Diffeomorphism on path connected components and applications

Journal Pre-proof Diffeomorphism on path connected components and applications

Hong Ding

PII:

S0022-247X(20)30091-3

DOI:

https://doi.org/10.1016/j.jmaa.2020.123929

Reference:

YJMAA 123929

To appear in:

Journal of Mathematical Analysis and Applications

Received date:

7 July 2019

Please cite this article as: H. Ding, Diffeomorphism on path connected components and applications, J. Math. Anal. Appl. (2020), 123929, doi: https://doi.org/10.1016/j.jmaa.2020.123929.

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Diffeomorphism on path connected components and applications Hong Ding College of Applied Sciences, Beijing University of Technology, Beijing, P.R. China E-mail: [email protected] Abstract In this paper, we prove the following: Let F = (F1 , F2 ) ∈ C ∞ (R2 , R2 ). Let R > 0. And suppose det(DF (x)) > 0, ∀x ∈ B(0, R). Suppose there exist K > 0, r ∈ (0, K] and a ∈ R2 satisfying F −1 (B(a, r)) ∩ B(0, R) = ∅ and F (∂B(0, R)) ⊂ B(a, K) \ B(a, r). Let H be any path connected component of F −1 (B(a, r)) ∩ B(0, R). Then F |H : H → B(a, r) is an onto diffeomorphism. We give some applications of this result.

Keywords: diffeomorphism, Hausdorff measure, escape lemma. AMS Subject Classification: 58C25, 57R50, 26B10. 1. Introduction A classical theorem in the area of global inversion states that a local diffeomorphism f : X → Rn is a diffeomorphism ( here X ⊂ Rn a compact set) if f |∂X is one-to-one ( see [9]). Naturally, we want to know what we can get if we don’t have the condition that f |∂X is one-to-one. In this paper, we get a conclusion in R2 without using the injective condition on the boundary. We ask the boundary of the disk ∂B(0, R) to be mapped into an annular region.(Here the closed disk B(0, R) is the domain of our local diffeomorphism.) And we can still get a result of diffeomorphism over what could be a quite large region. Actually, our Theorem 1 states the following: ” Let F = (F1 , F2 ) ∈ C ∞ (R2 , R2 ). Let R > 0. And suppose det(DF (x)) > 0, ∀x ∈ B(0, R). Suppose there exist K > 0, r ∈ (0, K] and a ∈ R2 satisfying F −1 (B(a, r)) ∩ B(0, R) = ∅ and F (∂B(0, R)) ⊂ B(a, K) \ B(a, r). Let H be any path connected component of F −1 (B(a, r)) ∩ B(0, R). Then F |H : H → B(a, r) is an onto C ∞ -diffeomorphism.” And after we get Theorem 1, we have two applications for Theorem 1. One of the applications is to give a proof of a version of the Hadamard’s global inverse function theorem in the plane for functions that map a closed disk centered at origin into another closed disk centered at origin. The other application is Theorem 2. 2. The lemmas Throughout this paper we use the definition: B(a, r) = {x ∈ R2 | x − a < r}. Here a ∈ R2 . The norm is the Euclidean norm. 1

In order to prove the main theorem, we have a few lemmas. In the proof of Lemma 1 and Lemma 2, we use the area formula in geometric measure theory. Let us first state the theorem. We copy the theorem from [10], which is Theorem 3.7 there. The Area Formula Consider a Lipschitz function f : Rm → Rn for m ≤ n. If A is an L m measurable set, then   Jm f (x)dL m x = N (f | A, y)dH m y  Rn

