# Math4202 Topology II (Lecture 21)

## Algebraic Topology

### Application of fundamental groups

Recall from last Friday, $j:S^1\to \mathbb{R}^2-\{0\}$ is not null homotopic

#### Hairy ball theorem

Given a non-vanishing vector field on $B^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}$, ($v:B^2\to \mathbb{R}^2$ continuous and $v(x,y)\neq 0$ for all $(x,y)\in B^2$) there exists a point of $S^1$ where the vector field points directly outward, and a point of $S^1$ where the vector field points directly inward.

<details>
<summary>Proof</summary>

By our assumption, then $v:B^2\to \mathbb{R}^2-\{0\}$ is a continuous vector field on $B^2$.

$v|_{S^1}:S^1\to \mathbb{R}^2-\{0\}$ is null homotopic.

We prove by contradiction.

Suppose $v:B^2\to \mathbb{R}^2-\{0\}$ and $v|_{S^1}:S^1\to \mathbb{R}^2-\{0\}$ is everywhere outward. (for everywhere inward, consider $-v$ must be everywhere outward)

Because $v|_{S^1}$ extends continuously to $B^2$, then $v|_{S^1}:B^2\to \mathbb{R}^2-\{0\}$ is null homotopic.

We construct a homotopy for functions between $v|_{S^1}$ and $j$. (Recall $j:S^1\to \mathbb{R}^2-\{0\}$ is not null homotopic)

Define $H:S^1\times I\to \mathbb{R}^2-\{0\}$ by affine combination

$$
H((x,y),t)=(1-t)v(x,y)+tj(x,y)
$$

we also need to show that $H$ is non zero.

Since $v$ is everywhere outward, $v(x,y)\cdot j(x,y)$ is positive for all $(x,y)\in S^1$.

$H((x,y),t)\cdot j(x,y)=(1-t)v(x,y)\cdot j(x,y)+tj(x,y)\cdot j(x,y)=(1-t)(v(x,y)\cdot j(x,y))+t$

which is positive for all $t\in I$, therefore $H$ is non zero.

So $H$ is a homotopy between $v|_{S^1}$ and $j$.
</details>

#### Corollary of the hairy ball theorem

$\forall v:B^2\to \mathbb{R}^2$, if on $S^1$, $v$ is everywhere outward/inward, there is $(x,y)\in B^2$ such that $v(x,y)=0$.

#### Brouwer's fixed point theorem

If $f:B^2\to B^2$ is continuous, then there exists a point $x\in B^2$ such that $f(x)=x$.

<details>
<summary>Proof</summary>

We proceed by contradiction again.

Suppose $f$ has no fixed point, $f(x)-x\neq 0$ for all $x\in B^2$.

Now we consider the map $v:B^2\to \mathbb{R}^2$ defined by $v(x,y)=f(x)-x$, this function is continuous since $f$ is continuous.

$forall x\in S^1$, $v(x)\cdot x=f(x)\cdot x-x\cdot x=f(x)\cdot x-1$.

Recall the cauchy schwartz theorem, $|f(x)\cdot x|\leq \|f(x)\|\cdot\|x\|\leq 1$, note that $f(x)\neq 0$ for all $x\in B^2$, $v(x)\cdot x<0$. This means that all $v(x)$ points inward.

This is a contradiction to the hairy ball theorem, so $f$ has a fixed point.
</details>
