 61, I, Theorem 5] IR^{2} \ (A ∪ B) has a bounded component G; but then f(G) is bounded, open, and with boundary contained in f(A∪B), hence in the line L = {y + tu; t ∈ IR }, which is impossible. Note that the proof actually shows that there is at most one direction u for which Γ_{a,u} ∩ Γ_{b,u} ≠ ∅.

Case II. Γ_{a,u} ∩ Γ_{b,u} = ∅ and Γ_{a,−u} ∩ Γ_{b,−u} ≠ ∅. Choose a segment D such that D ∩ Γ_{a,u} = {p}, D ∩ Γ_{b,u} = {q}, where ∥p − a∥ ≥ r/2, ∥q − b∥ ≥ r/2 and diam(D) < Ω^{−1}(\frac{r}{6R_{0}} ). Let A ⊂ Γ_{a,u} ∪ Γ_{a,−u} ∪ Γ_{b,−u} ∪ Γ_{b,u} be a minimal continuum containing p and q. Since A ∩ f^{−1}({y + tu; t > 0}) is disconnected, A ∩ f^{−1}({y + tu; t ≤ 0}) ≠ ∅. By Lemma 2.4, f(p) = y + su, where s ≥ \frac{r}{2R_{0}} , and we infer that diam(f(A)) ≥ \frac{r}{2R_{0}} .

From [Kur, ch. 10, § 62 V, Theorem 6] we infer that A ∪ D is the boundary of a bounded component, say G, of its complement. Since f is open, f(G) is open, and since its boundary is contained in L∪B_{Ω(∥q−p∥)}(f(p)) (recall that L = {y +tu; t ∈ IR }), we infer that f(G) ⊂ B_{Ω(∥q−p∥)}(f(p)) ⊂ B_{\frac{r}{6R_{0}}} (f(p)). But diam(f(G)) ≥ diamf(A) ≥ \frac{r}{2R_{0}} .

Case III. Γ_{a,u} ∩ Γ_{b,u} ≠ ∅ and Γ_{a,−u} ∩ Γ_{b,−u} = ∅. Symmetric to II.

Case IV. Γ_{a,u} ∩ Γ_{b,u} = ∅ and Γ_{a,−u} ∩ Γ_{b,−u} = ∅.

Choose segments D^{+}, D^{−} such that D^{+} ∩ Γ_{a,u} = {p^{+}}, D^{+} ∩ Γ_{b,u} = {q^{+}}, where ∥p^{+} − a∥ ≥ r/2, ∥q^{+} − b∥ ≥ r/2, D^{−} ∩ Γ_{a,−u} = {p^{−}}, D^{−} ∩ Γ_{b,−u} = {q^{−}}, where ∥p^{−} − a∥ ≥ r/2, ∥q^{−} − b∥ ≥ r/2 and diam(D^{+}) + diam(D^{−}) < Ω^{−1}(\frac{r}{6R_{0}} ). Note that Ω(∥p^{+} − p^{−}∥) ≥ r/R_{0} (otherwise ∥f(p^{+}) − f(p^{−})∥ < r/R_{0} so one of ∥f(p^{+}) − 0∥ or ∥f(p^{−}) − 0∥ is less than \frac{r}{2R_{0}} in contradiction to Lemmas 2.4, 2.5 and the choice of p^{+}, p^{−}), so D^{+} ∩ D^{−} = ∅. Let A ⊂ Γ_{a,u} ∪ Γ_{a,−u} be a minimal continuum containing p^{+}, p^{−} and let B ⊂ Γ_{b,u} ∪ Γ_{b,−u} be a minimal continuum containing q^{+}, q^{−}.

Clearly, diam(f(A)) ≥ r/R_{0}. Noting that A and D^{+} ∪ B ∪ D^{−} are minimal continua whose intersection is {p^{+}, p^{−}}, we infer from [Kur, ch. 10, § 62.V, Theorem 6] that A ∪ D^{+} ∪ B ∪ D^{−} is the boundary of a bounded component, say G, of its complement. Since f is open, f(G) is open, and since its boundary is contained in L ∪ B_{Ω(diam(D^{+}))}(f(p^{+})) ∪ B_{Ω(diam(D^{−}))}(f(p^{−})), we infer that f(G) ⊂ B_{Ω((diam(D^{+}))}(f(p^{+})) ∪ B_{Ω(diam(D^{−}))}(f(p^{−})). Since these two discs are disjoint and f(G) is connected, it is contained in one of them. But diam(f(G)) ≥ diamf(A) ≥ r/R_{0} is bigger than the diameter of either of these discs. This contradiction ends the proof. Lemma 2.7. Let f : IR^{2} → IR^{2} satisfy (2.1) then there is an N < ∞, depending only on Ω and ω, such that, for each y ∈ IR^{2}, the cardinality of the set of components of f^{−1}(y) is at most N.

Proof. Choose a ∈ f^{−1}(y). If s is large enough then, applying Lemma 2.6 with r = 4(R_{0} + s), we get that the number of components of f^{−1}(y) which meet a disc of radius s around a cannot be bigger than the largest number of elements of a set M ⊂ {x; ∥x − a∥ ≤ 5(R_{0} + s)} × {y; ∥y − a∥ ≤ 5(R_{0} + s)} which has all ℓ_{1} distances larger than or equal to Ω^{−1}(\frac{4(R_{0}+s)}{6R_{0}}).

