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
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
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 .