57 lines
1.9 KiB
Markdown
57 lines
1.9 KiB
Markdown
# Math4202 Topology II (Lecture 29)
|
|
|
|
## Algebraic Topology
|
|
|
|
### Fundamental Groups of Some Surfaces
|
|
|
|
Recall from previous lecture, we talked about figure 8 shape.
|
|
|
|
#### Lemma The fundamental group of figure-8 is not abelian
|
|
|
|
The fundamental group of figure-8 is not abelian.
|
|
|
|
<details>
|
|
<summary>Proof</summary>
|
|
|
|
Consider $U,V$ be two "fish shape" where $U\cup V$ is the figure-8 shape, and $U\cap V$ is $x$ shape.
|
|
|
|
The $x$ shape is path connected,
|
|
|
|
$\pi_1(U,x_0)$ is isomorphic to $\pi_1(S^1,x_0)$, and $\pi_1(V,x_0)$ is isomorphic to $\pi_1(S^1,x_0)$.
|
|
|
|
To show that is not abelian, we need to show that $\alpha*\beta\neq \beta*\alpha$.
|
|
|
|
We will use covering map to do this.
|
|
|
|
[Universal covering of figure-8](https://notenexta.trance-0.com/Math4202/universal-covering-of-figure-8.png)
|
|
|
|
However, for proving our result, it is sufficient to use xy axis with loops on each integer lattice.
|
|
|
|
And $\tilde{\alpha*\beta}(1)=(1,0)$ and $\tilde{\beta*\alpha}(1)=(0,1)$. By path lifting correspondence, the two loops are not homotopic.
|
|
|
|
</details>
|
|
|
|
#### Theorem for fundamental groups of double torus (Torus with genus 2)
|
|
|
|
The fundamental group of Torus with genus 2 is not abelian.
|
|
|
|
<details>
|
|
<summary>Proof</summary>
|
|
|
|
If we cut the torus in the middle, we can have $U,V$ is two "punctured torus", which is homotopic to the figure-8 shape.
|
|
|
|
But the is trick is not enough to show that the fundamental group is not abelian.
|
|
|
|
---
|
|
|
|
First we use quotient map $q_1$ to map double torus to two torus connected at one point.
|
|
|
|
Then we use quotient map $q_2$ to map two torus connected at one point to figure-8 shape.
|
|
|
|
So $q=q_2\circ q_1$ is a quotient map from double torus to figure-8 shape.
|
|
|
|
Then consider the inclusion map $i$ and let the double torus be $X$, we claim that $i_*:\pi_1(\infty,x_0)\to \pi_1(X,x_0)$ is injective.
|
|
|
|
If $\pi_1(X,x_0)$ is abelian, then the figure 8 shape is abelian, that is contradiction.
|
|
|
|
</details> |