We assume as we may that, for t ≥ 1, Ω(t) ≤ 2Ω(1)t. Homogeneity implies now that, if s is large enough, the number of elements of M is at most the number of couples of points in a disc of radius 1 whose mutual ℓ_{1} distances are not smaller than some positive number c (depending only on Ω(1) and R_{0}). This number is a bound for N. Proof of Theorem 2.3. Let n be the largest number for which one can find y with n components, say H_{1}, …, H_{n} of f^{−1}(y) of which at least one, say H_{1}, is non-trivial. Let y and H_{1}, …, H_{n} realize this maximum and let G_{i} ⊃ H_{i} be open, and with disjoint closures. For z sufficiently close to y, say ∥z − y∥ < δ, f^{−1}(z) meets each G_{i}; moreover, for δ sufficiently small, the component of f^{−1}(z) meeting G_{i} has to be contained in G_{i}, since otherwise we would get contradiction by taking the limit of such components as z → y. Denoting H_{i}(z) = f^{−1}(z) ∩ G_{i}, we thus have f^{−1}(z) ⊃ ⋃_{i=1}^{n} H_{i}(z), and H_{i}(z) are components of f^{−1}(z). The component H_{1}(z) is non-trivial for z close to y (otherwise, since H_{1} is non-trivial and f a uniform quotient mapping, G_{1} would contain an arbitrarily large number of components for z close to y in contradiction to Lemma 2.7). Also, the maximality of n implies that G_{1} contains only one component H_{1}(z). Let u, v be two different points of H_{1}, and let L be their perpendicular bisector. If ∥z − y∥ < δ_{1} = min{δ, ω(∥v − u∥ /4)}, then f^{−1}(z) meets L, so f(L) has non-empty interior. This not possible if lim_{r→0} Ω(r)/\sqrt{r} = 0; an easy way to see this is to compare the cardinality of a maximal ϵ-separated set of points in a segment of L and in its image.

It now follows from a deep theorem of Stoilow that uniform quotient mappings satisfying the assumptions of either Proposition 2.1 or Theorem 2.3 are topologically equivalent to polynomials.

Theorem 2.8. Let f : IR^{2} → IR^{2} satisfy one of the following three assumptions:

(i) f is uniformly continuous and co-Lipschitz, or, (ii) f is uniformly continuous with modulus of continuity Ω satisfying Ω(r)/\sqrt{r} → 0 as

r → 0 and f is also co-uniformly continuous, or, (iii) There are c, C, p, q with q < 1 + p so that f is uniformly continuous with modulus

of continuity Ω satisfying Ω(r) ≤ Cr^{p}, for 0 < r < 1, and f is also co-uniformly continuous with modulus of co-uniform continuity ω satisfying ω(r) ≥ cr^{q}, for 0 < r < 1. Then f = P ∘ h where h is a homeomorphism of IR^{2} and P is a polynomial.

Proof of 2.8(i) and 2.8(ii). By Stoilow’s Theorem [Sto, p.121], since f is discrete and open, f = P ∘h with h a homeomorphism of IR^{2} onto a (simply connected) domain G in IR^{2} and P an analytic function. (In the formulation of Stoilow’s Theorem in [Sto] the image is a Riemann surface but the uniformization theorem, see e.g., [FK, p. 195], implies that it must be a simply connected domain in the plane.) By Proposition 2.1 and Lemma 2.7, the inverse image of each point under f, satisfying assumption (i), is finite (even bounded independently of the point). The same is true also under the assumption (ii), by Theorem 2.3. Thus, also P^{−1}(y) ∩ G is finite for each point y. We shall show below that G is necessarily IR^{2} so that P is an entire function with the property that the inverse image of each point is finite. It then follows that P is a polynomial.

We now prove that G, the image of h, is the entire plane. First notice that f(z) → ∞ as z → ∞. Indeed, otherwise there would be a sequence z_{n} so that z_{n} → ∞ and f(z_{n}) → a. Since f is co-uniformly continuous, f will take the value a in the disc of radius 1 around z_{n} for all large enough n. This contradicts the fact that f^{−1}(a) is finite.

If f = P ∘ h with P analytic and h(IR^{2}) = G ≠ IR^{2} then, since G is simply connected, we may assume without loss of generality that G is the unit disc. It follows from the previous paragraph that P (z) tends to infinity as |z| → 1. Since P has only finitely many zeros in the unit disc, we get, by dividing the Blaschke product corresponding to the zeros by P, an analytic function in the disc tending to zero as |z| → 1. This is clearly impossible, by the maximum principle.

The proof of Theorem 2.8 under assumption (iii) is delayed to the end of Section 4.

Remark. Note that the homeomorphism h in the representation f = P ∘ h is determined up to a transformation of the form h → ah + b for some complex a and b (and then necessarily P is determined up to a change of variable z → az + b). Indeed if P ∘ h = Q ∘ g for polynomials P and Q and homeomorphisms h and g then P and Q must have the same degree (which is equal to the maximal cardinality of a preimage of a point under f). If w = gh^{−1}(z) and Q is invertible in a neighborhood of w then gh^{−1} is analytic in a neighborhood of z. It is then necessarily a polynomial of degree one; this follows easily from the equation P (z) = Q(gh^{−1}(z)). Since there are only finitely many exceptional points, the preimages under gh^{−1} of the zeroes of Q′, gh^{−1}, being an homeomorphism of the plane onto itself and analytic except at finitely points, must be a linear function.

We also have a converse statement to Theorem 2.8.

Proposition 2.9. Let P be a polynomial in one complex variable with complex coefficients. Then there is an homeomorphism h of the plane such that f = P ∘ h is a Lipschitz quotient mapping.