A

Some explanation of the notations: Here Jm f (x) is the m-dimensional Jacobian of f at x. Computationally, Jm f (x) equals the square root of the determinant of the m×m matrix (Df (x))t Df (x). Here Df (x) is the Jacobian matrix of f at x. N (f | A, y) = card{x ∈ A : f (x) = y} (card(X) = the number of elements in the set X. ) H m is the m-dimensional Hausdorff measure. L m is the Lebesgue measure. Now we have our first lemma. Lemma 1 Let f = (f1 , f2 ) ∈ C 1 (R2 , R2 ), i ∈ {1, 2}, K ⊂ R2 be a compact set and the Jacobian det(Df (x)) > 0 , x ∈ K. Then for every c ∈ fi (K), the 1-dimensional Hausdorff measure of fi−1 (c) ∩ K is finite. Proof. Fix i = 1 or 2. We can see that fi−1 (c) ∩ K is closed and bounded. ∂fi 2 ) + So it is compact. Let (x, y) ∈ fi−1 (c) ∩ K. Then fi (x, y) − c = 0 and ( ∂x 1 ∂fi 2 ∂fi ( ∂x2 ) = 0 at (x, y). Without loss of generality, suppose ∂x1 (x, y) = 0. By the implicit function theorem (Theorem 17.3 of [4]), there exists a neighborhood W = (y − r, y + r) of y, and a C 1 function g : W → R, so that fi (g(x2 ), x2 ) ≡ c on W . And the conditions: | x1 − x |< r, x2 ∈ W, and fi (x1 , x2 ) = c imply that x1 = g(x2 ). In case that g  is unbounded on W , we can find a smaller open interval W  = (y − 2r , y + 2r ) and let g¯ = g |W  , then g¯ is bounded on W  . g (x2 ), x2 ) ≡ c on W  and 2) The And the other two conditions, [i.e. : 1) fi (¯ r  conditions: | x1 − x |< 2 , x2 ∈ W , and fi (x1 , x2 ) = c imply that x1 = g¯(x2 )] are automatically satisfied. So here without loss of generality, we assume that g  is bounded on W . Let V = {(x1 , x2 ) ∈ R2 : | x1 − x |< r, x2 ∈ W }. Then V is an open neighborhood of (x, y) ∈ R2 . And fi−1 (c) ∩ V ⊂ {(g(x2 ), x2 ) ∈ R2 | x2 ∈ W }. Now define j : W → R2 , x2 → (g(x2 ), x2 ). Then j is one-to-one on W . By Theorem 3.7(the area formula) of [10],   J1 j(x2 )dx2 = N (j | W, y)dH 1 y = H 1 (j(W )) W

R2

≥ H 1 (fi−1 (c) ∩ V )

 Here H 1 is the 1-dimensional Hausdorff measure. Meanwhile W J1 j(x2 )dx2 =   (g  (x2 ))2 + 1 dx2 is finite. So we have shown for every (x, y) ∈ fi−1 (c) ∩ K, W 2

∃ an open neighborhood V of (x, y) so that H 1 (fi−1 (c) ∩ V ) is finite. A finite number of such V ’s cover fi−1 (c) ∩ K . Suppose they are V1 , · · · , Vp . Then fi−1 (c) ∩ K ⊂ V1 ∪ V2 ∪ · · · ∪ Vp p ⇒ fi−1 (c) ∩ K ⊂ fi−1 (c) ∩ (V1 ∪ · · · ∪ Vp ) = j=1 (fi−1 (c) ∩ Vj ), ⇒ H 1 (fi−1 (c) ∩ K) ≤



