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 ,
- .

Then .

**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 ,

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 ,

Then .

**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

and .

Notice that

and

Moreover, if then ; if then hence .

Thus .

Because and

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

and .

Then the second equality of our lemma follows easily.

For the first equality, let be fixed and let

and .

Thus, . Therefore

hence

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 ,

Therefore,