The aim of this note is to give an axiomatic definition of the Brouwer’s topological degree constructed in our previous post.

Definition: Let \mathcal{A} denote the set of tuples (f,\Omega,p) where \Omega \subset \mathbb{R}^n is a bounded open subspace, f : \overline{\Omega} \to \mathbb{R}^n be a continuous map and p \notin f(\partial \Omega). A topological degree is a map \deg : \mathcal{A} \to \mathbb{R} satisfying:

  • (normality) \deg(\mathrm{Id},\Omega,p)=1 if p \in \Omega,
  • (additivity) Let \Omega_1,\Omega_2 \subset \Omega be two disjoint open subspaces such that p \notin f(\overline{\Omega} \backslash (\Omega_1 \cup \Omega_2)). Then

\deg(f,\Omega,p)= \deg(f, \Omega_1,p)+ \deg(f,\Omega_2,p).

  • (homotopy invariance) Let H : [0,1] \times \overline{\Omega} \to \mathbb{R}^n be a homotopy and y : [0,1] \to \mathbb{R}^n be a continuous map satisfying y(t) \notin H(t,\partial \Omega) for all t \in [0,1]. Then

\deg(H(0,\cdot),\Omega,y(0))= \deg(H(1,\cdot),\Omega,y(1)).

First, let us consider some consequences of the definition:

Property 1: Let \deg : \mathcal{A} \to \mathbb{R} be a topological degree. Then

  • If \deg(f,\Omega,p) \neq 0 then there exists y \in \Omega such that f(y)=p
  • For all z \in \mathbb{R}^n, \deg(f,\Omega,y)= \deg(f-z,\Omega,y-z),
  • For all z \in \mathbb{R}^n, \deg(f,\Omega,p)= \deg( f( \cdot -z),z+ \Omega,p),
  • \deg(f,\Omega, \cdot) is constant on any connected component of \mathbb{R}^n \backslash f(\partial \Omega),
  • Let g : \overline{\Omega} \to \mathbb{R}^n be a continuous map and q \in \mathbb{R}^n satisfying

\sup\limits_{x \in \partial \Omega} \| f(x)-g(x) \| + \|p-q\| < d(p,f(\partial \Omega)).

Then \deg(f,\Omega,p)=\deg(g,\Omega,q).

Proof. First point: Suppose by contradiction that p \notin f^{-1}(\Omega). Then p \notin f^{-1} (\overline{\Omega} \backslash \emptyset \cup \emptyset) hence

\deg(f,\Omega,p)=2 \deg(f, \emptyset,p)

by additivity. But p \notin f^{-1}( \emptyset \backslash \emptyset \cup \emptyset) hence

\deg(f, \emptyset,p)= 2 \deg (f, \emptyset,p)

by additivity, ie. \deg(f,\emptyset,p)=0. We deduce that \deg(f,\Omega,p)=0.

Second point: Let H : (t,x) \mapsto f(x)-tz and y : t \mapsto p-tz. Notice that y(t) \in H(t, \partial \Omega) for some t \in [0,1] implies p \in f(\partial \Omega). Therefore, by homotopy invariance:

\deg(H(0, \cdot ), \Omega,y(0)) = \deg(H(1, \cdot), \Omega,y(1)),

that is

\deg(f,\Omega,p)= \deg(f-z,\Omega, p-z).

Fourth point: The fifth point shows that \deg(f, \Omega, \cdot) is locally constant. So it is sufficient to conclude by connectedness.

Fifth point: Let H : (t,x) \mapsto (1-t) f(x)-tg(z) and y : t \mapsto (1-t)p + tq. Suppose by contradiction that there exist t \in [0,1] and x \in \partial \Omega such that y(t) \in H(t,x), that is

p-f(x)= t(p-q)+ t(g(x)-f(x)).

