Computing $\pi _1(T_0\cup _{S^1} M)$

Question

$T_0$ be the torus minus an open disk in the boundary. $M$ is the Möbius band.And the union is made by identifying the boundary of the open disk with the boundary of the Möbius band. I want to compute $\pi _1$ and i have the following: I know $M\equiv _H S^1$ so $\pi _1 (M)=\mathbb{Z}$ and $\pi _1(T_0)=\pi _1(S^1 \vee S^1)=\mathbb{Z}*\mathbb{Z}$ so $\pi _1(T_0\cup _{S^1}M)=\mathbb{Z}*\mathbb{Z}*\mathbb{Z}/\langle{i_*}_1{i_*}_2^{-1}\rangle,$ and I see the wrap of $S^1$ in the torus as a degree $1$ map and the wrap of $S^1$ on $\delta M$ as a degree $2$ map which I see like $ac^{-2}$ if $a$ is generator of first circle of $S^1\vee S^1$ and $c$ the generator of middle circle of Möbius but I do not know how to proceed, hints and references are welcome.

Thanks in advance

Answer 1

You obviously applied Seifert-van Kampen with $U:=T_0$ plus a little bit and $V:=M$ plus a littel bit, s.t. they are open. By deformation retraction you observe $M\simeq S^1$ and $T_0\simeq S^1\vee S^1$, because $T_0$ has a 'hole' in its boundary.

Now $U\cap V\simeq S^1$ by deformation retraction. By Seifert-van Kampen $i_\alpha\colon U\cap V\to U$ and $i_\beta\colon U\cap V\to V$ induce $(i_\alpha)_\ast\colon \pi_1(U\cap V)\to \pi_1(U)$ and $(i_\beta)_\ast\colon \pi_1(U\cap V)\to \pi_1(V)$, respectively. So a generator in the intersection gets mapped in $U$ to one time the generators of $\pi_1(U)$ by deformation retraction. A generator in the intersection gets mapped in $V$ to two times a generator of $\pi_1(V)$ (just imagine the corresponding loops).

The normal subgroup $N$ in Seifert-van Kampen is generated by $(i_\alpha)_\ast(\gamma)(i_\beta)_\ast^{-1}(\gamma)$ $\forall \gamma\in \pi_1(U\cap V)$.

This gives us $\pi_1(\pi _1(T_0\cup _{S^1}M)=\pi_1(U)\ast\pi_1(V)/N=\langle a,b, d|adb^2\rangle$, because $c\mapsto ad$ and $c\mapsto b^2$ for $a,d$ generators in $U$, $b$ generator in $V$ and $c$ generator in $U\cap V$. If you don't know this notation look at Presentation of a group.

But $adb^2=1\Leftrightarrow ad=b^{-2}\Leftrightarrow b^2=a^{-1}d^{-1}$. Now one can guess which group this is. To get an intuition just write down words of the free product and replace letters by the relation above.