We conclude our series of notes about Brouwer’s topological degree by a final application, namely Borsuk theorem and its corollaries.

Theorem: (Borsuk) Let $\Omega \subset \mathbb{R}^n$ be a bounded open subspace symmetric with respect to the origin and $f : \overline{\Omega} \to \mathbb{R}^n$ be a continuous function satisfying $0 \notin f(\partial \Omega)$. If $f$ is odd then $\deg(f, \Omega,0)$ is odd.

Proof. Let $(g_n)$ be a sequence of $C^1$ functions converging uniformely to $f$. If $h_n : x \mapsto \frac{1}{2}(g_n(x)-g_n(-x))$ for each $n \geq 1$, then $(h_n)_n$ is a sequence of odd $C^1$ functions converging uniformely to $\frac{1}{2}( f( \cdot)- f(- \cdot))=f$. Because $\deg(h_n, \Omega,0)= \deg(f,\Omega,0)$ when $n$ is large enough, we may suppose without loss of generality that $f$ is $C^1$.

Suppose first that $0$ is a regular value of $f$. Because $f(0)=0$,

$\displaystyle \deg(f, \Omega,0)= \mathrm{sign}(J_f(0)) + \sum\limits_{x \in f^{-1}(0),x \neq 0} \mathrm{sign}(J_f(x))$.

Since $f$ is odd, the group $\mathbb{Z}_2$ acts on $f^{-1}(0) \backslash \{0\}$ by $x \mapsto -x$ and if $K$ is a fundamental domain for the action,

$\begin{array}{ll} \deg(f, \Omega,0) & \displaystyle = \mathrm{sign} (J_f(0)) + \sum\limits_{x \in K} \mathrm{sign} (J_f(x)) + \sum\limits_{x \in K} \mathrm{sign} (J_f(-x)) \\ \\ & \displaystyle = \mathrm{sign} (J_f(0))+ 2 \sum\limits_{x \in K} \mathrm{sign} (J_f(x)) \end{array}$

because $f$ odd implies $df$ even. Therefore, the theorem is proved in our case.

From now on, suppose that $0$ is not a regular value of $f$. Thanks to the previous case, it is enough to show that $f$ is arbitrarily closed to an odd $C^1$ function whose $0$ is a regular value. For this purpose, we use the following lemma:

Lemma: Let $f : \Omega \to \mathbb{R}^n$ be an odd $C^1$ function. For every $\epsilon>0$, there exists $y \in \mathbb{R}^n$ such that $\|y\|< \epsilon$ and such that $g : x \mapsto f(x)-y x_i^3$ is an odd $C^1$ function satisfying:

• $0$ is a regular value of $g$ on $\{x \in \Omega \mid x_i \neq 0\}$,
• $dg(0)=df(0)$,
• If $x_i=0$ and $g(x)=0$ then $f(x)=0$ and $df(x)=dg(x)$,
• $\|f-g\|_{\overline{\Omega}} := \sup\limits_{x \in \overline{\Omega}} \|f(x)-g(x)\| \leq \epsilon R^3$ where $\Omega \subset B(0,R)$.

First, because there exists an arbitrarily small $\delta>0$ not in the spectrum of $f$, we may suppose without loss of generality that $df(0)$ is invertible (otherwise replace $f$ with $f- \delta \mathrm{Id}$).

Let $f_1$ be the function given by our lemma with $f$ and $i=1$ and $f_2$ be the function given by our lemma with $f_1$ and $i=2$.

By construction, $0$ is a regular value of $f_2$ on $\{x \in \Omega \mid x_2 \neq 0\}$. Moreover, if $f_2(x)=0$, $x_2=0$ and $x_1 \neq 0$, then $f_1(x)=0$ and $df_1(x)=df_2(x)$; in particular, $df_2(x)$ is invertible. Therefore, $0$ is a regular value of $f_2$ on $\{ x \in \Omega \mid x_1 \neq 0 \ \text{or} \ x_2 \neq 0\}$.

By induction, let $f_3, \dots, f_n$ be defined in the same way. Then $f_n : \Omega \to \mathbb{R}^n$ is an odd $C^1$ function whose $0$ is a regular value on

$\{x \in \Omega \mid x_1 \neq 0 \ \text{or} \ x_2 \neq 0 \ \text{or} \ \dots \ \text{or} \ x_n \neq 0\} = \Omega \backslash \{0\}$.

