Van Kampen's Theorem

Suppose we have a topology space $X=A\cup B$ where $A$ and $B$ are open, path-connected subsets of $X$, and $A\cap B$ is also path-connected

Then we have

$${\pi }_1(X)\cong \frac{{\pi }_1(A)\ast {\pi }_1(B)}{ker(\Phi )}=\frac{{\pi }_1(A)\ast {\pi }_1(B)}{⟨⟨i_A(k)i_B^{-1}(k)|\forall k:{\pi }_1(A\cap B)⟩⟩}=\frac{{\pi }_1(A)\ast {\pi }_1(B)}{\text{rel identifying loops from }A\cap B}$$

(the above fundamental groups should all have a chosen base point $x_0$ in $A\cap B$) Notation:

• $\ast$ is the free product

• $\ll S\gg$ represents the normal subgroup generated by set S

  • given by $⟨gs{g}^{-1}|\forall s\in S,g\in G⟩$ where $G={\pi }_1(A)\ast {\pi }_1(B)$

• $i_A:{\pi }_1(A\cap B,x_0)\to {\pi }_1(A,x_0)$ induced by the inclusion map

  • elements of ${\pi }_1(A,x_0)$ naturally sit inside ${\pi }_1(A)\ast {\pi }_1(B)$

• $i_B:{\pi }_1(A\cap B,x_0)\to {\pi }_1(B,x_0)$ induced by the inclusion map

• $\Phi :{\pi }_1(A,x_0)\ast {\pi }_1(B,x_0)\to {\pi }_1(X,x_0)$ (a surjective homomorphism)

Explanation

Imagine $\nu$ is a loop based at $x_0$ and contained in $A\cap B$. Then, there are elements $a=[\nu {]}_A\in {\pi }_1(A,x_0)\text{and}b=[\nu {]}_B\in {\pi }_1(B,x_0),$ which, as words in ${\pi }_1(A,x_0)\ast {\pi }_1(B,x_0),$ are different. However, in $X$ they represent the same loop. So we must have $\Phi ([\nu {]}_A)=\Phi ([\nu {]}_B)=[\nu {]}_X\in {\pi }_1(X,x_0)$

$$⟹\Phi ([\nu {]}_A)\cdot \Phi ([\nu {]}_B{)}^{-1}=e$$ $$⟹\Phi ([\nu {]}_A[\nu {]}_B^{-1})=e\text{(by group homomorphism)}$$

So, $[\nu {]}_A[\nu {]}_B^{-1}$ is contained in the kernel of $\Phi$ (see definition of kernel) and the Van Kampen’s theorem says that they generate the entire kernel. And the quotient forces the relation $[\nu {]}_A=[\nu {]}_B$

trivial case

When $A\cap B$ is simply connected (or even simpler: $A$ and $B$ are connected by a point) then we have the simplified equation: (because ${\pi }_1(A\cap B,x_0)=0$)

$${\pi }_1(X,x_0)\cong {\pi }_1(A,x_0)\ast {\pi }_1(B,x_0)$$

Application