Sketch of Proof. Assume without loss of generality that P (z) = z^{n} + a_{n−1}z^{n−1} + a_{n−2}z^{n−2}+…+a_{0}. We first show how to find an homeomorphism h which makes f = P ∘h Lipschitz and co-uniformly continuous. Fix a (large) R > 0 and define h by

h(z) = \{ \begin{matrix}|z|^{1/n}e^{i arg(z)}, & if 2R ≤ |z|; \\ (\frac{2R−|z|}{R}|z| + \frac{|z|−R}{R}|z|^{1/n}) e^{i arg(z)}, & if R ≤ |z| < 2R; \\ z, & if |z| ≤ R. \end{matrix}

It is easy to see that h is an homeomorphism of IR^{2} onto itself. Also, h is Lipschitz on the ball of radius 3R about 0 and is co-uniformly continuous on the same ball in the sense that B_{ω(r)}(h(x)) ⊂ h(B_{r}(x)) for an appropriate ω(r) > 0 and all x in the ball of radius 3R about 0. Since P is Lipschitz on the image of that ball (as it is on any compact domain), f = P ∘ h is Lipschitz on the ball of radius 2R about 0. Outside that ball

f(z) = |z|e^{in arg(z)} + a_{n−1}|z|^{(n−1)/n}e^{i(n−1) arg(z)} + a_{n−2}|z|^{(n−2)/n}e^{i(n−2) arg(z)} + … + a_{0}

which is checked to be Lipschitz in this domain.

Since any polynomial is an open mapping, a simple compactness argument, mentioned in the introduction, shows that any polynomial is co-uniformly continuous when restricted to any bounded domain. It follows that f is co-uniformly continuous when restricted to the ball of radius 3R about 0. The special form of f outside the ball of radius 2R about 0 shows that, if R is large enough, f is even co-Lipschitz there.

Assume now that R is such that, in addition to the implicit requirements on its size above, also all the zeros of P′ are in a ball of radius R/4 about zero. We now show how to adjust h on a ball of radius R/2 about zero as to remain Lipschitz and be also co-Lipschitz.

Let z_{1}, …, z_{m} be the distinct zeros of P′. Let r > 0 be such that B_{3r}(z_{j}), j = 1, …, m, are pairwise disjoint and all contained in B_{R/2}(0). There is no problem with the co-Lipschitzity of h outside the union of these balls. Fix any j = 1, …, m. Then, by taking an even smaller r > 0, we may also assume that in B_{3r}(z_{j}) one can write P (z) = (z − z_{j})^{k}Q(z) + a where Q does not vanish on B_{3r}(z_{j}). Modify the definition of h on B_{3r}(z_{j}) as follows. For z = z_{j} + se^{iθ},

h(z) = \{ \begin{matrix}z_{j} + s^{1/k}e^{iθ}, & if 0 ≤ s ≤ r; \\ z_{j} + (\frac{2r−s}{r}s^{1/k} + \frac{s−r}{r} s) e^{iθ}, & if r < s < 2r; \\ z, & if 2r ≤ s ≤ 3r. \end{matrix}

We do that on each of the balls B_{3r}(z_{j}), leave h as it was outside the union of the balls.

3. The example

Here we give an example showing that without some restrictions on the moduli of uniform continuity or co-uniform continuity of a uniform quotient mapping of the plane to itself the conclusions of Theorems 2.3 and 2.8 no longer hold.

Lemma 3.1. Given d ≥ c > 0 and a ∈ IR^{2} with ∥a∥ ≤ d, there is a mapping g : IR^{2} → IR^{2} such that:

(i) g([0, a]) ⊂ [0, g(a)] and ∥g(a)∥ ≤ c/4, (ii) the Lipschitz constants of g and of g^{−1} are less than or equal to 36(d/c)^{2}, (iii) for all z ∈ IR^{2} and r ≥ 2c, g(B_{r}(z)) is 6c dense in B_{r+d}(z), (iv) for all z ∈ IR^{2} and r ≥ 2c, g(B_{r}(z)) is 6c dense in B_{r}(g(z)), (v) for all z ∈ R^{2}, ∥g(z) − z∥ ≤ 2d.

Proof. Rotate the coordinate system so that a is a positive multiple of (c, −4d). Define continuous 1-periodic η : R → R by η(0) = η(1/2) = 0, η(1/4) = 1, η(3/4) = −1, and η is affine on every component of R \ (1/4)Z. Let φ(x, y) = (x, y + dη(x/c)), ψ(x, y) = (x + dη(y/c), y), and g = ψ ∘ φ.

(i) On the segment I = {(τc, −4τd) : 0 ≤ τ ≤ 1/4} we have φ(τc, −4τd) = (τc, 0), and on the segment J = {(τc, 0) : 0 ≤ τ ≤ 1/4} we have ψ(τc, 0) = (τc, 0). Hence I is mapped affinely on J. Since a ∈ I, this shows (i).

(ii) Obvious.

(iii, iv) Let z = (u, v) and k, l be the integer parts of u/c, v/c, respectively. For every m = 0, 1, …, the g image of P_{m} = [(k − m)c, (k + 1 + m)c] × [(l − m)c, (l + 1 + m)c] is contained in Q_{m} = [(k − m)c − d, (k + 1 + m)c + d] × [(l − m)c − d, (l + 1 + m)c + d] and meets every square [pc, (p + 1)c] × [qc, (q + 1)c] that lies inside Q_{m}, so it is 4c dense in Q_{m}. Choosing the largest m such that P_{m} ⊂ B_{r}(z), i.e., m is the integer part of (r/2c) − 1, we have that Q_{m} ⊃ B_{r+d−2c}(z), which proves (iii). To prove (iv), we note that g(z) ∈ Q_{0}, so Q_{m} ⊃ B_{r−2c}(g(z)).

