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

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