Math4202 Topology II (Lecture 29)

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 

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.

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.