(v) Obvious.

To illustrate the complexity of the seemingly simple mapping of Lemma 3.1, here is a sketch of the image under such a g of the boundary of the square with vertices (0, 0), (1, 0), (1, 1), (0, 1) (here c = 1, d = 1.5).

In the example below we denote B_{t}(x) = B(x, t).

Example 3.2. There is a mapping f : IR^{2} → IR^{2} such that f^{−1}(0) contains a segment and for every z ∈ IR^{2} and r > 0,

B(f(z), C_{1} min(r^{β}, r)) ⊂ f(B(z, r)) ⊂ B(f(z), C_{2} max(r^{α}, r)),

where α, β, C_{1}, C_{2} are positive constants.

Proof. Let c_{0} = 1/8 and c_{k+1} = 48^{−4k−1}c_{k}^{5}. Let a_{1} = (0, 1) and let g_{1} be the function obtained by Lemma 3.1 with c = c_{1}, d = 8c_{0}, a = a_{1}. Recursively put a_{k+1} = g_{k}(a_{k}), note that ∥a_{k+1}∥ ≤ c_{k}, and let g_{k+1} be the function obtained by Lemma 3.1 with c = c_{k+1}, d = 8c_{k}, a = a_{k+1}.

Define f_{1} = g_{1} and f_{k+1} = g_{k+1} ∘ f_{k}. Then ∥f_{k+1} − f_{k}∥ ≤ 16c_{k}, So the sequence f_{k} converges uniformly to a continuous f : IR^{2} → IR^{2}. In particular, f(z) = 0 for all z ∈ [0, a_{1}], so f^{−1}(0) contains [0, a_{1}]. By (ii) of Lemma 3.1, for each k the Lipschitz constants of f_{k} and of its inverse do not exceed 48^{2k}/c_{k}^{2}. In particular, f_{k}(B(z, r)) ⊃ B(f_{k}(z), 48^{−2k}c_{k}^{2}r).

Let z ∈ R^{2} and 0 < r < c_{1}. Find the least k such that r ≥ 48^{−k}c_{k}, and let s = 48^{−2k}c_{k}^{2}r. Then f_{k}(B(z, r)) ⊃ B(f_{k}(z), s) and s ≥ 48^{−3k}c_{k}^{3} ≥ 2c_{k+1}.

We prove that for every n > k, f_{n}(B(z, r)) is 6c_{n} dense in B(f_{k+1}(z), s): For n = k +1 this follows from (iv) of Lemma 3.1. If it holds for some n and if y ∈ B(f_{k+1}(z), s), choose x ∈ B(z, r) such that ∥y − f_{n}(x)∥ ≤ 6c_{n}. Let t = 48^{−2n}c_{n}^{3} and let u ∈ B(z, r) be such that x ∈ B(u, t) ⊂ B(z, r). Then ∥f_{n}(u) − f_{n}(x)∥ ≤ c_{n} and f_{n}(B(u, t)) ⊃ B(f_{n}(u), 48^{−4n}c_{n}^{5}) ⊃ B(f_{n}(u), 2c_{n+1}). Hence f_{n+1}(B(z, r)) ⊃ g_{n+1}(B(f_{n}(u), 2c_{n+1})) is 6c_{n+1} dense in B(f_{n}(u), 8c_{n}). Since y ∈ B(f_{n}(u), 8c_{n}), the set f_{n+1}(B(z, r)) contains a point 6c_{n+1} close to y.

Using that ∥f(z) − f_{k+1}(z)∥ ≤ 16 ∑_{j=k+1}^{∞} c_{j} ≤ s/2, we conclude that f(B(z, r)) ⊃ B(f_{k+1}(z), s) ⊃ B(f(z), s/2). Moreover, s/2 ≥ r^{β} if β > 11 and r is sufficiently small.

Given any x, y and any k, we have ∥f(x) − f(y)∥ ≤ 32 ∑_{j=k}^{∞} c_{j} + Lip(f_{k})∥x − y∥ ≤ 64c_{k}+48^{2k}∥x−y∥/c_{k}^{2}. If c_{k+1}^{3} ≤ ∥x−y∥ ≤ c_{k}^{3}, this gives ∥f(x)−f(y)∥ ≤ 48^{2k+2}c_{k} ≤ ∥x−y∥^{α} if α < 1/15 and ∥x − y∥ is sufficiently small.

Remark 3.3. We now show how to modify any non-trivial uniform quotient mapping f of the plane onto itself to obtain a simple construction of an example of a continuous open monotone mapping of the plane onto itself which is not an homeomorphism. (Monotone means that the inverse image of each point is a continuum. Such a mapping was first constructed by Anderson [And].) To this end we first observe that the complement of the inverse image under f of any open disc is connected; indeed, recalling that inverse images of discs are bounded, the opposite would allow us to find first a bounded component C of the complement of f^{−1}(B_{r}(y)), then a bounded open set V ⊃ C whose boundary would lie entirely in f^{−1}(B_{r}(y)), and then conclude that f, being continuous and mapping the boundary of V to B_{r}(y), cannot be open. Next we observe that the argument from the beginning of the proof of Theorem 2.3 provides us with an open disc B_{r}(y) and a bounded open set G containing a non-trivial component of f^{−1}(y) such that for every z ∈ B_{r}(y) there is exactly one component H_{z} of f^{−1}(z) meeting \overline{G}, and that this component is, in fact, contained in G. Then U = ⋃_{z∈B_{r}(y)} H_{z} is a component of f^{−1}(B_{r}(y)), so, by our first observation, it is homeomorphic to the whole plane. So it suffices to point out that f is clearly a non-trivial monotone map of U to B_{r}(y).

