# 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.
Proof 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.
#### Theorem for fundamental groups of double torus (Torus with genus 2) The fundamental group of Torus with genus 2 is not abelian.
Proof 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.