Math4202 Topology II (Lecture 31)

Algebraic Topology

Before this chapter, we have a big chapter about free groups, be

Seifert-Van Kampen Theorem

The Seifert-Van Kampen Theorem (classical version)

There is an isomorphism between $\pi_1(U,x_0)* \pi_1(V,x_0)/N$ and $\pi_1(U\cup V,x_0)=\pi_1(X,x_0)$ .

Note

H is a normal subgroup of $G$ if $\forall g\in G$ , $h\in H, ghg^{-1}\in H$

Consider the element $i_1(g)^{-1}i_2(g)$ , where $g\in g\in \pi_(U\cap V,x_0)$ . Since $i_1(g)\in \pi_1(U,x_0)$ and $i_2(g)\in \pi_1(V,x_0)$ , then $i_1(g)^{-1}i_2(g)\in \pi_1(U,x_0)* \pi_1(V,x_0)$ . (the element is in the free product)

Let $N$ be the smallest normal subgroup containing all elements of the form $i_1(g)^{-1}i_2(g)$ for $g\in \pi_1(U\cap V,x_0)$ . Such group is constructed by taking the intersection of $N_\alpha$ s.

$\pi_1(U,x_0)* \pi_1(V,x_0)$ is a normal subgroup satisfying $i_1(g)^{-1}i_2(g)\in N$ for all $g\in \pi_1(U\cap V,x_0)$ .
Take the intersection of all such normal subgroups to get $N$ .

Proof

Let ~~$H=\pi_1(U,x_0)* \pi_1(V,x_0)$~~ $H=\pi_1(U,x_0)* \pi_1(V,x_0)/N$ (we must quotient by $N$ ; in the raw free product $i_1(g)^{-1}i_2(g)$ is in general a non-trivial word, so the relation we want to enforce below only holds after passing to the quotient).

$\phi_1:\pi_1(U,x_0)\to H$ and $\phi_2:\pi_1(V,x_0)\to H$ are defined as the inclusion into the free product followed by the quotient map $q:\pi_1(U,x_0)* \pi_1(V,x_0)\to H$ .

We need to verify that $\phi_1\circ i_1=\phi_2\circ i_2$ .

That is $i_1(g)=i_2(g)\in H$ for all $g\in \pi_1(U\cap V,x_0)$ . This is equivalent to show that $i_1(g)^{-1}i_2(g)=e\in H$ .

Motivation: By construction, $N$ is the normal closure of $\{i_1(g)^{-1}i_2(g):g\in\pi_1(U\cap V,x_0)\}$ , so every such element lies in $\ker q$ . Hence $q(i_1(g)^{-1}i_2(g))=e_H$ , giving the commutativity needed to invoke the universal property.

By Seifert-Van Kampen Theorem, $\exists \Phi:\pi_1(X,x_0)\to H$ .

And $\phi_1(U,x_0)\to \pi_1(X,x_0)$ by $j_1$ and $\phi_2(V,x_0)\to \pi_1(X,x_0)$ by $j_2$ .

And $j:\pi_1(U,x_0)* \pi_1(V,x_0)\to \pi_1(X,x_0)$ .

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Incentive to continue: To obtain the classical isomorphism we must exhibit a two-sided inverse to $\Phi$ . The natural candidate is the map $\tilde j:H\to\pi_1(X,x_0)$ induced by $j$ ; it is well-defined because $j_1\circ i_1=j_2\circ i_2$ forces $j(N)=\{e\}$ . Verifying $\tilde j\circ\Phi=\mathrm{id}_{\pi_1(X,x_0)}$ (by uniqueness in the universal property) and $\Phi\circ\tilde j=\mathrm{id}_H$ (by checking on classes coming from $\pi_1(U)$ and $\pi_1(V)$ , which generate $H$ ) completes the proof — this is carried out in Lecture 32.

Corollary of the Seifert-Van Kampen Theorem (classical version)

Assume $U,V,U\cap V$ are path connected, and $U\cap V$ is simply connected.

Then $\pi_1(U\cup V,x_0)=\pi_1(U,x_0)* \pi_1(V,x_0)$ . ( $N$ is the trivial subgroup)

Example

The figure eight shape have intersection with trivial subgroup.

So $\pi_1(U\cup V,x_0)=\pi_1(U,x_0)* \pi_1(V,x_0)=\pi_1(S^1,x_0)*\pi_1(S^1,x_0)=\mathbb Z*\mathbb Z$ (Free product of $\mathbb Z$ ). $st\neq ts$ .

Another Corollary of the Seifert-Van Kampen Theorem (classical version)

Assume $U,V,U\cap V$ are path connected, and $V$ is simply connected.

Then $\pi_1(U\cup V,x_0)=\pi_1(U,x_0)/N$ .

Example

Consider the punctured torus $U$ union with a spherical cap $V$ .

$\pi_1(V,x_0)=\{e\}$ .

$U\cap V\cong S^1$ .

And $\pi_1(U\cap V,x_0)=\mathbb Z=\langle g\rangle$ .

Consider $i_1(g)^{-1}$ in $\mathbb Z*\mathbb Z$ .

$i_1(g)^{-1}=st^{-1}s^{-1}t=1$ . So in the quotient, $st=ts$ .

$\pi_1(U\cup V,x_0)=\pi_1(U,x_0)/N=\mathbb Z\times \mathbb Z$ .