H 1 (fi−1 (c) ∩ Vj ) < +∞

j

 Lemma 2 Let f = (f1 , f2 ) : R2 → R2 , f ∈ C ∞ (R2 , R2 ). Let R > 0. And suppose det(Df (x)) > 0, ∀x ∈ B(0, R). Suppose there exist K > 0, r ∈ (0, K] so that f (∂B(0, R)) ⊂ B(0, K) \ B(0, r). Here ∂B(0, R) represents the boundary of the set B(0, R). Suppose B(0, R) ∩ f −1 (B(0, r)) = ∅. Let ξ ∈ B(0, R) ∩ f −1 (B(0, r)), c = f2 (ξ). Let Γ be the connected component of f2−1 (c) ∩ B(0, R) containing ξ. Then f (Γ) ⊃ {(x, c) ∈ R2 | x2 + c2 < r2 }. Similar result exists for f1 . Proof. Let f = (f1 , f2 ). Define f˜ = f |B(0,R) , f˜i = fi |B(0,R) , i = 1, 2. We have rank(f˜i ) ≡ 1 on B(0, R), i = 1, 2. f˜i : B(0, R) → R is C 1 . Let −1 ξ ∈ B(0, R)∩f −1 (B(0, r)), c = f˜2 (ξ). By Theorem 3.2 of Chapter 1 of [7], f˜2 (c) −1 is a regular submanifold of class C 1 of B(0, R). And dim(f˜2 (c)) = 1. Let Γ −1 be the connected component of f˜2 (c) containing ξ. Then Γ is a connected 1-manifold of class C 1 . By Corollary 3.4.6 of [1], we can always have a global parametrization by arc length of Γ. Suppose the parametrization of Γ is:  Γ = {h0 (s) = (h01 (s), h02 (s)) ∈ B(0, R) | (h01 (s))2 + (h02 (s))2 ≡ 1, s ∈ (aΓ , bΓ )} There exists s0 ∈ (aΓ , bΓ ) so that ξ = h0 (s0 ). It can be seen that h0 (s), s ∈ (aΓ , bΓ ) is the maximal integral curve of the initial value problem (IVP): ⎧ ∂ f˜2 ⎪ dx1 ⎪ ∂x2 ⎪  = ⎪ ⎪ ⎪ ds ∂ f˜2 2 ∂ f˜2 2 ⎪ ( ∂x ) + ( ∂x ) ⎪ ⎨ 1 2 ˜

∂ f2 − ∂x dx2 1 ⎪ ⎪ = ⎪ ⎪ ds ∂ f˜2 2 ∂ f˜2 2 ⎪ ⎪ ( ∂x1 ) + ( ∂x ) ⎪ 2 ⎪ ⎩ (x1 (s0 ), x2 (s0 )) = ξ

(1)

For the argument see [3]. So df˜1 (h0 (s)) det(Df˜) = (h0 (s)) > 0, ∀s ∈ (aΓ , bΓ ) ds ∂ f˜2 2 ∂ f˜2 2 ( ∂x ) + ( ) ∂x 1 2

3

(2)

If Γ is diffeomorphic to S1 , then f˜1 (h0 (s)) should be a periodic function of s. This contradicts (2). So Γ can only be diffeomorphic to R.(By the classification of connected 1-manifolds.) So there is a one-to-one correspondence between parameters s and points of Γ. By Lemma 1, (3) H 1 (f˜2−1 (c)) < +∞ Then the domain of (h01 , h02 ) must be finite by (3) and Theorem 3.7(the area formula ) of [10]. So both aΓ , bΓ are finite. By escape lemma (Lemma 5.1 of chapter IV of [2]), for an increasing sequence {tn } converging to bΓ , {(h01 (tn ), h02 (tn ))} cannot converge to a limit in B(0, R). But because {(h01 (tn ), h02 (tn ))} is a Cauchy sequence in R2 , it must converge to a point in R2 . So the limit is on the boundary of B(0, R). So lim distance((h01 (tn ), h02 (tn )), ∂B(0, R)) = 0 n→∞

⇒ lim distance( f (h01 (tn ), h02 (tn )), f (∂B(0, R)) = 0 n→∞

Define Γ+ = {(h01 (s), h02 (s)) ∈ B(0, R) | s0 ≤ s < bΓ }. Then we have distance( f (Γ+ ), ∂B(0, K) ) ≤ K − r

(4)

for f (∂B(0, R)) ⊂ B(0, K) \ B(0, r). If we let Γ− = {h0 (s) | aΓ < s ≤ s0 }, then distance(f (Γ− ), ∂B(0, K)) ≤ K − r. And we know by the intermediate value theorem of continuous function f (Γ+ ) = {(x, c) ∈ R2 | x ∈ [f1 (ξ), lim f1 (h0 (s)) )} sbΓ

