Now that Brouwer’s topological degree has been completely caracterized in our two previous notes, we may give the first applications.

**Theorem 1:** Let be a bounded open subspace and be a continuous map satisfying . Then is onto.

**Proof.** Let ; in particular, . According to the fifth point of property 1 in our previous note, depends only on , hence

.

Thus, there exists such that .

**Theorem 2:** Let be a continuous map satisfying as . Then is onto.

**Proof.** Let and be large enough so that and when .

Let be a homotopy between and . Suppose by contradiction that there exists such that . Then

,

hence

a contradiction. Therefore, by homotopy invariance

,

So there exists such that .

**Theorem 3:** *(Brouwer’s fixed point theorem)* Let be a nonempty compact convex subset and be a continuous map. Then has a fixed point.

**Lemma:** Let be a continuous map satisfying for all . Then has a fixed point.

**Proof.** Let be a homotopy between and and let and be such that . Then

,

so and . Therefore, either has a fixed point on or by homotopy invariance

and there exists such that .

**Proof of theorem 3.** Let be large enough so that and let be a retraction.

(Considering as a convex subset of , we may suppose that has a nonempty interior. Let . Then, for any , the half-line starting at and passing through meet at only one point . Finally, is an example of retraction.)

Then apply the previous lemma to get a fixed point of . Notice that so and is in fact a fixed point of .

**Theorem 4:** *(Fundamental theorem of algebra)* Every nonconstant complex polynomial has a root.

**Proof.** Let be a complex polynomial of degree . We prove more generally that when is large enough.

Let (with ) and of degree smaller than such that . Let be large enough so that for all .

By contradiction, suppose that the homotopy between and has a zero . Then

,

a contradiction. Therefore, by homotopy invariance, .

Let be small enough so that . Notice that

.

Moreover, any nonzero complex number has distincts th roots, hence

.

**Theorem 5:** *(Rouché)* Let be a bounded open subspace and be two holomorphic functions on a neighborhood of . Suppose that has no zero on and for all . Then and have the same number of zeros in (with multiplicity).

**Lemma:** Let be a bounded open subspace and be a holomorphic function on a neighborhood of such that for all . Then coincides with the number of zeros of (with multiplicity).

**Proof.** If has no zero in , then . From now on, suppose that has a zero in . Let (with ) be the zeros of in (there are only finitely many zeros because they are isolated and that is compact).

Let such that and when . Clearly, , hence by additivity

.

Therefore, it is sufficient to show that if is a zero of with multiplicity , then for small enough.

We can write where (with ) and . Let be small enough so that for all .

Suppose by contradiction that the homotopy between and has a zero . Then

,

a contradiction. Therefore, by homotopy invariance, . But we already proved in the previous proof that .

**Proof of theorem 5. **First, notice that for all , otherwise . So it is sufficient to show that according to our previous lemma.

Suppose by contradiction that the homotopy between and has a zero . Then

,

a contradiction. Therefore, by homotopy invariance, .

**Corollary:** Let be a bounded open subspace and be a map holomorphic on a neighborhood of without zero on . Let denote the zeros of and their multiplicities. Then for all there exists such that for all map holomorphic on a neighborhood of satisfying , then the zeros of on are in with exactly in with multiplicity.

**Proof. **Let and be the zeros of in . Let such that implies , and if .

Set .

Let be a holomorphic function on a neighborhood of satisfying . In particular, for all ,

.

According to theorem 5, we deduce that and have the same number of zeros in each .

**Theorem 6:** *(Hedgehog)* Let be an odd integer, be an open subspace containing and be a continuous map. Then there exist and such that .

**Proof.** If , we can chose and a zero of . From now on, suppose that . Let be a continuous extension of ; in particular, .

Suppose that . By homotopy invariance, the homotopy between and has a zero . Moreover, because , then and , hence

.

Suppose that . In the same way, we find such that , ie.

.

**Theorem 7:*** (Hairy ball theorem)* In even dimension, any tangent vector field of a sphere has a zero.

**Proof.** Let be such a vector field; in particular, for all . According to Hedgehog theorem, there exsit and such that . However

.

Therefore, .