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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
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.