If [ lim f1 (h0 (s))]2 + c2 < r2 , then we have distance(f (Γ+ ), ∂B(0, K)) > K − r. sbΓ

That contradicts (4). Therefore [ lim f1 (h0 (s))]2 + c2 ≥ r2 sbΓ

Similar result holds for f (Γ− ). So we have f (Γ) = f (Γ+ ) ∪ f (Γ− ) ⊃ {(x, c) ∈ R2 | x2 + c2 < r2 }.  Lemma 3 Let f = (f1 , f2 ) ∈ C ∞ (R2 , R2 ). Let R > 0. And suppose det(Df (x)) > 0, ∀x ∈ B(0, R) . Suppose there exist K > 0, r ∈ (0, K], and f (∂B(0, R)) ⊂ B(0, K) \ B(0, r). Let λ ∈ f (B(0, R)) ∩ B(0, r) = ∅, μ ∈ f −1 (λ) ∩ B(0, R). Then there exists a continuous function L : B(0, r) → B(0, R), so that f (L(u)) = u, ∀u ∈ B(0, r) and L(λ) = μ. Proof. Fix λ ∈ f (B(0, R)) ∩ B(0, r) and μ ∈ f −1 (λ) ∩ B(0, R). Let Γ1 be the connected component of f1−1 (f1 (μ))∩B(0, R) containing μ. Then Γ1 is a regular submanifold of B(0, R) of dimension 1 (Theorem 5.8 of chapter III of [2]). And 4

it is connected and of class C 1 . By Corollary 3.4.6 of [1], there exists a global parametrization by arc length of Γ1 : Γ1 = {h1 (s) = (h11 (s), h12 (s)) ∈ R2 : s ∈ (aΓ1 , bΓ1 )} Suppose h1 (0) = μ. Then we can see s → h1 (s) is the maximal integral curve of the IVP of ODE: ⎧ ∂f1 dx1 2 ⎪ =  ∂f1 ∂x ⎪ ∂f ⎪ ds ⎪ ( ∂x )2 +( ∂x1 )2 ⎨ 1 2 ∂f

⎪ ⎪ ⎪ ⎪ ⎩

So

dx2 ds

=

− ∂x1 1  ∂f1 2 ∂f ( ∂x ) +( ∂x1 )2 1

here,(x1 , x2 ) ∈ B(0, R)

2

(x1 (0), x2 (0)) = μ

− det(Df ) df2 (h1 (s)) = (h1 (s)) < 0, ∀s ∈ (aΓ1 , bΓ1 ). And by Lemma ∂f1 2 ∂f1 2 ds ( ∂x1 ) + ( ∂x ) 2

2, f (Γ1 ) ⊃ {(f1 (μ), y) ∈ R2 : (f1 (μ))2 + y 2 < r2 }. So, there exists a unique p1 ∈ Γ1 so that f2 (p1 ) = 0. Let u = (u1 , u2 ) ∈ B(0, r). Let Γ2 be the connected component of f2−1 (0) ∩ B(0, R) containing p1 . For Γ2 is a connected C 1 manifold of dimension 1, there exists a global parametrization by arc length of Γ2 (Corollary 3.4.6, [1] ): Γ2 = {h2 (s) = (h21 (s), h22 (s)) ∈ R2 : s ∈ (aΓ2 , bΓ2 )}, (h2 (0) = p1 ). Then h2 (s) is the maximal integral curve of the IVP of ODE: ⎧ dx1 ⎪ ⎪ ⎪ ds = ⎪ ⎨ dx2 ⎪ ⎪ ds

⎪ ⎪ ⎩

=

∂f2 ∂x2  ∂f ∂f ( ∂x2 )2 +( ∂x2 )2 1

∂f

− ∂x2

2

here,(x1 , x2 ) ∈ B(0, R)

