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# If a mapping f : K → Cn is continuous, the restriction f| Int K is holomorphic and f|∂K is injective, then f is injective

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ANNALES

POLONICI MATHEMATICI 55 (1991)

Some criteria for the injectivity of holomorphic mappings by Stanis law Spodzieja ( L´od´z)

Abstract. We prove some criteria for the injectivity of holomorphic mappings.

Let K ⊂ Cn be a bounded and closed domain such that (1) no closed proper subset of ∂K disconnects Cn.

Theorem 1. If a mapping f : K → Cn is continuous, the restriction f| Int K is holomorphic and f|∂K is injective, then f is injective.

P r o o f. The proof will be carried out in three steps:

1. f| Int K : Int K → Cn is an open mapping. By the assumption, for any y ∈ Cn, f−1(y) ∩ ∂K has at most one point. Consequently, from the Remmert–Stein theorem on removable singularities, f has isolated ﬁbres.

So, by Remmert’s theorem on open mappings, f| Int K is an open mapping.

2. f (∂K) ∩ f (Int K) = ∅. It is known (see [1], Cor. in Sec. 12, p. 248) that if A, B ⊂ Rm are compact and homeomorphic, and A disconnects Rm, then so does B. Hence and from (1), f (∂K) disconnects Cn, but no closed proper subset of f (∂K) does. Since f (Int K) is open, f (Int K) ⊂ Int f (K). Consequently, ∂f (K) ⊂ f (∂K). Since ∂f (K) disconnects Cn, we get ∂f (K) = f (∂K), and so f (∂K) ∩ f (Int K) = ∅.

3. f is injective. Let V = {(x, y) ∈ K × K : f (x) = f (y)}. Then each irreducible component of V ∩ Int(K × K) has a positive dimension. Deﬁne

gi: V ∋ (x1, . . . , xn, y1, . . . , yn) 7→ xi− yi∈ C, i = 1, . . . , n.

By the maximum principle for holomorphic functions on analytic sets, there exist (xi0, y0i) ∈ ∂(K × K) ∩ V , i = 1, . . . , n, such that

|gi(xi0, y0i)| = max

(x,y)∈V|gi(x, y)|, i = 1, . . . , n.

1991 Mathematics Subject Classification: Primary 32H99.

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322 S. Spodzieja

From the deﬁnition of V we have f (xi0) = f (yi0), thus, by step 2, xi0, y0i

∂K. Hence and from the injectivity of f|∂K we have xi0 = yi0, that is, gi(xi0, yi0) = 0, and thus gi≡ 0 for i = 1, . . . , n. Hence V = {(x, x) : x ∈ K}, therefore f is injective.

The proof is complete.

R e m a r k. In the case n = 1, this theorem is known (see [2], §11, Ch. IV, p. 209).

We shall now give another criterion in which we weaken the assumption on the boundary of the domain at the cost of strengthening the assumption on the mapping.

Let D ⊂ Cn be a bounded and closed domain with connected boundary.

Theorem 2. If f : D → Cn is a continuous mapping, f| Int D is holo- morphic, f|∂D is injective, and

(2) each x ∈ ∂D has a neighbourhood U ⊂ C such that f|U ∩Dis injective, then f is injective.

P r o o f. The proof will be carried out in three steps:

1. f| Int D : Int D → Cn is an open mapping. This is proved in the same way as step 1 in the proof of Theorem 1.

2. f (∂D) ∩ f (Int D) = ∅. Assume to the contrary that f (∂D) ∩ f (Int D) 6= ∅. By step 1, f (∂D) ∩ f (Int D) is open in f (∂D). Take any sequence yn ∈ f (∂D) ∩ f (Int D) such that lim yn = y0. Then there ex- ist sequences zn ∈ ∂D, xn ∈ Int D such that f (zn) = yn, f (xn) = yn. Passing to subsequences if necessary, we may assume that lim xn = x0, lim zn = z0. From (2) we have z0 6= x0. So, from the injectivity of f|∂D

we get z0 ∈ ∂D, x0∈ Int D. In consequence, y0∈ f (∂D) ∩ f (Int D). Thus f (∂D) ∩ f (Int D) is closed in f (∂D), that is, by the connectedness of ∂D, f (∂D) = f (∂D) ∩ f (Int D). To sum up, f (D) = f (Int D), which is impos- sible because f (D) is compact and f (Int D) open.

3. f is injective. This is proved in the same way as step 3 in the proof of Theorem 1.

The proof is complete.

Corollary. If f : D → Cn is a holomorphic mapping such that f|∂D

is injective and the Jacobian of f does not vanish anywhere in D, then f is injective.

We shall now give an example illustrating the fact that the assumptions (1) in Theorem 1 and (2) in Theorem 2 cannot be omitted.

Example 1. Let B = exp({z ∈ C : |z| < π}). Take a homography h such that h(πi) = 2i, h(−πi) = −2i, h(−π) = −1, and a function f : B → C

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Injectivity of holomorphic mappings 323

deﬁned by

f (z) = [h(Log z)]2, z ∈ B \ {−1},

−4, z = −1.

B πi

2i z2

h Log

−1 −4

π −1

−2i

πi

Then f and B have the following properties:

1) f is injective on ∂B, 2) ∂B does not satisfy (1),

3) f does not satisfy (2) at the point −1, 4) f is not injective in B.

It is easy to show, using the Osgood–Brown theorem, that we need not assume the connectedness of the boundary of the domain in Theorem 2 for n ≥ 2. In the case n = 1, this assumption is essential, which is shown by the following example.

Example 2. Let D = {z ∈ C : 1/5 ≤ |z| ≤ 4} and f : D → C, f (z) = z + 1/z. It is easy to see that f|∂D is injective. Since f(z) = 0 for z = 1 and z = −1, condition (2) in Theorem 2 is also satisﬁed. But f (3) = f (1/3), thus f is not injective.

References

[1] K. B o r s u k, ¨Uber Schnitte der n-dimensionalen Euklidischen R¨aume, Math. Ann.

106 (1932), 239–248.

[2] S. S a k s and A. Z y g m u n d, Analytic Functions, PWN, Warszawa 1965.

INSTITUTE OF MATHEMATICS L ´OD´Z UNIVERSITY

BANACHA 22

90-238 L ´OD´Z, POLAND

Re¸cu par la R´edaction le 12.9.1990

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