Following our previous note, we give here another application to Brouwer’s topological degree, namely Jordan curve theorem. More precisely, our main theorem is:

Theorem: Let $\Omega_1,\Omega_2 \subset \mathbb{R}^n$ be two homeomorphic compact subspaces. Then $\mathbb{R}^n \backslash \Omega_1$ and $\mathbb{R}^n \backslash \Omega_2$ has the same number of connected components.

As a corollary, we immediatly get:

Corollary: (Jordan curve theorem) A non-intersecting continuous loop divides the plane into two connected components.

Lemma: (Product formula) Let $\Omega \subset \mathbb{R}^n$ be a bounded open subspace, $f : \overline{\Omega} \to \mathbb{R}^n$ and $g : \mathbb{R}^n \to \mathbb{R}^n$ be two continuous maps. Let $\{U_i \mid i \in I\}$ denote the bounded connected components of $\mathbb{R}^n \backslash f( \partial \Omega)$. If $p \notin g \circ f (\partial \Omega)$ then

$\displaystyle \deg(g \circ f, \Omega,p)= \sum\limits_{i \in i} \deg(f, \Omega,U_i) \deg(g, U_i,p)$.

Proof. First, we show that all but finitely many terms of the previous sum are zero.

Let $r>0$ be such that $f( \overline{\Omega}) \subset B(0,r)$ and let $M= \overline{B}(0,r) \cap g^{-1}(p)$. Because $M$ is compact and

$\displaystyle M \subset \mathbb{R}^n \backslash f( \partial \Omega)= \coprod\limits_{i \in I} U_i \coprod U_{\infty}$,

where $U_{\infty}$ is the unbounded connected component of $\mathbb{R}^n \backslash f( \partial \Omega)$, there exists a finite subset $K \subset I$ such that $M \subset \coprod\limits_{i \in K} U_i \coprod U_{\infty}$.

Notice that if $i \in I$ then $U_i \subset f(\overline{\Omega})$, so if $i \notin K$ then $U_i \cap g^{-1}(p)= \emptyset$ hence $\deg(g,U_i,p)=0$.

Now, we prove the lemma when $f$ and $g$ are $C^1$ and when $p$ is a regular value of $g \circ f$. We have:

$\begin{array}{ll} \deg( g \circ f, \Omega,p) & \displaystyle = \sum\limits_{x \in (g \circ f)^{-1}(p)} \mathrm{sign} (J_{g \circ f}(x)) \\ \\ & \displaystyle = \sum\limits_{x \in f^{-1}(z)} \sum\limits_{z \in g^{-1}(p) \cap f(\Omega)} \mathrm{sign} (J_g( \underset{=z}{\underbrace{ f(x) }} )) \cdot \mathrm{sign}(J_f(x)) \\ \\ & \displaystyle = \sum\limits_{z \in g^{-1}(p) \cap f(\Omega)} \mathrm{sign} (J_g(z)) \left( \sum\limits_{x \in f^{-1}(z)} \mathrm{sign} (J_f(x)) \right) \\ \\ & \displaystyle = \sum\limits_{z \in g^{-1}(p) \cap f(\Omega)} \mathrm{sign} (J_g(z)) \cdot \deg(f, \Omega,z) \\ \\ & \displaystyle = \sum\limits_{i \in I} \sum\limits_{z \in g^{-1}(p) \cap U_i} \mathrm{sign} (J_g(z)) \cdot \deg(f, \Omega,U_i) \\ \\ & \displaystyle = \sum\limits_{i \in I} \deg(f,\Omega,U_i) \left( \sum\limits_{z \in g^{-1}(p) \cap U_i} \mathrm{sign} (J_g(f)) \right) \\ \\ & \displaystyle = \sum\limits_{i \in I} \deg(f,\Omega,U_i) \cdot \deg(g,U_i,p) \end{array}$

From now on, we suppose that $f$ and $g$ are continuous. Let

$V_m= \{ z \in B(0,r) \backslash f( \partial \Omega) \mid \deg(f, \Omega,p)= m\}$

and

$N_m= \{ i \in I \mid \deg(f, \Omega,U_i)=m \}$.

