Here is the first note of a series of posts devoted to Brouwer’s topological degree. Our aim is to give a construction of Brouwer’s topological degree.
Lemma 1: Let be a bounded open subspace, be a continuous map on , and be a function satisfying:
- is continuous on and vanishes on a neighborhood of and on for some ,
Then does not depend on , where is the jacobian determinant of .
Lemma 2: Let be a bounded open subspace, be a continuous map on and be a continuous function. Suppose that:
- for all ,
- vanishes on a neighborhood of and on ,
Proof. Let and for . Let denote the -cofactor of the jacobian matrix of .
First, notice that vanishes on a neighborhood of and on a neighborhood of since when . Therefore, is and can be extended on by zero. Then
We deduce that
Notice that we used a property on cofactors of a jacobian matrix whose a proof will be found at the end of this note for convenience:
Lemma 3: Let be a -map and let denote the -cofactor of the facobian matrix of . Then for all and ,
Proof of lemma 1. Let be the vector space of applications satisfying the first point of our lemma and consider the following linear functional:
Applying lemma 2 to and , we deduce that implies .
Let satisfying . Because , we deduce that hence ie. .
Now we are able to define Brouwer’s degree of a map.
Definition: Let be a bounded open subspace, be a continuous map on and . With the previous notations, we define the degree of at with respect to by
In fact, it is possible to simplify the definition when is a regular value of .
Lemma 4: Let be a bounded open subspace, be a continuous map on and be a regular value of . Then there exists such that for all continuous map satisfying
- while ,
we have .
Proof. Let . There exists such that defines a diffeomorphism from and for . Let such that et let .
We show that any works. Let be a function satisfying the two points of our lemma.
Notice that if then so . Thus
Because has the same sign for all ,
Corollary 1: Let be a bounded open subspace, be a continuous map on and be a regular value of . Then
Now we want to extend our definition for continuous functions. The key observation is that in fact is not a differential property:
Lemma 5: Let be a bounded open subspace, be two continuous maps on , and such that:
- for all ,
Proof. Because , we may suppose without loss of generality that .
Let be a -map satisfying and let .
In particular, .
Let be two continuous maps vanishing on a neighborhood of and satisfying
Moreover, if then ; if then hence .
for all , we deduce by integrating the equalities above that .
Now we are ready to introduce our definition (we implicitely use Stone-Weierstrass approximation theorem):
Definition: Let be a bounded open subspace, be a continuous map and . We define
where is any sequence of continuous maps on converging uniformely to and satisfying for all .
The main properties of Brouwer’s degree are summarized in the following property:
Property 6: Brouwer’s degree satisfies:
- (normality) if ,
- (additivity) Let be two disjoint bounded open subspaces such that . Then
- (homotopy invariance) Let be a homotopy and be a continuous map satisfying for all . Then
Proof. Normality and additivity are consequences of the definition and corollary 1.
Let be a continuous map and for all .
Because is compact, there exists such that for all and ; because is compact, is uniformely continuous so there exists such that for all , implies for all .
Let be a partition of satisfying for all . For all , let be a sequence of continuous maps on converging uniformely to .
Notice that for all and for large enough,
and for all and for large enough,
According to lemma 5, we deduce that . Thus
To conclude the proof, it is sufficient to prove that is constant on any connected component of . For that, we show that is locally constant.
Let . Let be two points satisfying and . Then and we conclude thanks to lemma 5.
Proof of lemma 3. Let be the jacobian matrix of . It is known that for any matrix . Now notice that
Then the second equality of our lemma follows easily.
For the first equality, let be fixed and let
Thus, . Therefore
where and . Noticing that defines a bijection from to , we finally get
Let . Notice that
- If then ,
- If then ,
- If then ,
- Because and using Schwarz’ lemma, .
We deduce the general formula:
Thus, if ,
and if ,