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 $i$th 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 $\deg(w,C_{\delta},0)=0$.

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

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

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.