 ∂f ∂f ( ∂x2 )2 +( ∂x2 )2 1

1

2

(x1 (0), x2 (0)) = p1

Define θ2 (p1 , s) = h2 (s). By Theorem 17.19 of [8], θ2 ∈ C ∞ . Because det(Df ) df1 (h2 (s)) = (h2 (s)) > 0, ∀s ∈ (aΓ2 , bΓ2 ). ∂f2 2 ∂f2 2 ds ( ∂x ) + ( ) ∂x2 1 And by Lemma 2, f (Γ2 ) ⊃ {(x, 0) ∈ R2 : x2 + 02 < r2 }. So there exists a unique s1 ∈ (aΓ2 , bΓ2 ) so that f1 (h2 (s1 )) = u1 . So s1 = s1 (u1 ) is a function of u1 . By the implicit function theorem (Theorem 17.3 of [4] ), s1 ∈ C 1 . Let p2 = h2 (s1 ) = θ2 (p1 , s1 (u1 )). Now let Γ3 be the connected component of f1−1 (f1 (p2 )) ∩ B(0, R) containing p2 . ⇒ Γ3 is a connected C 1 1-manifold. Let the global parametrization by arc length of Γ3 be: Γ3 = {h3 (s) = (h31 (s), h32 (s)) ∈ R2 : s ∈ (aΓ3 , bΓ3 )}, (h3 (0) = p2 )

5

Then h3 (s) is the maximal integral curve of the IVP: ⎧ dx1 ⎪ ⎪ ⎪ ds = ⎪ ⎨ dx2 ⎪ ⎪ ds

⎪ ⎪ ⎩

=

∂f1 ∂x2  ∂f ∂f ( ∂x1 )2 +( ∂x1 )2 1

∂f

2

− ∂x1 1  ∂f1 2 ∂f ( ∂x ) +( ∂x1 )2 1

here,(x1 , x2 ) ∈ B(0, R)

(5)

2

(x1 (0), x2 (0)) = p2

Define θ1 (p2 , s) = h3 (s). By Theorem 17.19 of [8], θ1 ∈ C ∞ . From (5) we get, − det(Df ) df2 (h3 (s)) = (h3 (s)) < 0, ∀s ∈ (aΓ3 , bΓ3 ). ∂f1 2 ∂f1 2 ds ( ∂x ) + ( ) ∂x 1 2 By Lemma 2, f (Γ3 ) ⊃ {(u1 , y) ∈ R2 : u21 + y 2 < r2 }. So u = (u1 , u2 ) ∈ f (Γ3 ). And ∃ a unique s2 ∈ (aΓ3 , bΓ3 ) so that f2 (h3 (s2 )) = u2 . So s2 = s2 (u1 , u2 ) is a function of (u1 , u2 ). And by the implicit function theorem (Theorem 17.6 of [4]), s2 ∈ C 1 . Now define L(u1 , u2 ) = h3 (s2 ) = θ1 (p2 , s2 (u1 , u2 )) = θ1 (θ2 (p1 , s1 (u1 )), s2 (u1 , u2 )) ∈ B(0, R) Clearly we have L ∈ C 1 (B(0, r), B(0, R)). Furthermore, we have f (L(u1 , u2 )) = (f1 (h3 (s2 )), f2 (h3 (s2 ))) = (u1 , u2 ), ∀(u1 , u2 ) ∈ B(0, r).  3. The main theorem and applications. Proposition 1 Let f = (f1 , f2 ) ∈ C ∞ (R2 , R2 ). Let R > 0. And suppose det(Df (x)) > 0, ∀x ∈ B(0, R). Suppose there exist K > 0, r ∈ (0, K] satisfying f −1 (B(0, r)) ∩ B(0, R) = ∅ and f (∂B(0, R)) ⊂ B(0, K) \ B(0, r). Let H be any path connected component of f −1 (B(0, r)) ∩ B(0, R). Then f |H : H → B(0, r) is an onto C ∞ -diffeomorphism. Remark Please note that when using the proposition, there is no need to know that f |∂B(0,R) is one-to-one . Proof. By C ∞ -diffeomorphism, we use the definition from the book [2](on p.41). So the map needs to be from an open set in R2 onto an open set in R2 . B(0, r) is certainly open in R2 . So we need to prove that H is open in R2 . Because f is continuous, we see that f −1 (B(0, r)) ∩ B(0, R) is open in R2 . By Theorem 25.4 of [11], H is open in R2 , since R2 is locally path connected. (Path connected component is called path component in [11].) 6

