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 be two homeomorphic compact subspaces. Then and 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 be a bounded open subspace, and be two continuous maps. Let denote the bounded connected components of . If then

.

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

Let be such that and let . Because is compact and

,

where is the unbounded connected component of , there exists a finite subset such that .

Notice that if then , so if then hence .

Now, we prove the lemma when and are and when is a regular value of . We have:

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

and

.

Because , we deduce that

Let and be two maps and let be a regular value of such that

.

If , we get using the previous argument

.

If is small enough, and , hence

Then, notice that

Therefore, if is small enough, . Moreover, because is uniformely continuous on , we get if is small enough. Finally,

**Proof of theorem.** Let (resp. ) be the bounded connected components of (resp. ). Let be a homeomorphism. We may extend (resp. ) into a continuous map (resp. ).

Let and be fixed and denote the bounded connected components of . Noticing that for all , we deduce that

Because , each can be written as a disjoint union of ‘s. More precisely, if then . We deduce that

and for all .

Thus, noticing that ,

Moreover, because and , we have

,

hence

.

With a symmetric argument, replacing and with and respectively, we get (where is fixed):

.

To conclude, it is sufficient to write

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

**Theorem:** Let be an even integer. Then is the only group acting freely by homeomorphisms on the -dimensional sphere .

**Proof.** For all continuous map , let us define where is a continuous extension of on the closed unit ball of . Notice that does not depend on the extension .

According the product formula, if is a group of automorphisms of , then for all . Therefore, defines a homomorphism.

Using the homotopy between and , we deduce by homotopy invariance that if has no fixed point. However, if the action of on is free, no has a fixed point, so the kernel of is trivial.

Consequently, is a subgroup of .