Now that Brouwer’s topological degree has been completely caracterized in our two previous notes, we may give the first applications.

Theorem 1: Let \Omega \subset \mathbb{R}^n be a bounded open subspace and f : \overline{\Omega} \to \overline{\Omega} be a continuous map satisfying f_{| \partial \Omega}= \mathrm{Id}. Then f is onto.

Proof. Let x \in \Omega; in particular, x \notin f(\partial \Omega). According to the fifth point of property 1 in our previous note, \deg(f,\Omega,p) depends only on f_{| \partial \Omega}, hence

\deg(f, \Omega,x)= \deg( \mathrm{Id},\Omega,x)=1.

Thus, there exists y \in \overline{\Omega} such that f(y)=x. \square

Theorem 2: Let f : \mathbb{R}^n \to \mathbb{R}^n be a continuous map satisfying \displaystyle \frac{f(x) \cdot x}{\|x\|} \to + \infty as \|x\| \to + \infty. Then f is onto.

Proof. Let p \in \mathbb{R} and r>0 be large enough so that p \in B(0,r) and \displaystyle \frac{f(x) \cdot x}{\|x\|} > \|p\| when \|x\| \geq \|p\|.

Let H : (t,x) \mapsto (1-t) f(x)+ tx be a homotopy between f and \mathrm{Id}. Suppose by contradiction that there exists x \in \partial B(0,r) such that p=H(t,x). Then

\|p \| \cdot \|x\| \geq p \cdot x = (1-t) f(x) \cdot x + t \|x\|^2,

hence

\displaystyle \|p\| \geq (1-t) \frac{f(x) \cdot x}{\|x\|} + t\|x\| > (1-t) \|p\| + t\|p\| = \|p\|

a contradiction. Therefore, by homotopy invariance

\deg(f,B(0,r),p)= \deg( \mathrm{Id}, B(0,r),p)=1,

So there exists x \in B(0,r) such that f(x)=p. \square

Theorem 3: (Brouwer’s fixed point theorem) Let C \subset \mathbb{R}^n be a nonempty compact convex subset and f : C \to C be a continuous map. Then f has a fixed point.

Lemma: Let f : \overline{B}(0,r) \to \overline{B}(0,r) be a continuous map satisfying \|f(x\| \leq r for all x \in \partial B(0,r). Then f has a fixed point.

Proof. Let H : (t,x) \mapsto t(x-f(x)) + (1-t)x be a homotopy between \mathrm{Id}-f and \mathrm{Id} and let x \in \partial B(0,r) and t \in [0,1] be such that H(t,x)=0. Then

\displaystyle r \geq \|f(x)\| = \frac{\|x\|}{t} = \frac{r}{t},

so t=1 and f(x)=x. Therefore, either f has a fixed point on \partial B(0,r) or by homotopy invariance

\deg( \mathrm{Id}-f, B(0,r), 0) = \deg (\mathrm{Id}, B(0,r),0)=1

and there exists x \in B(0,r) such that f(x)=0. \square

Proof of theorem 3. Let r>0 be large enough so that C \subset B(0,r) and let \nu : \overline{B}(0,r) \to C be a retraction.

(Considering C as a convex subset of \mathrm{span}(C), we may suppose that C has a nonempty interior. Let p \in \mathrm{int}(C). Then, for any x \in \overline{B}(0,r) \backslash \{p\}, the half-line starting at p and passing through x meet \partial C at only one point h_x. Finally, \nu : x \mapsto \left\{ \begin{array}{cl} x & \text{if} \ x \in C \\ h_x & \text{otherwise} \end{array} \right. is an example of retraction.)

Then apply the previous lemma to get a fixed point y \in \overline{B}(0,r) of f \circ \nu. Notice that y \in f \circ \nu (\overline{B}(0,r)) \subset C so \nu(y)=y and y is in fact a fixed point of f. \square

Theorem 4: (Fundamental theorem of algebra) Every nonconstant complex polynomial has a root.

Proof. Let P \in \mathbb{C}[X] be a complex polynomial of degree n \geq 1. We prove more generally that \deg(P,B(0,r),0)=n when r is large enough.

