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 $<a,b:[a,b]=1>$).
Any hint to continue is appreciated!
 A: $L=\vee_{i=1}^2\Bbb{S}_i^1\cup\Bbb{S}^1$ 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.