Then

d(p, \partial \Omega) \leq \|p-f(x)\| \leq \|p-q\| + \sup\limits_{x \in \partial \Omega} \|g(x)-f(x)\|,

a contradiction. Therefore, by homotopy invariance

\deg(f, \Omega,p)= \deg( H(0, \cdot), \Omega,y(0)) = \deg (H(1, \cdot), \Omega,y(1)) = \deg(g, \Omega,q).

Third point: Let T_{f,\Omega,y} : z \mapsto \deg( f( \cdot-z),z+ \Omega,y); the map is well-defined since y \notin f(\partial \Omega) = f(\partial (z+ \Omega)-z).

We shall prove that T_{f,\Omega,y} is locally constant at 0. Then, we will be able to deduce that T_{f,\Omega,y} is locally constant at any point from the equality

T_{f, \Omega,y}(z+z_0)= T_{f( \cdot-z_0),z_0+ \Omega,y}(z), \ \forall z

and so constant by connectedness of \mathbb{R}^n, hence

\deg(f( \cdot -z),z+ \Omega,y)= T_{f, \Omega,y}(z)= T_{f,\Omega,y}(0)= \deg(f, \Omega,y).

First, let \Omega_s= \{x \in \Omega \mid d(x, \partial \Omega)>s \} and suppose by contradiction that y \in f( \overline{\Omega} \backslash \Omega_s) for every s>0.

By compactness of \overline{\Omega}, there exists a decreasing sequence (r_n) converging to 0 so that (x_{r_n}) converge to some x \in \overline{\Omega}. So

d(x, \partial \Omega) = \lim\limits_{n \to + \infty} d(x_{r_n},\partial \Omega) \leq \lim\limits_{ n \to + \infty} r_n =0,

that is x \in \partial \Omega. Thus y= \lim\limits_{n \to + \infty} f(x_{r_n})=f(x) \in \partial \Omega, a contradiction. So let s>0 so that y \notin f(\overline{\Omega} \backslash \Omega_{2s}).

In particular, y \notin f( \overline{\Omega} \backslash \Omega_s \cup \emptyset) hence by additivity

\deg(f, \Omega,y)= \deg(f, \Omega_s,y)+ \underset{=0}{\underbrace{ \deg(f, \emptyset,y) }}= \deg( f, \Omega_s,y) \hspace{1cm} (1).

Afterwards, if z \in B(0,s) then \Omega_s \subset z+ \Omega, hence

\overline{z+ \Omega} \backslash \Omega_s -z= \overline{\Omega} \backslash (\Omega_s-z) \subset \overline{\Omega} \backslash \Omega_{2s}.

 Therefore, y \notin f( \overline{z+ \Omega} \backslash \Omega_s -z) because y \notin f( \overline{\Omega} \backslash \Omega_{2s}), hence by additivity

\deg(f( \cdot -z),z+ \Omega,y) = \deg(f( \cdot -z), \Omega_s,y), \ \forall z \in B(0,s) \hspace{1cm} (2)

For z \in B(0,s) fixed, let h : \left\{ \begin{matrix} [0,1] \times \Omega_s & \to & \mathbb{R}^n \\ (t ,x) & \mapsto & f(x-tz) \end{matrix} \right. be a homotopy between f and f( \cdot -z). Notice that for all t \in [0,1] and x \in \partial \Omega_s,

d(x-tz, \partial \Omega) \leq d(x, \partial \Omega)+ |tz| \leq 2s,

so x-tz \in \overline{\Omega} \backslash \Omega_{2s}; consequently, y \neq h(t,x) since y \notin f( \overline{\Omega} \backslash \Omega_{2s}), and by homotopy invariance

\deg(f, \Omega_s,y)= \deg(f( \cdot -z), \Omega_s,y) \hspace{1cm} (3)

We deduce from (1), (2) and (3) that for all z \in B(0,s)

\begin{array}{ll} \deg(f, \Omega,y) & = \deg( f, \Omega_s,y) = \deg( f( \cdot -z), \Omega_s,y) \\ & = \deg( f( \cdot -z),z+ \Omega,y)= T_{f, \Omega,y}(z). \hspace{1cm} \square \end{array}

Now we are able to prove our main theorem:

Theorem 2: There exists only one topological degree \deg : \mathcal{A} \to \mathbb{R}.