Because $V_m= \coprod\limits_{i \in I} U_i$, we deduce that

$\begin{array}{ll} \sum\limits_{i \in I} \deg(f, \Omega,U_i) \deg(g,U_i,p) & \displaystyle = \sum\limits_{m \geq 0} \sum\limits_{ i \in N_m} m \deg(g,U_i,p) \\ \\ & \displaystyle = \sum\limits_{m \geq 0} m \deg(g,V_m,p) \end{array}$

Let $g_0$ and $f_0$ be two $C^1$ maps and let $p_0$ be a regular value of $g_0 \circ f_0$ such that

$\sup\limits_{ x \in \overline{B}(0,r) } \| g_0(x) - g(x) \|, \ \sup\limits_{x \in \overline{B}(0,r)} \| f_0(x)-f(x) \|, \ \|p-p_0\|< \epsilon$.

If $V_m' = \{z \in B(0,r) \backslash f_0(\partial \Omega) \mid \deg(f_0,\Omega,z)=m \}$, we get using the previous argument

$\displaystyle \deg(g_0 \circ f_0, \Omega,p_0) = \sum\limits_{m \geq 0} m \deg(g_0,V_m',p_0)$.

If $\epsilon$ is small enough, $V_m \cap g_0^{-1}(p)= V_m' \cap g_0^{-1}(p)$ and $\deg(g_0,V_m',p_0)= \deg(g_0,V_m',p)$, hence

$\begin{array}{ll} \deg(g_0 \circ f_0, \Omega,p_0) & \displaystyle = \sum\limits_{m \geq 0} m \deg(g_0,V_m',p) \\ \\ & \displaystyle = \sum\limits_{m \geq 0} \deg(g_0,V_m,p) \\ \\ & \displaystyle = \sum\limits_{m \geq 0} m \deg(g,V_m,p) \end{array}$

Then, notice that

$\begin{array}{ll} \displaystyle \sup\limits_{x \in \overline{\Omega}} \| g_0 \circ f_0(x) - g \circ f_0(x) \| & \displaystyle = \sup\limits_{x \in f_0(\overline{\Omega})} \| g_0(x)-g(x) \| \\ \\ & \displaystyle \leq \sup\limits_{x \in \overline{B}(0,r)} \| g_0(x)-g(x) \| \leq \epsilon \end{array}$

Therefore, if $\epsilon$ is small enough, $\deg(g_0 \circ f_0, \Omega,p)= \deg(g \circ f_0 , \Omega,p)$. Moreover, because $g$ is uniformely continuous on $\overline{B}(0,r)$, we get $\deg(g \circ f, \Omega,p)= \deg(g \circ f_0, \Omega,p)$ if $\epsilon$ is small enough.  Finally,

$\begin{array}{ll} \deg(g \circ f, \Omega,p) & = \deg(g \circ f_0, \Omega,p) = \deg(g_0 \circ f_0, \Omega,p) \\ \\ & \displaystyle = \deg(g_0 \circ f_0, \Omega,p_0) = \sum\limits_{ m \geq 0} m \deg(g,V_m,p) \\ \\ & \displaystyle = \sum\limits_{i \in I} \deg(f, \Omega,U_i) \cdot \deg(g,U_i,p) \hspace{1cm} \square \end{array}$

Proof of theorem. Let $\{ U_i \mid i \in I \}$ (resp. $\{ V_j \mid j \in J\}$) be the bounded connected components of $\mathbb{R}^n \backslash \Omega_1$ (resp. $\mathbb{R}^n \backslash \Omega_2$). Let $h : \Omega_1 \to \Omega_2$ be a homeomorphism. We may extend $h : \Omega_1 \to \Omega_2$ (resp. $h^{-1} : \Omega_2 \to \Omega_1$) into a continuous map $f: \mathbb{R}^n \to \mathbb{R}^n$ (resp. $g : \mathbb{R}^n \to \mathbb{R}^n$).

