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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!

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    $\begingroup$ $L$ 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 $\Bbb{Z}^4$. $\endgroup$
    – Kevin.S
    May 5 '20 at 0:30
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$L=\vee_{i=1}^2\Bbb{S}_i^1\cup\Bbb{S}^1$ where $\vee_{i=1}^2\Bbb{S}_i^1\cap\Bbb{S}^1=(0,-1)$.

Apply Van Kampen's Thm,

enter image description here

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.

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  • $\begingroup$ 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 $\endgroup$
    – sopach96
    May 5 '20 at 7:31
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    $\begingroup$ @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 $a=d$. $\endgroup$
    – Kevin.S
    May 5 '20 at 10:16
  • $\begingroup$ Thank you very much ! @Kevin.S $\endgroup$
    – sopach96
    May 5 '20 at 10:19

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