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 $n$th 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, $f(x)=0$. $\square$