4. Local results

If one relaxes the assumptions of Theorem 2.3 by changing the domain of f from IR^{2} to a bounded domain the conclusion fails in a very strong sense. Example 4.1. Define f : IR^{2} → IR^{2} by f(x, y) = y^{2}(cos(\frac{x}{y}), sin(\frac{x}{y})) when y ≠ 0 and f(x, 0) = 0. Then f is Lipschitz on bounded sets and for each M there exists δ = δ(M) > 0 so that if z is in IR^{2} and |z| ≤ M, then for all r ≤ 1, f(B_{r}(z)) ⊃ B_{δr^{3}}(f(z)). Proof. That f is Lipschitz on bounded sets follows by taking partial derivatives. To check the second statement assume that f(x_{0}, y_{0}) = s_{0}(cos θ_{0}, sin θ_{0}), and without loss of generality that y_{0} ≥ 0. Assume first that r ≤ \frac{y_{0}}{2} ∧ 1.

We would like to show that, for an appropriate δ, f(B_{r}(x_{0}, y_{0})) contains the set S = {s(cos θ, sin θ); |s − s_{0}| < δr^{3}, |θ − θ_{0}| < δr^{2}/s} and thus a ball of radius δ′r^{3} around s_{0}(cos θ_{0}, sin θ_{0}). We shall actually show that, for an appropriate δ, f([x_{0} − r, x_{0} + r] × [y_{0} − \frac{r^{2}}{4M} , y_{0} + \frac{r^{2}}{4M}]) ⊃ S which is clearly enough.

Notice that

(1) for every 0 < t ≤ y_{0},

[(y_{0} − t)^{2}, (y_{0} + t)^{2}] ⊃ [s_{0} − ty_{0}, s_{0} + ty_{0}].

(2) For 0 < t ≤ \frac{y_{0}}{2} and |y − y_{0}| ≤ t,

|\frac{x_{0}}{y} − \frac{x_{0}}{y_{0}} | ≤ t\frac{|x_{0}|}{yy_{0}} ≤ t\frac{2|x_{0}|}{y^{2}} ≤ \frac{2Mt}{y^{2}}.

(3) for a fixed y and any positive u,

\{arg(f(x, y)); |x − x_{0}| ≤ u\} ⊃ [\frac{x_{0}}{y} − \frac{u}{y}, \frac{x_{0}}{y} + \frac{u}{y}] (mod 2π).

Taking t = \frac{r^{2}}{4M} in (2) and u = r in (3) we get, for a fixed y and for s = y^{2},

\{arg(f(x, y)); |x − x_{0}| ≤ r\} ⊃ [\frac{x_{0}}{y_{0}} − \frac{r^{2}}{2s}, \frac{x_{0}}{y_{0}} + \frac{r^{2}}{2s}] (mod 2π).

Finally, applying (1) for t = \frac{r^{2}}{4M}, we get

[(y_{0} − \frac{r^{2}}{4M})^{2}, (y_{0} + \frac{r^{2}}{4M})^{2}] ⊃ [s_{0} − \frac{r^{3}}{4M} , s_{0} + \frac{r^{3}}{4M}].

This settles the case r ≤ \frac{y_{0}}{2} ∧ 1. If r ≤ (20y_{0}) ∧ 1 then f(B_{r}(z)) ⊃ f(B_{r/40}(z)) ⊃ B_{δ′r^{3}}(f(z)), so we are left only with the case 20y_{0} < r ≤ 1. In this case for every y with |y − y_{0}| < r/20 the set \{\frac{x}{y}; |x − x_{0}| < r\} (mod 2π) contains all possible arguments. It follows that f(B_{r}(x_{0}, y_{0})) contains B_{y_{0}^{2}+ \frac{r^{2}}{400}} (0) and in particular B_{\frac{r^{2}}{400}} (f(x_{0}, y_{0})).

Remark. One can generalize the example above. For β > 0 and α ≥ 1 let f(x, y) = |y|^{β}(cos(\frac{x}{|y|^{α}}), sin(\frac{x}{|y|^{α}})). One can check that, when restricting f to a bounded domain, the modulus of uniform continuity of f is bounded by Cr (i.e., f is Lipschitz), if β ≥ 1+α, and by Cr^{β/(1+α)}, if β ≤ 1 + α. The modulus of co-uniform continuity is bounded from below by cr^{β∨(1+ \frac{β}{α})}. In particular, minimizing over β = 1 + α we get a function which is Lipschitz on bounded domains and has modulus of co-uniform continuity bounded from below by cr^{2.62} on bounded domains.

In spite of the example above one can show that, under some restriction concerning the relation between the modulus of uniform continuity of f and its modulus of co-uniform continuity, a local form of Theorem 2.3 still holds.

Proposition 4.2. Suppose that p < 1 + q, G ⊂ IR^{2} is open, and f : G → IR^{2} is such that B_{cr^{p}}(f(z)) ⊂ f(B_{r}(z)) ⊂ B_{Cr^{q}}(f(z)) whenever B_{r}(z) ⊂ G and r ≤ 1. Then the inverse images of points under f are discrete.

Proof. It suffices to assume that c = C = 1, p > q, 0 ∈ G, f(0) = 0 and that for some r_{0} > 0, B_{3r_{0}}(0) = G, and to show that B_{r_{0}}(0) ∩ f^{−1}(0) is finite.