Proof. The existence was proved in our previous post, so it is sufficient to show the uniqueness. In order to make the proof clear, let us proceed step by step.

  • Step 1: Let (f , \Omega,p) \in \mathcal{A}.

According to the last point of our property and to Stone-Weierstrass approximation theorem, we may suppose that f is in fact C^1 on \Omega. Then, according to the last point of our property and to Sard’s lemma, we may suppose that p is a regular value of f.

  • Step 2: Let (f, \Omega,p) \in \mathcal{A} where f is C^1 on \Omega and p is a regular value of f.

According to the inverse function theorem, f is a local diffeomorphism at any point of f^{-1}(p). Therefore, f^{-1}(p) is a closed discrete subset of the compact \overline{\Omega}, so f^{-1}(p) is finite.

Let f^{-1}(p)= \{ x_1, \dots, x_m\} and \delta >0 be such that B(x_i, \delta) \subset \Omega and B(x_i, \delta) \cap f^{-1}(p) = \{x_i\}. By additivity and our previous property:

\begin{array}{ll} \deg(f, \Omega,p) & \displaystyle = \sum\limits_{i=1}^m \deg(f, B(x_i,\delta),p) = \sum\limits_{i=1}^m \deg(f-p, B(x_i, \delta),0) \\ \\ & \displaystyle = \sum\limits_{i=1}^m \deg (f( \cdot +x_i)-p,B(0,\delta),0) \end{array}

Notice that the previous equality still holds if we replace \delta with any element of (0,\delta).

Let i be fixed. Suppose by contradiction that for any \eta < \delta, the homotopy H : (t,x) \mapsto t(f(x+x_i)-p) + (1-t) f'(x_i) \cdot x between f(\cdot +x_i)-p) and f'(x_i) has a zero (t_{\eta},x_{\eta}) \in [0,1] \times \partial B(0,\eta). Because

\begin{array}{ll} \|f'(x_i) \cdot x_{\eta} \| & = t_{\eta} \| f(x_{\eta}+x_i)-p - f'(x_i) \cdot x_{\eta} \| \\ & \leq \|f(x_{\eta}+x_i)-f(x_i)-f'(x_i) \cdot x_{\eta} \| \end{array},

we deduce that \|f'(x_i) \cdot x_{\eta} \| = o(\eta). On the other hand, f'(x_i) is invertible so there exists C>0 such that \|f'(x_i) \cdot y \| \geq C for all y \in \mathbb{R}^n satisfying \|y\|=1. Therefore, \|f'(x_i) \cdot x_{\eta} \| \geq C \|x_{\eta}\| = C \eta, hence a contradiction since C does not depend on \eta.

By homotopy invariance, we get

\deg(f( \cdot +x_i)-p, B(0,\eta),0) = \deg ( f'(x_i), B(0,\eta),0),

for some \eta \in (0, \delta). Therefore we may suppose that f is a linear isomorphism, \Omega has the form B(0,\delta) and p=0.

  • Step 3: Let (A, B(0,\delta),0) \in \mathcal{A} where A is a linear isomorphism.

By polar decomposition, there exist O \in O_n(\mathbb{R}) and S \in S_n^{++}(\mathbb{R}) such that A=OS.

In particular, there exist \lambda_1,\dots, \lambda_n>0 and P \in O_n(\mathbb{R}) such that

S= P \left( \begin{matrix} \lambda_1 & & 0 \\ & \ddots & \\ 0 & & \lambda_n \end{matrix} \right)P^{-1},

and there exist \theta_1,\dots, \theta_m \in [0,2\pi) and Q \in O_n(\mathbb{R}) such that

O= Q \left( \begin{matrix} \mathrm{sign}(\det(0)) & & & & & & 0 \\ & 1 & & & & & \\ & & \ddots & & & & \\ & & & 1 & & & \\ & & & & R(\theta_1) & & \\ & & & & & \ddots & \\ 0 & & & & & & R(\theta_n) \end{matrix} \right) Q^{-1}.