Let g = f |H : H → B(0, r). Pick any point μ ∈ H. Let λ = g(μ) = f (μ). Then λ ∈ B(0, r) ∩ f (B(0, R)). ⇒ μ ∈ f −1 (λ) ∩ B(0, R). By Lemma 3, there exists a continuous L : B(0, r) → B(0, R), so that f (L(u)) = u, ∀u ∈ B(0, r). And L(λ) = μ. We know that L(B(0, r)) is path connected. For the reason f (L(u)) = u, ∀u ∈ B(0, r), we have f (L(B(0, r))) ⊂ B(0, r). So L(B(0, r)) ⊂ f −1 (B(0, r)). Certainly, L(B(0, r)) ⊂ f −1 (B(0, r)) ∩ B(0, R). Also we have L(λ) = μ ∈ H, (λ ∈ B(0, r).) So we get L(B(0, r)) ⊂ H (which is a path connected component of f −1 (B(0, r)) ∩ B(0, R). ) Also, g(L(B(0, r))) = f (L(B(0, r))) = B(0, r). We get from L(B(0, r)) ⊂ H, the conclusion that g(H) ⊃ B(0, r). So g is surjective. Next we prove that g is injective. The proof is essentially the same as in the proof of Theorem 7 of [3]. Suppose g is not injective. Then ∃z1 , z2 ∈ H, z1 = z2 , so that f (z1 ) = f (z2 ). By Lemma 3, we can have a continuous L : B(0, r) → B(0, R) such that L(f (z1 )) = z1 . There exists a path δ : [0, 1] → H, so that δ(0) = z1 and δ(1) = z2 . Then by property of L, f (L(f (δ(t)))) = f (δ(t)), ∀t ∈ [0, 1]. So, both δ and L ◦ f ◦ δ are liftings of f ◦ δ : t → f (δ(t)) with respect to f . Because f is a local homeomorphism, by the uniqueness of lifting for local homeomorphism (Theorem 4.8 of [6] ), δ(0) = z1 , L ◦ f ◦ δ(0) = L(f (z1 )) = z1 , ⇒ δ(1) = L ◦ f ◦ δ(1). That is z2 = δ(1) = L ◦ f ◦ δ(1) = L(f (z2 )) = L(f (z1 )) = z1 . This contradicts the assumption z1 = z2 . So we have proven that g is injective. So the inverse g −1 exists. And by the inverse function theorem (Theorem 6.4 of Chapter II of [2]), g : H → B(0, r) is a homeomorphism. What is left is to show that the inverse is C ∞ . Again by the inverse function theorem, this is done. So g : H → B(0, r) is an onto C ∞ -diffeomorphism.  Now we get our main theorem. Theorem 1 Let F = (F1 , F2 ) ∈ C ∞ (R2 , R2 ). Let R > 0. And suppose det(DF (x)) > 0, ∀x ∈ B(0, R). Suppose there exist K > 0, r ∈ (0, K] and a ∈ R2 satisfying F −1 (B(a, r)) ∩ B(0, R) = ∅ and F (∂B(0, R)) ⊂ B(a, K) \ B(a, r). Let H be any path connected component of F −1 (B(a, r)) ∩ B(0, R). Then F |H : H → B(a, r) is an onto C ∞ -diffeomorphism. Proof. Let f (x) = F (x) − a, ∀x ∈ R2 . So we can apply Proposition 1 to f. Then we get the conclusion.  Using Theorem 1, we can prove a version of Hadamard’s global inverse function theorem in the plane for functions that map a closed disk centered at origin into another closed disk centered at origin. In order to prove the theorem, we need a lemma. Lemma 4 Let f = (f1 , f2 ) : R2 → R2 , f ∈ C 1 (R2 , R2 ). And suppose det(Df (x)) > 0, ∀x ∈ B(0, 1). Furthermore suppose f (∂B(0, 1)) ⊂ ∂B(0, 1). Then we have: f (B(0, 1)) ⊂ B(0, 1). 7

