# Seifert Van Kampen Example

I'm stuck in the explicit calculation of the fundamental group of one space.

I have the space that is three copies of $$\mathbb{S}^{1}$$ disposed in vertical, for example, $$L=\partial(B[(0,0),1])\cup\partial(B[(0,2),1])\cup\partial(B[(0,-2),1])$$ (where the symbol $$\partial$$ denotes the boundary).

What I tought is, well, let be $$U=X-\{(0,3)\}$$ and $$V=X-\{(0,-3)\}$$ both are path connected and open with the usual topology of $$\mathbb{R}^{2}$$ and both are homotopical to the 8 figure, that has fundamental group $$\mathbb{Z}*\mathbb{Z}$$. Their intersection is $$U\cap V=X-\{(0,3),(0,-3)\}$$ (connected by paths) that is homotopic to $$\mathbb{S}^{1}$$ that has $$\mathbb{Z}$$ as fundamental group.

My question is, the fundamental group is $$(\mathbb{Z}*\mathbb{Z})*(\mathbb{Z}*\mathbb{Z})$$? And if it is, how can I show that? Also I would like to know how to express it in terms of presentations of groups (for example, $$\mathbb{Z}*\mathbb{Z}$$ I know that is isomorphic to $$$$).

Any hint to continue is appreciated!

• $L$ is actually the union of $\vee_{i=1}^2S_i^1$ and a circle (they have one point in common). we know $\pi_1(\vee_{i=1}^2S_i^1,x_0)\approx\Bbb{Z}\times\Bbb{Z}$ and $\pi_1(S^1,x_0)\approx\Bbb{Z}$ then apply Van Kampen's theorem. Let $\vee_{i=1}^2S_i^1\subset A$, $S^1\subset B$ where $A,B$ open. then $A\cap B$ is contractible and so the fundamental group should be $\Bbb{Z}^3$ instead of $\Bbb{Z}^4$. May 5 '20 at 0:30

$$L=\vee_{i=1}^2\Bbb{S}_i^1\cup\Bbb{S}^1$$ where $$\vee_{i=1}^2\Bbb{S}_i^1\cap\Bbb{S}^1=(0,-1)$$.

Apply Van Kampen's Thm,

Let the space in the blue part be an open set $$A$$ s.t. $$\vee_{i=1}^2\Bbb{S}\subset A$$ and the red part be another open set $$B$$ s.t. $$\Bbb{S}^1\subset B$$. We can see that $$A\simeq \vee_{i=1}^2\Bbb{S}_i^1\implies \pi_1(\vee_{i=1}^2\Bbb{S},(0,-1))\approx\Bbb{Z}\ast\Bbb Z$$, and $$B\simeq\Bbb{S}^1\implies\pi_1(\Bbb{S}^1,(0,-1))\approx\Bbb{Z}$$, and $$A\cap B\simeq(0,-1)$$ (i.e. contractible) which implies that $$\pi_1(A\cap B,(0,-1))=\{1\}$$. So, finally let $$x_0=(0,-1)$$, we conclude that $$\pi_1(L,x_0)=\langle a,b,c \rangle \approx\Bbb{Z}\ast\Bbb{Z}\ast\Bbb{Z}$$. (No amalgamated relation from the intersection)

The problem in your solution is that you didn't consider the amalgamated relation. we see that $$U\simeq V\simeq \vee_{i=1}^2\Bbb{S}_i^1$$ and $$U\cap V\simeq\Bbb{S}^1$$. If we let $$a,b$$ be the generators of $$\pi_1(U,x_0)$$ and $$c,d$$ be the generators of $$\pi_1(V,x_0)$$ and $$e$$ be the generator of $$\pi_1(U\cap V,x_0)$$, then the induced homomorphism $$i_*:\pi_1(U\cap V)\to \pi_1(U)$$ gives us $$i_*(e)=a$$, similarly the other induced homomorphism gives us $$j_*(e)=d$$, which means you should get $$\pi_1(L,x_0)=\langle a,b,c,d|a=d \rangle$$, where the relation is given by the amalgamation. After simplification, $$\pi_1(L,x_0)=\langle a,b,c \rangle \approx \Bbb{Z}\ast\Bbb{Z}\ast \Bbb{Z}$$.

I would choose the first method since the intersection is contractible which means I don't have to consider the amalgamated relation.

• I only would like ask you a favour... Can you write down how to proceed in the second paragraph of your answer? (where you work with the induced homomorphism) ... My problem is that I haven't worked this of amalgamated relations enough... Thank you ! @Kevin.S May 5 '20 at 7:31
• @sopach96 Well, Van Kampen's Thm states that $f:(\pi_1(U)\times\pi_1(V))\to \pi_1(L)$ induces an epimorphism $\Phi_*:(\pi_1(U)\times \pi_1(V))/N\to \pi_1(L)$ where $N=ker(f)$. Here, $ker(f)$ in this case is generated by $i(e)^{-1}j(e)=a^{-1}d$ (because $\pi_1(U\cap V)$ has an identity and a generator $e$) and then, $\Phi(a^{-1}d)=1$, hence we get the amalgamated relation $a=d$. May 5 '20 at 10:16
• Thank you very much ! @Kevin.S May 5 '20 at 10:19