Moreover,

$df_n(0)=df_{n-1}(0)= \dots = df_1(0)=df(0)$

is invertible, so $0$ is a regular value of $f_n$. Finally, we conclude the proof by noticing that:

$\|f_n-f\|_{\overline{\Omega}} \leq \|f_n-f_{n-1}\|_{\overline{\Omega}}+ \dots + \|f_2-f_1\|_{\overline{\Omega}}+ \|f_1-f\|_{\overline{\Omega}} \leq n \epsilon R^3$. $\square$

Proof of the lemma. Let $x \in \Omega$ be a point such that $g(x)=0$ and $x_i \neq 0$. Then $\displaystyle y = \frac{f(x)}{x_i^3}$ and $dg(x)=df(x)-3y (dx_i)^2= x_i^3 dF(x)$ where $\displaystyle F : x \mapsto \frac{f(x)}{x_i^3}$.

Because $F(x)=y$, we deduce that $dg(x)$ is invertible if $y$ is a regular value of $F$. According to Sard’s theorem, there exists such value with an arbitrarily small norm, concluding the proof of the first point. The other points are straightforward. $\square$

Theorem 1: Let $n >p \geq 1$ be two integers and $f : \mathbb{S}^{n-1} \to \mathbb{R}^p$ be a continuous map. Then there exists $x \in \mathbb{S}^{n-1}$ such that $f(x)=f(-x)$.

Proof. Let $g : \overline{B} \to \mathbb{R}^p \times \{0\} \subset \mathbb{R}^n$ be an odd continuous extension of $f( \cdot)-f(- \cdot)$ on the unit closed ball $\overline{B} \subset \mathbb{R}^n$. If $0 \notin g(\partial B)$ we deduce that $\deg(gB,0) \neq 0$ from Borsuk theorem.

On the other hand, there exists a point $x \in \mathbb{R}^n \backslash \mathbb{R}^p \times \{0\}$ with an arbitrarily small norm, so $\deg(g,B,x)=0$ because $g(y) \neq x$ for all $y \in B$ and $\deg(g,B,x)= \deg(g,B,0)$. A contradiction.

Therefore, $0 \in g(\partial B)$ ie. there exists $x \in \partial B= \mathbb{S}^{n-1}$ such that $f(x)-f(-x)=g(x)=0$. $\square$

Theorem 2: (open mapping theorem) Let $U \subset \mathbb{R}^n$ be open subspace and $f : U \to \mathbb{R}^n$ be an injective continuous map. Then $f$ is open.

Proof. It is sufficient to prove that for all $a \in U$ and $r>0$ such that $\overline{B}(a,r) \subset U$, $f(B(a,r))$ contains an open neighborhood of $f(a)$. For this purpose, we show that $\deg(f,B(a,r),f(a)) \neq 0$. Indeed, in this case there exists an open neighborhood $V$ of $f(a)$ such that $\deg(f,B(a,r),v) \neq 0$ for all $v \in V$, so there exists $W \subset B(a,r)$ such that $f(W)=V$ hence $V \subset f(B(a,r))$.

Because $\deg(f,B(a,r),f(a))= \deg(f( \cdot +a)-f(a),B(0,r),0)$, we may suppose without loss of generality that $a=f(a)=0$.

Let $\displaystyle H : (t,x) \mapsto f \left( \frac{x}{1+t} \right)- f \left( - \frac{tx}{1+t} \right)$ be a homotopy. Because $f$ is injective, if $H(t,x)=0$ then $x=0$ hence $0 \notin H(t,\partial B(0,r))$ for all $t \in [0,1]$. So by homotopy invariance,

$\deg(f,B(0,r),0)= \deg(H(0, \cdot),B(0,r),0)= \deg(H(1, \cdot),B(0,r),0)$,

where $\displaystyle H(1, \cdot) : x \mapsto f \left( \frac{x}{2} \right)- f \left( - \frac{x}{2} \right)$ is odd. Now, according to Borsuk theorem, $\deg(H(1,\cdot),B(0,r),0) \neq 0$, which concludes the proof. $\square$

Theorem 3: (Invariance of the domain) Let $U \subset \mathbb{R}^n$ be an open subspace and $n>m \geq 1$ be two integers. There does not exist a continuous injection $U \hookrightarrow \mathbb{R}^m$.

Proof. Suppose by contradiction that there exists such an injection

$f : U \to \mathbb{R}^m \simeq \mathbb{R}^m \times \{0\} \subset \mathbb{R}^n$.

According to theorem 2, $f(U) \subset \mathbb{R}^m \times \{0\}$ is an open subspace of $\mathbb{R}^n$. Because the projection $p : \mathbb{R}^m \times \mathbb{R}^{n-m} \to \mathbb{R}^{n-m}$ is open, we deduce that $p(f(U))= \{0\}$ is open in $\mathbb{R}^{n-m}$ (with $n-m>0$): a contradiction. $\square$