For all t \in [0,1], let

S(t)= P \left( \begin{matrix} (1-t) \lambda_1+t & & 0 \\ & \ddots & \\ 0 & & (1-t) \lambda_n +t \end{matrix} \right)P^{-1}

and

O(t)= Q \left( \begin{matrix} \mathrm{sign}(\det(A)) & & & & & & 0 \\ & 1 & & & & & \\ & & \ddots & & & & \\ & & & 1 & & & \\ & & & & R((1-t)\theta_1) & & \\ & & & & & \ddots & \\ 0 & & & & & & R((1-t)\theta_n) \end{matrix} \right) Q^{-1}.

Using t \mapsto O(t)S(t), we can define a homotopy of isomorphisms between A and

J_A= Q\left( \begin{matrix} \mathrm{sign}(\det(A)) & & & 0 \\ & 1 & & \\ & & \ddots & \\ 0 & & & 1 \end{matrix} \right) Q^{-1},

hence

\deg(A,B(0,\delta),0)= \deg(J_A,B(0,\delta),0).

If \det(A)>0 then J_A= \mathrm{I}_n and \deg(J_A,B(0,\delta),0)=1. From now on, suppose \det(A)<0.

Without loss of generality, by additivity we may replace B(0,\delta) with (-\delta,\delta)^n in the coordinates associated to Q. Let

C_{\delta} = (-3\delta, \delta) \times ( - \delta, \delta)^{n-1}, \ w(x)=(- \delta,x_2, \dots,x_n),

and

g(x)= ( \mathrm{1}_{[-3\delta,- \delta)}(x_1)(x_1+2\delta)- \mathrm{1}_{[-\delta, \delta]} (x_1)x_1,x_2, \dots ,x_n).

Let H : (t,x) \mapsto (1-t) g(x)+t w(x) be a homotopy between g and w. Notice that:

  • If x_1= -3 \delta then the first coordinate of H(t,x) is -(1-t) \delta-t \delta=- \delta \neq 0,
  • If x_1= \delta then the first coordinate of H(t,x) is -(1-t) \delta -t \delta=- \delta \neq 0,
  • If x_i= \pm \delta with 2 \leq i \leq n then the ith coordinate of H(t,x) is \pm(1- t ) \delta \mp t \delta= \pm \delta \neq 0.

Therefore, 0 \notin H(t, \partial C_{\delta}) for all t \in [0,1] hence \deg(g, C_{\delta},0) = \deg(w,C_{\delta},0) by homotopy invariance. Moreover, w(x) \neq 0 for all x \in C_{\delta} so \deg(w,C_{\delta},0)=0.

Then by additivity

\deg(g,C_{\delta},0)= \deg(g, (-3\delta,- \delta) \times (- \delta,\delta)^{n-1},0) + \deg(g,(- \delta,\delta)^n,0),

\deg (g,(-\delta,\delta)^n,0)= \deg(J_A, (-\delta, \delta)^n,0)

since g and J_A coincide on (-\delta, \delta)^n, and

\begin{array}{ll} \deg(g, (-3\delta,\delta) \times (-\delta, \delta)^{n-1},0) & = \deg( \mathrm{Id}+(2 \delta,0, \dots,0), (-3 \delta,- \delta) \times (-\delta,\delta)^{n-1},0) \\ & = \deg( \mathrm{Id}, (-\delta,\delta) \times(- \delta, \delta)^{n-1},0)=1 \end{array}

Therefore, we get

\deg(J_A,(- \delta, \delta)^n,0)=- \deg(g,(-3 \delta,- \delta) \times (- \delta, \delta)^{n-1},0)=-1.

Finally, we deduce that \deg(A, B(0,\delta),0)= \mathrm{sign}(\det(A)). \square

Corollary: A topological degree has integer values.

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