Lemma 4.3. There are a positive constant a and a strictly increasing function h : [0, a] → [0, ∞] such that, whenever x ∈ B_{r_{0}}(0)∩f^{−1}(0) and u ∈ IR^{2} is a unit vector, then there is a curve ϕ : [0, a] → B_{2r_{0}}(x) with Lipschitz constant one such that ϕ(0) = x, ∥f(ϕ(t))∥ ≥ h(t) and f(ϕ(t)) ∈ ⋃_{s>0} B_{s/4}(su) for t ∈ (0, a].

Proof. Choose p−1 < α < q/(p−q). Fix a sufficiently large m and choose x_{m} ∈ B_{m^{−α/p}}(x) such that f(x_{m}) = m^{−α}u and recursively choose, for k < m, x_{k} ∈ B_{α^{1/p}k^{−(α+1)/p}}(x_{k+1}) ∩ B_{2r_{0}}(x) such that f(x_{k}) = k^{−α}u; the construction stops when either k = 1 or no such x_{k} exists. If x_{k} is defined, we use that (α + 1)/p > 1 to estimate ∑_{j=k}^{m−1} ∥x_{j+1} − x_{j}∥ ≤ c_{1}k^{1−(α+1)/p}. Noting that k^{−α} −(k +1)^{−α} ≤ αk^{−(α+1)}, we infer that there is an integer k_{0} independent of x such that x_{k_{0}} is defined (as long as m is large enough). Since α < q/(p−q), we can enlarge k_{0} if necessary to ensure that α^{q/p}k^{−(α+1)q/p} ≤ k^{−α}/4 for k ≥ k_{0}.

Let ϕ_{m}(t) (t ∈ [0, a_{m}]) be the arc-length parameterization of the path obtained by joining x, x_{m}, x_{m−1}, …, x_{k_{0}} (in this order) by the linear segments [x, x_{m}] and [x_{k+1}, x_{k}], k = 0, …, m − 1. Then a_{m} ≤ m^{−α/p} + c_{1}k_{0}^{1−(α+1)/p}, so we may find a subsequence of a_{m} converging to a ≤ c_{1}k_{0}^{1−(α+1)/p} such that the corresponding subsequence of ϕ_{m} converges to a Lipschitz curve ϕ : [0, a] → IR^{2}. Then ϕ(0) = x and f(ϕ(a)) = k_{0}^{−α}u, so a ≥ a_{0} > 0 where a_{0} is independent of x and u. For any 0 < t ≤ a denote s_{m} = min{a_{m}, t} and for any sufficiently large m choose k_{m} ≥ k_{0} such that ϕ_{m}(s_{m}) ∈ [x_{k_{m}+1}, x_{k_{m}}] and note first that s_{m} ≤ c_{1}k_{m}^{1−(α+1)/p}, hence a suitable subsequence of k_{m} has a limit k ≤ (t/c_{1})^{p/(p−1−α)}. Moreover ∥f(ϕ_{m}(s_{m})) − k_{m}^{−α}u∥ ≤ ∥ϕ_{m}(s_{m}) − x_{k_{m}}∥^{q} ≤ α^{q/p}k_{m}^{−(α+1)q/p} ≤ k_{m}^{−α}/4, which, upon taking the limit as m → ∞, gives that ∥f(ϕ(t)) − k^{−α}u∥ ≤ k^{−α}/4. Hence, f(ϕ(t)) ∈ ⋃_{s>0} B(su, s/4) and ∥f(ϕ(t))∥ ≥ (t/c_{1})^{−pα/(p−1−α)}/2 for any t ∈ (0, a].

Let r = h(a)^{1/q}/4 and assume, as we may, that 3r ≤ 1. We may also assume that t^{q} ≥ h(t) for 0 < t ≤ a, in particular r ≤ a/4. Denoting d = (h(r)/5)^{1/q}, we show that M = B_{r_{0}}(0) ∩ f^{−1}(0) has at most N = ((4r_{0} + d)/d)^{6} elements. Assume that M has more than N elements.

Let u_{k} = e^{kπi/3}. For each x ∈ M and k = 1, 3, 5 choose ϕ_{k,x} : [0, a] → B_{2r_{0}}(x) with Lipschitz constant one such that ϕ_{k,x}(0) = x, and ∥f(ϕ_{k,x}(t))∥ ≥ h(t) and f(ϕ_{k,x}(t)) ∈ ⋃_{s>0} B_{s/4}(su_{k}) for t > 0. Note that the last statements also show that f(ϕ_{k,x}(t)) ≠ 0 if t ≠ 0.

The triples (ϕ_{1,x}(r), ϕ_{3,x}(r), ϕ_{5,x}(r)), x ∈ M, belong to the product of discs of radius 2r_{0}; since N > ((2r_{0} + d/2)/(d/2))^{6}, we infer that there are x, y ∈ M, x ≠ y such that ∥ϕ_{k,x}(r) − ϕ_{k,y}(r)∥ < d for k = 1, 3, 5.

Whenever z ∈ [ϕ_{k,x}(r), ϕ_{k,y}(r)], then ∥f(z) − f(ϕ_{k,x}(r))∥ ≤ d^{q} ≤ h(r)/5. Finding s > 0 such that f(ϕ_{k,x}(r)) ∈ B_{s/4}(su_{k}), we infer from ∥f(ϕ_{k,x}(r))∥ ≥ h(r) that h(r) ≤ 5s/4, and we conclude that f(z) ∈ B_{s/4+h(r)/5}(su_{k}) ⊂ B_{s/2}(su_{k}). We also note that this implies that f(z) ≠ 0.

