Math4202 Topology II (Lecture 30)

Algebraic Topology

We skipped a few chapters about Jordan curve theorem, which will be your final project soon. LOL, I will embedded the link once I'm done.

Seifert-Van Kampen Theorem

The Seifert-Van Kampen Theorem

Let X=U\cup V be a union of two open subspaces. Suppose that U\cap V, U,V are path connected. Fix x_0\in U\cap V.

Let H be a group (arbitrary). And now we assume \phi_1,\phi_2 be a group homomorphism, and \phi_1:\pi_1(U,x_0)\to H, and \phi_2:\pi_1(V,x_0)\to H.

Let i_1,i_2,j_1,j_2,i_{12} be group homomorphism induced by the inclusion maps.

Assume this diagram commutes.


\phi_1\circ i_1=\phi_2\circ i_2

There is a group homomorphism \Phi:\pi_1(X,x_0)\to H making the diagram commute. \Phi\circ j_1=\phi_1 and \Phi\circ j_2=\phi_2.

We may change the base point using conjugations.

Side notes about free product of two groups

Consider arbitrary group G_1,G_2, then G_1\times G_2 is a group.

Note that the inclusion map i_1:G_1\to G_1\times G_2 is a group homomorphism and the inclusion map i_2:G_2\to G_1\times G_2 is a group homomorphism. The image of them commutes since (e,g_2)(g_1,e)=(g_1,g_2)=(g_1,e)(e,g_2).

The universal property

Then we want to have a group G such that for all group homomorphism \phi:G_1\to H and G_2\to H, such that there always exists a map \Phi: G\to H such that:

\Phi\circ i_1=\phi_1
\Phi\circ i_2=\phi_2

How to construct the free group?

We consider


G_1*G_2=S=\{g_1h_1g_2h_2:g_1,g_2\in G_1,h_1,h_2\in G_2\}/\sim

And we set g_ie_{G_2}g_{i+1}\sim g_ig_{i+1} for g_i\in G_1 and g_{i+1}\in G_2.

And h_je_{G_1}h_{j+1}\sim h_jh_{j+1} for h_j\in G_2 and h_{j+1}\in G_1.

And we define the group operation


(g_1 h_1\cdots g_k h_k)*(h_1' g_1'\cdots h_l' g_l')=g_1 h_1\cdots g_k h_k g_1' h_2'\cdots h_l' g_l'

And the inverse is defined


(g_1 h_1\cdots g_k h_k)^{-1}=h_k^{-1} g_k^{-1}\cdots h_1^{-1} g_1^{-1}

And G=S is a well-defined group.

The homeomorphism G\to H is defined as


\Phi((g_1 h_1\cdots g_k h_k))=\phi_1(g_1)\circ \phi_2(h_1)\circ \cdots \circ \phi_1(g_k)\circ \phi_2(h_k)

Note \circ is the group operation in H.

Group with such universal property is unique, so we don't need to worry for that too much.

Back to the Seifert-Van Kampen Theorem:

Let H=\pi_1(U,x_0)* \pi_1(V,x_0).

Let N be the least normal subgroup in the free product H, containing i_1(g)i_2(g)^{-1}, \forall g\in \pi_1(U\cap V,x_0).

Note i_1(g)\in \pi_1(U,x_0) and i_2(g)\in \pi_1(V,x_0). You may think of them as G_1,G_2 in the free group descriptions.

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).

3.0 KiB Raw Blame History