Proof. Because f is continuous on B(0, 1), so f assumes a maximum on B(0, 1). Because f is a local diffeomorphism on B(0, 1), the maximum point of f cannot be in B(0, 1). So the maximum point of f must be on the boundary ∂B(0, 1). Because f (∂B(0, 1)) ⊂ ∂B(0, 1), f = 1 on ∂B(0, 1), f ≤ 1 on B(0, 1). And furthermore we have ∀x ∈ B(0, 1), f (x) < 1. Because if f (x) = 1 for some x ∈ B(0, 1), the point x will be a maximum point of f inside B(0, 1). That contradicts what we have above. So f (B(0, 1)) ⊂ B(0, 1).  Next we prove the version of the Hadamard’s global inverse function theorem in the plane via Theorem 1. Corollary 1 ( A version of the Hadamard’s global inverse function theorem in the plane ) Let f = (f1 , f2 ) ∈ C ∞ (R2 , R2 ). Let R > 0. And suppose det(Df (x)) > 0, ∀x ∈ B(0, R). Let K > 0 so that f (∂B(0, R)) ⊂ ∂B(0, K). Then f |B(0,R) : B(0, R) → B(0, K) is an onto C ∞ -diffeomorphism. Proof. Let g = f |B(0,R) . By Lemma 4, g(B(0, R)) ⊂ B(0, K). First we prove that g : B(0, R) → B(0, K) is injective. Suppose there exist a, b ∈ B(0, R), a = b, and p ∈ B(0, K), so that p = f (a) = f (b). Let l be the line segment connecting a and b. Let a, b also represent the vectors which are starting from 0 = (0, 0) and ending at points a and b. Then any point on the line segment l can be represented as l(t) = (1 − t)a + tb, t ∈ [0, 1]. We know that no point f (l(t)) can be on ∂B(0, K). So the distance(f (l(t)), (K cos θ, K sin θ)) > 0, ∀t ∈ [0, 1], θ ∈ [0, 2π]. Because the distance function is a continuous function, so there is a positive minimum α on the compact set [0, 1] × [0, 2π]. So there exists an α > 0, so that distance(f (l), ∂B(0, K)) ≥ α So f (l) ⊂ B(0, K − α). Now choose an R1 ∈ (0, R) close to R enough, so that it satisfies (a) a, b ∈ B(0, R1 ), and (b) f (∂B(0, R1 )) ⊂ B(0, K) \ B(0, r1 ), where r1 ∈ (K − α2 , K). Then we get, a, b ∈ f −1 (B(0, r1 )) ∩ B(0, R1 ) and

l ⊂ f −1 (B(0, r1 )) ∩ B(0, R1 ).

So there is a path connected component H of f −1 (B(0, r1 )) ∩ B(0, R1 ) so that a, b ∈ H. By Theorem 1, (Please note that we do not know that f |∂B(0,R1 ) is one-to-one, but we can still use Theorem 1. ) f |H : H → B(0, r1 ) is an onto C ∞ -diffeomorphism. That contradicts the fact f (a) = f (b). So there can’t be two points a, b ∈ B(0, R), so that f (a) = f (b). So g is injective. To prove g is surjective, pick any point q ∈ B(0, K). Pick a point p ∈ B(0, R), then f (p) ∈ B(0, K). Certainly there exists an r2 ∈ (0, K), so that q, f (p) ∈ B(0, r2 ). Choose R2 ∈ (0, R) close to R enough, so that (1) p ∈ B(0, R2 ) (2) f (∂B(0, R2 )) ⊂ B(0, K) \ B(0, r2 ). 8