Let L_{k} be a simple curve joining x and y and lying in the set [ϕ_{k,x}(r), ϕ_{k,y}(r)] ∪ ϕ_{k,x}[0, r] ∪ ϕ_{k,y}[0, r]. By the theorem on θ-curves [Kur, ch. 10, § 61, II, Theorem 2], one of these curves, say L_{k}, lies, with the exception of its end points, entirely in the bounded component C of the complement of the remaining two. By connectedness, ϕ_{k,x}(0, a] ⊂ C. Since the ϕ’s have Lipschitz constant one and diam(C) ≤ 2r + d ≤ 3r, (3r)^{q} ≥ ∥f(ϕ_{k,x}(a)) − f(ϕ_{k,x}(0))∥ ≥ h(a) > (3r)^{q}. A contradiction.

As a simple corollary we now get The proof of Theorem 2.8(iii). If f satisfies the assumptions of Theorem 2.8(iii) then applying first Lemma 2.7 and then Proposition 4.2 with G = IR^{2} we get that, for some N < ∞ and for all y ∈ IR^{2}, f^{−1}(y) is a set consisting of at most N elements. The proof of the other two cases of Theorem 2.8 can now be carried over also for this case.

5. Nonlinear quotient mappings from IR^{2} to IR

Notice that there is no uniform quotient mapping from IR^{k} to IR^{n} for k < n. One way to see this is to notice that such a mapping would be Lipschitz and co-Lipschitz for large distances which leads to a contradiction when looking at the maximal number of disjoint balls of a certain radius contained in a ball of a larger radius. Thus, the simplest case of Lipschitz and uniform quotient mappings between Euclidean spaces is that of mappings from IR^{2} to IR (since from IR to IR they are all one to one). In this section we discuss briefly this case which, as we shall see, it is not entirely trivial. The main result is that, for Lipschitz quotient mappings, the inverse image of a point has finitely many components. Before we start consider the following two examples of Lipschitz quotient mappings from IR^{2} to IR . In both cases the mapping f is the distance from the solid lines multiplied, in each component of the complement of the solid lines, by the sign indicated.

Notice that f^{−1}(0) has one component in the first example and two in the second. It is easy to draw examples with an arbitrary finite number of components. Notice also that IR^{2} \ f^{−1}(0) has six components in the first example and three in the second.

Proposition 5.1. Let f : IR^{n} → IR be a uniform quotient mapping satisfying (2.1). Then for each t ∈ IR the number of components of IR^{n} \ f^{−1}(t) is finite and bounded by a function of n, ω(·) and Ω(·) only.

Proof. According to the remarks made in the introduction, each component of IR^{n}\f^{−1}(0) is mapped by f onto a component of IR \ {0}, i.e., onto either (0, ∞) or (−∞, 0). Recall that f is Lipschitz and co-Lipschitz for large distances and let L and δ, depending only on the moduli of uniform and co-uniform continuity, be such that for all z ∈ IR^{n} and all r ≥ 1, B_{δr}(f(z)) ⊂ f(B_{r}(z)) ⊂ B_{Lr}(f(z)).

Let D_{1}, …, D_{k} be distinct components of IR^{n} \ f^{−1}(0) intersecting B_{r}(0) for some r ≥ 1. Increasing r we may also assume that there are x_{i} ∈ D_{i} ∩ B_{r}(0) with |f(x_{i})| > 1 for i = 1, …, k. Note that each D_{i} intersect ∂B_{2r}(0) . Moreover, there is a y_{i} ∈ D_{i} ∩ ∂B_{2r}(0) such that |f(y_{i})| ≥ δr. Indeed, assuming f(x_{i}) > 1, there is an x_{i}^{1} ∈ B_{1}(x_{i}) so that f(x_{i}^{1}) > 1 + δ. Note that, if L ≤ 1 as we may assume, B_{1}(x_{i}) ⊂ D_{i}. There is an x_{i}^{2} ∈ B_{1}(x_{i}^{1}) so that f(x_{i}^{2}) > 1 + 2δ. Again B_{1}(x_{i}^{1}) ⊂ D_{i}. Continuing this way at least [r + 1] times (and interpolating between the last two points) we get a y_{i} ∈ D_{i} ∩ ∂B_{2r}(0) with f(y_{i}) ≥ 1 + δ[r] ≥ δr. It follows that B_{δr/L}(y_{i}) ⊂ D_{i} and we get k disjoint balls of radius δr/L included in a ball of radius 3r. Consequently, k ≤ (3L/δ)^{n}.

We now aim at proving, in Proposition 5.4, that for each t ∈ IR every component of f^{−1}(t) separates the plane. We first need two lemmas.

Lemma 5.2. Let f : IR^{n} → IR be a continuous open mapping. Then, for every t ∈ IR , no component of f^{−1}(t) is bounded.

Proof. Assume A is a compact component of f^{−1}(0). Let U be an open bounded connected set containing A whose boundary does not meet f^{−1}(0). One way to get such a set is to let r be such that A ⊂ B_{r}(0) and let B be the union of IR^{n} \ B_{r}(0) with all the components of f^{−1}(0) meeting IR^{n} \B_{r}(0) (which is a closed set). B and A are components of B ∪ f^{−1}(0) so there is and open set V ⊂ IR^{n} \ B which contains A and whose boundary does not intersect B ∪ f^{−1}(0). Now let U ⊂ V be the component containing A.