Let $i \in I$ and $p \in U_i$ be fixed and $\{W_k \mid k \in K\}$ denote the bounded connected components of $\mathbb{R}^n \backslash h( \partial U_i)$. Noticing that $g \circ f(x)=x$ for all $x \in \partial U_i$, we deduce that

$\begin{array}{ll} 1 & = \deg( \mathrm{Id},U_i,p) = \deg(g \circ f,U_i,p) \\ \\ & \displaystyle = \sum\limits_{k \in K} \deg(f, U_i,W_k) \cdot \deg(g, W_k,p) \end{array}$

Because $h(\partial \Omega_1) \subset \partial \Omega_2$, each $W_k$ can be written as a disjoint union of $V_j$‘s. More precisely, if $N_k= \{ j \in J \mid V_j \subset W_k$ then $W_k= \coprod\limits_{j \in N_k} V_j$. We deduce that

$\displaystyle \deg(g,W_k,p) = \sum\limits_{j \in N_k} \deg(g,V_j,p)$ and $\deg(f,U_i,W_k)= \deg(f,U_i,V_j)$ for all $j \in N_k$.

Thus, noticing that $J= \coprod\limits_{k \in K} N_i$,

$\begin{array}{ll} 1 & \displaystyle = \sum\limits_{k \in K} \deg(f,U_i,W_k) \cdot \deg(g, W_k,p) \\ \\ & \displaystyle = \sum\limits_{k \in K} \sum\limits_{j \in N_k} \deg(f,U_i,V_j) \cdot \deg(g,V_j,p) \\ \\ & \displaystyle = \sum\limits_{j \in J} \deg(f,U_i,V_j) \cdot \deg(g,V_j,p) \end{array}$

Moreover, because $p \in U_i$ and $g(\partial V_j) \subset g(\partial \Omega_2) \subset \partial \Omega_1$, we have

$\deg(g,V_j,p)= \deg(g,V_j,U_i)$,

hence

$\displaystyle 1= \sum\limits_{j \in J} \deg(f,U_i,V_j) \cdot \deg(g,V_j,U_i)$.

With a symmetric argument, replacing $f$ and $\{U_i \mid i \in I\}$ with $g$ and $\{ V_j \mid j \in J\}$ respectively, we get (where $j$ is fixed):

$\displaystyle 1= \sum\limits_{i \in I} \deg(g,V_j,U_i) \cdot \deg(f,U_i,V_j)$.

To conclude, it is sufficient to write

$\begin{array}{ll} |J| & \displaystyle = \sum\limits_{j \in J} \sum\limits_{i \in I} \deg(g,V_j,U_i) \cdot \deg (f,U_i,V_j) \\ \\ & \displaystyle = \sum\limits_{i \in I} \sum\limits_{j \in J} \deg(f,U_i,V_j) \cdot \deg(g, V_j,U_i) \\ \\ & = |I| \hspace{1cm} \square \end{array}$

Just for pleasure, let us conclude with an application of the product formula:

Theorem: Let $n \geq 1$ be an even integer. Then $\mathbb{Z}_2$ is the only group acting freely by homeomorphisms on the $n$-dimensional sphere $\mathbb{S}^n$.

Proof. For all continuous map $f: \mathbb{S}^n \to \mathbb{S}^n$, let us define $d(f)= \deg( \tilde{f},B^{n+1},0)$ where $\tilde{f} : \overline{B}^{n+1} \to \mathbb{R}^{n+1}$ is a continuous extension of $f$ on the closed unit ball of $\mathbb{R}^{n+1}$. Notice that $d(f)$ does not depend on the extension $\tilde{f}$.

According the product formula, if $G$ is a group of automorphisms of $\mathbb{S}^n$, then $d(g \circ f)= d(g)d(f)$ for all $f,g \in G$. Therefore, $d : G \to \mathbb{Z}_2$ defines a homomorphism.

Using the homotopy $H : (t,x) \mapsto (1-t) \tilde{f}(x)-tx$ between $\tilde{f}$ and $- \mathrm{Id}$, we deduce by homotopy invariance that $d(f)= \deg(- \mathrm{Id}, B^{n+1},0)=(-1)^n=-1$ if $f$ has no fixed point. However, if the action of $G$ on $\mathbb{S}^n$ is free, no $f \in G$ has a fixed point, so the kernel of $d$ is trivial.

Consequently, $G$ is a subgroup of $\mathbb{Z}_2$. $\square$