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

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