Next we would like to make sure that the boundary of U is connected. To do that look at the complement, in IR^{n}, of the unbounded component of IR^{n} \ U. Replace U with the component of this set containing A. By [Kur, ch. 8, § 57, II, Theorem 6] the boundary of this set is connected. We now have an open bounded connected set containing A whose connected boundary does not meet f^{−1}(0). The boundary of such a set and thus also the set itself is mapped by f into either (0, ∞) or (−∞, 0), a contradiction.

Lemma 5.3. Let f : IR^{2} → IR be a Lipschitz quotient mapping. Then, for each t ∈ IR and for each ball B, the number of components of f^{−1}(t) intersecting B is finite.

Proof. Assume B_{r}(f(x)) ⊂ f(B_{r}(x)) ⊂ B_{Lr}(f(x)) for all x ∈ IR^{2} and all r > 0 and assume that the number of components of f^{−1}(0) intersecting B_{r}(0) is infinite. Fix 0 < ϵ < r/(2+L), then there are infinitely many components A_{i} such that the distance between any two of A_{i} ∩ B_{r}(0) is less than ϵ. Since all the A_{i} are unbounded, we can find two of them, say A_{1} and A_{2} such that the distance between A_{1} ∩ ∂B_{3r}(0) and A_{2} ∩ ∂B_{3r}(0) is less than ϵ.

Let y ∈ A_{1} ∩ B_{r}(0) and z ∈ A_{1} ∩ ∂B_{3r}(0) be such that A_{2} ∩ B_{ϵ}(y) ≠ ∅ ≠ A_{2} ∩ B_{ϵ}(z). Arguing similarly to Case IV of the proof of Lemma 2.6, we get a bounded connected open set G which meets both discs B_{ϵ}(y) and B_{ϵ}(z) and whose boundary is contained in A_{1} ∪ A_{2} ∪ B_{ϵ}(y) ∪ B_{ϵ}(z). The latter property of G gives that |f(x)| < Lϵ on ∂G, while the former gives that {x ∈ G : 2r − ϵ < |x| < 2r + ϵ} is nonempty and open, hence it contains a point u with |u| = 2r. By Lemma 2.2 we may find a curve ϕ : [0, ∞) → IR^{2} with Lipschitz constant one, ϕ(0) = u, and such that f(ϕ(t)) = f(u) + tsign(f(u)). Since this curve is clearly unbounded, there is τ > 0 such that ϕ(τ) lies on the boundary of G; then ϕ(τ) ∈ B_{ϵ}(y) ∪ B_{ϵ}(z), because f(ϕ(τ)) ≠ 0 and f is zero on A_{1} ∪ A_{2}. Hence Lϵ > |f(ϕ(τ))| ≥ τ ≥ ∥ϕ(τ) − ϕ(0)∥ ≥ r − 2ϵ, which contradicts the choice of ϵ.

Proposition 5.4. Let f : IR^{2} → IR be a Lipschitz quotient mapping. Then, for each t ∈ IR , each component of f^{−1}(t) separates the plane.

Proof. Let A be a component of f^{−1}(t). By Lemma 5.3 f^{−1}(t) \ A is closed. Let G be the component of IR^{2} \(f^{−1}(t)\A) containing A. G is an open and connected set. Assume now that A does not separate the plane then we claim that A also does not separate G.

Indeed, G\A = G∩(IR^{2} \A), both sets in the intersection are connected and we can apply [Kur, ch. 8, § 57, II, Theorem 2] to deduce that G \ A is connected. But it is impossible that A separates G: if G \ A is connected, f maps it to either (0, ∞) or (−∞, 0) but near any point of A there are points whose images are positive and points whose images are negative. This contradiction finishes the proof.

Propositions 5.1 and 5.4 now imply

Corollary 5.5. Let f : IR^{2} → IR be a Lipschitz quotient mapping. Then, for each t ∈ IR , f^{−1}(t) has a bounded number of components. The upper bound of the number of components depends only on the Lipschitz and co-Lipschitz constants of f.

There are two unsettled problems related to the material of this section. One is whether one can weaken the assumptions of Lipschitz quotient to uniform quotient in the appropriate places. The other is to what extent the number of components of f^{−1}(t) or of IR^{2} \ f^{−1}(t) is independent of t. An examination of the examples above shows that these numbers may depend on t but leaves the possibility that after excluding finitely many t they are constants.

References

[And] R. D. Anderson, On monotone interior mappings in the plane, Trans AMS 73 (1952),

211–222.

[BJLPS] S. Bates, W. B. Johnson, J. Lindenstrauss, D. Preiss, and G. Schechtman, Affine

approximation of Lipschitz functions and nonlinear quotients,

[FK] H. M. Farkas and I. Kra, Riemann surfaces, Springer (1992).

[Jam] I. M. James, Introduction to uniform spaces, Cambridge University Press (1990).

[Kur] C. Kuratowski, Topology II, Academic Press (1968), translation from French of Topologie, volume II, Warsawa, (1961).

[Sto] S. Stoilow, Lecons sur les principes topologiques de la théorie des fonctions analytiques, 2nd ed., Paris, Gauthier-Villars (1956).

W.B. Johnson  J. Lindenstrauss
Department of Mathematics  Institute of Mathematics
Texas A&M University  The Hebrew University of Jerusalem
College Station, TX 77843–3368 U.S.A.  Jerusalem, Israel

email:  johnson@math.tamu.edu        email:  joram@math.huji.ac.il

D. Preiss  G. Schechtman
Department of Mathematics  Department of Mathematics
University College London  The Weizmann Institute of Science
London, Great Britain  Rehovot, Israel

email:  dp@math.ucl.ac.uk            email:  gideon@wisdom.weizmann.ac.il