Let f : z \mapsto az^n (with a \neq 0) and  Q \in \mathbb{C}[X] of degree smaller than n such that P=f+Q. Let r>0 be large enough so that |f(z)| > |Q(z)| for all z \in \partial B(0,r).

By contradiction, suppose that the homotopy H : (t,x) \mapsto f(z) +t Q(z) between P and f has a zero (t,x) \in [0,1] \times \partial B(0,r). Then

|Q(z)| < |f(z)| = t |Q(z)| \leq |Q(z)|,

a contradiction. Therefore, by homotopy invariance, \deg(P,B(0,r),0)= \deg(f,B(0,r),0).

Let \epsilon \in \mathbb{C}^* be small enough so that \deg(f,B(0,r),0)= \deg (f,B(0,r) , \epsilon). Notice that

J_f(x)= \left| \begin{matrix} \mathrm{Re}(f'(x)) & - \mathrm{Im}(f'(x)) \\ \mathrm{Im}(f'(x)) & \mathrm{Re}(f'(x)) \end{matrix} \right| = |f'(x)|^2= n^2 |x|^{2(n-1)}.

Moreover, any nonzero complex number has n distincts nth roots, hence

\deg(P,B(0,r),0)= \deg(f,B(0,r),0)= \deg(f, B(0,r), \epsilon)= n. \square

Theorem 5: (Rouché) Let \Omega \subset \mathbb{C} be a bounded open subspace and f,g be two holomorphic functions on a neighborhood of \overline{\Omega}. Suppose that f has no zero on \partial \Omega and |f(z)-g(z)| < |f(z)| for all z \in \partial \Omega. Then f and g have the same number of zeros in \Omega (with multiplicity).

Lemma: Let \Omega \subset \mathbb{C} be a bounded open subspace and f be a holomorphic function on a neighborhood of \overline{\Omega} such that f(z) \neq 0 for all z \in \partial \Omega. Then \deg(f, \Omega, 0) coincides with the number of zeros of f (with multiplicity).

Proof. If f has no zero in \Omega, then \deg(f,\Omega, 0). From now on, suppose that f has a zero in \Omega. Let \{z_1,\dots, z_k \} (with k \geq 1) be the zeros of f in \Omega (there are only finitely many zeros because they are isolated and that \overline{\Omega} is compact).

Let \epsilon>0 such that B(z_i,\epsilon) \subset \Omega and B(z_i, \epsilon) \cap B(z_j , \epsilon) = \emptyset when i \neq j. Clearly, 0 \notin f \left( \overline{\Omega} \backslash \bigcup\limits_{i=1}^k B(z_i,\epsilon) \right), hence by additivity

\displaystyle \deg(f, \Omega,0) = \sum\limits_{i=1}^k \deg(f,B(z_i,\epsilon),0).

Therefore, it is sufficient to show that if 0 is a zero of f with multiplicity n, then \deg(f,B(0,\epsilon),0)=n for \epsilon small enough.

We can write f(z)= g(z)(1+ \omega(z)) where g : z \mapsto a z^n (with a \neq 0) and \omega(z) \underset{z \to 0 }{\longrightarrow} 0. Let \epsilon>0 be small enough so that |\omega(z)|<1 for all z \in \partial B(0,\epsilon).

Suppose by contradiction that the homotopy H : (t,z) \mapsto g(z)(1+t\omega(z)) between f and g has a zero (t,z) \in [0,1] \times \partial B(0,\epsilon). Then

\displaystyle 1> |\omega(z)| = \frac{1}{t} \geq 1,

a contradiction. Therefore, by homotopy invariance, \deg(f, B(0,\epsilon),0)= \deg(g,B(0,\epsilon),0). But we already proved in the previous proof that \deg(g,B(0,\epsilon),0)=n. \square

Proof of theorem 5. First, notice that g(z) \neq 0 for all z \in \partial \Omega, otherwise |f(z)| = |f(z)-g(z)| < |f(z)|. So it is sufficient to show that \deg(f, \Omega,0)= \deg(g, \Omega,0) according to our previous lemma.

