We conclude our series of notes about Brouwer’s topological degree by a final application, namely Borsuk theorem and its corollaries.
Theorem: (Borsuk) Let be a bounded open subspace symmetric with respect to the origin and be a continuous function satisfying . If is odd then is odd.
Proof. Let be a sequence of functions converging uniformely to . If for each , then is a sequence of odd functions converging uniformely to . Because when is large enough, we may suppose without loss of generality that is .
Suppose first that is a regular value of . Because ,
Since is odd, the group acts on by and if is a fundamental domain for the action,
because odd implies even. Therefore, the theorem is proved in our case.
From now on, suppose that is not a regular value of . Thanks to the previous case, it is enough to show that is arbitrarily closed to an odd function whose is a regular value. For this purpose, we use the following lemma:
Lemma: Let be an odd function. For every , there exists such that and such that is an odd function satisfying:
- is a regular value of on ,
- If and then and ,
- where .
First, because there exists an arbitrarily small not in the spectrum of , we may suppose without loss of generality that is invertible (otherwise replace with ).
Let be the function given by our lemma with and and be the function given by our lemma with and .
By construction, is a regular value of on . Moreover, if , and , then and ; in particular, is invertible. Therefore, is a regular value of on .
By induction, let be defined in the same way. Then is an odd function whose is a regular value on
is invertible, so is a regular value of . Finally, we conclude the proof by noticing that:
Proof of the lemma. Let be a point such that and . Then and where .
Because , we deduce that is invertible if is a regular value of . According to Sard’s theorem, there exists such value with an arbitrarily small norm, concluding the proof of the first point. The other points are straightforward.
Theorem 1: Let be two integers and be a continuous map. Then there exists such that .
Proof. Let be an odd continuous extension of on the unit closed ball . If we deduce that from Borsuk theorem.
On the other hand, there exists a point with an arbitrarily small norm, so because for all and . A contradiction.
Therefore, ie. there exists such that .
Theorem 2: (open mapping theorem) Let be open subspace and be an injective continuous map. Then is open.
Proof. It is sufficient to prove that for all and such that , contains an open neighborhood of . For this purpose, we show that . Indeed, in this case there exists an open neighborhood of such that for all , so there exists such that hence .
Because , we may suppose without loss of generality that .
Let be a homotopy. Because is injective, if then hence for all . So by homotopy invariance,
where is odd. Now, according to Borsuk theorem, , which concludes the proof.
Theorem 3: (Invariance of the domain) Let be an open subspace and be two integers. There does not exist a continuous injection .
Proof. Suppose by contradiction that there exists such an injection
According to theorem 2, is an open subspace of . Because the projection is open, we deduce that is open in (with ): a contradiction.