Merge branch 'main' of https://git.trance-0.com/Trance-0/NoteNextra-origin

content/Math4202/Math4202_L21.md

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+# 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>

content/Math4202/_meta.js

@@ -24,4 +24,5 @@ export default {
     Math4202_L18: "Topology II (Lecture 18)",
     Math4202_L19: "Topology II (Lecture 19)",
     Math4202_L20: "Topology II (Lecture 20)",
+    Math4202_L21: "Topology II (Lecture 21)",
 }