Suppose by contradiction that the homotopy H : (t,z) \mapsto (1-t)f(z)+tg(z) between f and g has a zero (t,z) \in [0,1] \times \partial \Omega. Then

|f(z)- g(z)| < |f(z)| = t |f(z)-g(z) | \leq |f(z)-g(z)|,

a contradiction. Therefore, by homotopy invariance, \deg(f,\Omega,0)= \deg(g, \Omega,0). \square

Corollary: Let \Omega \subset \mathbb{C} be a bounded open subspace and f be a map holomorphic on a neighborhood of \overline{\Omega} without zero on \partial \Omega. Let z_1,\dots,z_k denote the zeros of f and n_1,\dots,n_k their multiplicities. Then for all \epsilon>0 there exists \delta >0 such that for all map g holomorphic on a neighborhood of \overline{\Omega} satisfying \sup\limits_{z \in \overline{\Omega}} |f(z)-g(z)|<\delta, then the zeros of g on \Omega are in \bigcup\limits_{i=1}^k B(z_i,\epsilon) with exactly n_i in B(z_i,\epsilon) with multiplicity.

Proof. Let \epsilon>0 and \{ z_1, \dots, z_k \} be the zeros of f in \Omega. Let \eta>0 such that z_i \in B(z_j,\epsilon) implies B(z_i, \eta ) \subset B(z_j, \epsilon), B(z_i, \eta) \subset \Omega and B(z_i, \eta) \cap B(z_j, \eta) = \emptyset if i \neq j.

Set \delta = \inf \left\{ |f(z)| \mid z \in \overline{\Omega} \backslash \bigcup\limits_{i=1}^k B(z_i, \eta) \right\}>0.

Let g be a holomorphic function on a neighborhood of \overline{\Omega} satisfying \sup\limits_{z \in \overline{\Omega}} |f(z)-g(z)| < \delta. In particular, for all z \in \partial B(z_i, \eta),

|f(z)-g(z)| \leq \sup\limits_{z \in \overline{\Omega}} |f(z)-g(z)| < \delta \leq |f(z)|.

According to theorem 5, we deduce that f and g have the same number of zeros in each B(z_i,\eta). \square

Theorem 6: (Hedgehog) Let n \geq 1 be an odd integer, \Omega \subset \mathbb{R}^n be an open subspace containing 0 and f : \partial \Omega \to \mathbb{R}^n be a continuous map. Then there exist \lambda \in \mathbb{R} and x \in \partial \Omega such that f(x)=\lambda x.

Proof. If 0 \in f(\partial \Omega), we can chose \lambda=0 and x a zero of f. From now on, suppose that 0 \notin f(\partial \Omega). Let \tilde{f} : \overline{\Omega} \to \mathbb{R}^n be a continuous extension of f; in particular, 0 \notin \tilde{f} (\partial \Omega).

Suppose that \deg( \tilde{f}, \Omega,0) \neq -1= \deg( - \mathrm{Id}, \Omega,0). By homotopy invariance, the homotopy H : (t,x) \mapsto (1-t) \tilde{f}(x) -tx between \tilde{f} and - \mathrm{Id} has a zero (t,x) \in [0,1] \times \partial \Omega. Moreover, because 0 \in \Omega, then 0 \notin \partial \Omega and t \neq 1, hence

\displaystyle f(x)= \tilde{f}(x)= \frac{t}{1-t} x.

Suppose that \deg(\tilde{f}, \Omega,0) \neq 1 = \deg( \mathrm{Id},\Omega,0). In the same way, we find (t,x) \in [0,1] \times \partial \Omega such that (1-t) \tilde{f}(x)+tx=0, ie.

\displaystyle f(x)= \tilde{f}(x)= \frac{t}{t-1}x. \square

Theorem 7: (Hairy ball theorem) In even dimension, any tangent vector field of a sphere has a zero.

Proof. Let f : \mathbb{S}^n \to \mathbb{R}^{n+1} be such a vector field; in particular, f(x) \cdot x =0 for all x \in \mathbb{S}^n \subset \mathbb{R}^{n+1}. According to Hedgehog theorem, there exsit \lambda \in \mathbb{R} and x \in \mathbb{S}^n such that f(x)=\lambda x. However

\lambda= \lambda \|x\|^2= \lambda x \cdot x = f(x) \cdot x =0.

Therefore, f(x)=0. \square

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