Math4202 Topology II (Lecture 33)

Algebraic Topology

Let $X$ be the wedge of $n$ circles.

We claim that $\pi_1(X,x_0)=F_n$ . (Free group generated by $n$ generators.)

We proceed by induction on $n$ ,

When $n=1$ , then $\pi_1(X,x_0)=\langle a\rangle$ .

Let $X=S_1\cup S_2\cup\cdots\cup S_n$ , where $S_i$ is the $i$ -th circle.

Let $W_i=S_1\\{p_i\}$ where $p_i\neq p$ , where we wedged the circle at $p$ .

Let

U=W_1\cup W_2\cup\cdots\cup W_{n-1}\cup S_n

V=S_1\cup S_2\cup\cdots\cup S_{n-1}\cup W_n

And $U\cap V$ is path connected, and simply connected, therefore $\pi_1(U\cap V,x_0)=\mathbb Z$ .

By Seifert-Van Kampen Theorem, $\pi_1(U,x_0)*\pi_1(V,x_0)=\pi_1(X,x_0)$ .

Since $\pi_1(U,x_0)=F_1$ and $\pi_1(V,x_0)=F_{n-1}$ , then $\pi_1(U,x_0)*\pi_1(V,x_0)=F_1*F_{n-1}=F_n$ .

Let $X$ be a Hausdorff space, $A\subseteq X$ be a closed, path connected subspace. Suppose there is a continuous map

h:B^2\to X

Such that:

Choose $p\in S^1$ , $a=h(p)\in A$

Then the homomorphism induced by inclusion $:A\to X$

i_*:\pi_1(A,a)\to\pi_1(X,a)

is surjective and the kernel is the least normal subgroup of $\pi_1(A,a)$ containing the image of

k_*:\pi_1(S^1,p)\to\pi_1(A,a)

That is $\pi_1(X,a)=\pi_1(A,a)/{\langle\langle\operatorname {im}(k_*)\rangle\rangle}$ .

Proof will not be discussed here.

Consider $X=T^2$ , take $A$ be the wedged-2 circle.

Then the fundamental group of $A$ is

\pi_1(A,a)=F_2/\langle\langle ab a^{-1}b^{-1}\rangle\rangle \simeq \mathbb Z\times \mathbb Z

Let $r:S^1\to S^1$ rotation through angle $2\pi/n$ .

r(\cos\theta, \sin \theta)=(\cos(\theta+2\pi/n), \sin(\theta+2\pi/n))

Then the $n$ folds dunce cap is

X_n=B^2/\{x\in S^1=\partial B^2, r(x)\sim r^2(x)\sim\cdots\sim r^{n-1}(x)\}

(Use quotient map to glue the boundary)

Note that $\pi:B^2\to X_n$ be a quotient map, then it is a closed map, and $X$ is hausdorff and normal.

How to compute the fundamental group of $X_n$ ?

$k_*:S^1\to A$ and $a\mapsto a^n$

\pi_1(X_n,a)=\pi_1(A,a)/\langle\langle\operatorname {im}(k_*)\rangle\rangle=\mathbb Z/n\mathbb Z