So, p ∈ f −1 (B(0, r2 )) ∩ B(0, R2 ). Let H1 be the path connected component of f −1 (B(0, r2 ))∩B(0, R2 ) containing p. By Theorem 1, f |H1 : H1 → B(0, r2 ) is an onto C ∞ -diffeomorphism. So there q ) = q. So g is surjective. Because we also exists q˜ ∈ H1 ⊂ B(0, R) so that f (˜ have that g is injective. By the similar argument at the end of the proof of Proposition 1, g : B(0, R) → B(0, K) is an onto C ∞ -diffeomorphism.  Another application is the following theorem. Theorem 2 Let f = (f1 , f2 ) ∈ C ∞ (R2 , R2 ). Suppose the Jacobian satisfies 0 < det(Df (x)) ≤ 1 on R2 . Let R > 0, r > 0, a ∈ R2 , and f −1 (B(a, r)) ∩ B(0, R) = ∅. Suppose f (∂B(0, R)) ⊂ R2 \B(a, r). If r > √12 R, then card f −1 (y) ∩ B(0, R) = 1, ∀y ∈ B(a, r). (Here card(X) represents the number of elements in the set X. ) Proof. Let H be any path connected component of f −1 (B(a, r)) ∩ B(0, R). We know from the proof of Proposition 1 that H is open in R2 . Because f is continuous, there exists a K > 0 such that f (∂B(0, R)) ⊂ B(a, K). So f (∂B(0, R)) ⊂ B(a, K) \ B(a, r). Then by Theorem 1, f |H : H → B(a, r) is an onto diffeomorphism. Then by Theorem (2.47) of [5],  | det(Df (x)) | dx = m(B(a, r)) H

m is the Lebesgue measure on R2 . So   1dx ≥ | det(Df (x)) | dx = m(B(a, r)) H

If r >

H

√1 R, 2

1 2 πR (6) 2 Now let y ∈ B(a, r). Because f |H : H → B(a, r) is an onto diffeomorphism, and H ⊂ B(0, R), there exists at least one x ∈ B(0, R) such that f (x) = y. If ∃x1 , x2 ∈ B(0, R), x1 = x2 such that f (x1 ) = f (x2 ) = y. Then x1 , x2 cannot be in the same path connected component of f −1 (B(a, r)) ∩ B(0, R), because f |H : H → B(a, r) is one-to-one. So there exist at least two path connected components in f −1 (B(a, r))∩B(0, R). Denote them by H1 , H2 . Then by (6), m(Hi ) > 12 πR2 , i = 1, 2. For H1 ∩ H2 = ∅, we have m(H1 ∪ H2 ) = 2 m(H 1 ) + m(H2 ) > πR

. This contradicts the fact H1 ∪ H2 ⊂ B(0, R). So −1 card f (y) ∩ B(0, R) = 1.  ⇒ m(H) ≥ πr2 >

Acknowledgements: The author would like to thank the reviewers and editors for their careful reading of the manuscript and constructive suggestions.

References [1] Berger,M. , Gostiaux,B. , Differential geometry: manifolds, curves and surfaces, Springer, New York, 1988. 9

[2] Boothby,W. M. , An introduction to differentiable manifolds and Riemannian geometry, second edition, Academic Press, San Diego, 1986. [3] Ding, H. , Behavior on level sets and global inversion, Applicable Analysis, vol.94(2015), No.9, 1838-1850. [4] Fitzpatrick, P. M., Advanced calculus, second edition, PWS Publishing Company, 2006. [5] Folland, G. B., Real analysis, John Wiley & Sons, 1984. [6] Forster, O. , Lectures on Riemann surfaces, Springer, New York, 1981. [7] Hirsch, M. W. , Differential topology, Springer, New York, 1976. [8] Lee, John M. , Introduction to smooth manifolds, Springer, New York, 2006. [9] Meisters, G. H., Olech, C., Locally one-to-one mappings and a classical theorem on schlicht functions, Duke Math. J. 30 1963 63-80 [10] Morgan, F. , Geometric measure theory: a beginner’s guide, Academic Press, 1988 [11] Munkres, James R., Topology, second edition, Prentice Hall, 2000

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