# Fundamental group of torus with 2-cell attached

Let $( \mathbb{S}^{1} × \mathbb{S}^{1} )∪e^2$ be the adjunction space of the torus and a 2-cell, where the attachment is given by the attaching map $f:\mathbb{S}^{1} \to ( \mathbb{S}^{1} × \mathbb{S}^{1} )$ such that$f(ζ)= (ζ^{2},ζ^{3})$. Compute the fundamental group of this space.

By a consequence of Seifert-Van Kampen theorem I know that the fundamental group of this space is the fundamental group of the quotient space of the torus (which I know is $\mathbb{Z}$x$\mathbb{Z}$), divided by its kernel (which is the normal subgroup generated by $\gamma$ where $\gamma=f_{*} (\alpha)$ and $\alpha$ is the generator of the infinitely cyclic group of the fundamental group of $\mathbb{S}^{1}$).

I'm having trouble with determining what is this last kernel, or at least how to simplify this last expression. Also, I don't know what am I supposed to obtain as a result, so any help would be appreciated.

Thank you so much!

Use Seifert van Kampen. You get by your chosen division $\pi_1(S^1\times S^1)=\Bbb{Z}\times\Bbb{Z}$, $\pi_1(e^2)=0$ and the fundamental group of the intersection $\pi_1(U_1\cap U_2)=\pi_1(S^1)=\Bbb{Z}$ where $U_1=S^1\times S^1+\mbox{ a little bit}$ and $U_2=e^2+\mbox{ a little bit}$. You should recognize, that $f(z)=(z^2,z^3)$ gives you a torus knot, which is homotopy equivalent to $S^1$. The knot goes $2$ times latitudinal around the torus and $3$ times longitudinal around the torus through the hole of the torus.
Now Seifert van Kampen gives you the following: $N$ is a normal subgroup of $\pi_1(S^1\times S^1)\ast \pi_1(e^2)=\Bbb{Z}\times\Bbb{Z}$ and $$\pi_1((S^1\times S^1)\cup e^2)=(\pi_1(S^1\times S^1)\ast \pi_1(e^2))/N=\Bbb{Z}\times\Bbb{Z}/N,$$ where $N$ is generated by all $\iota_{12}(\omega)\iota_{21}(\omega)$ with $\omega\in \pi_1(U_1\cap U_2)=\Bbb{Z}.$ Now $\iota_{12}$ is induced by $\iota_1\colon U_1\cap U_2\hookrightarrow U_1$ and $\iota_{21}$ is induced by $\iota_2\colon U_1\cap U_2\hookrightarrow U_2$.
We have to look how the generator of $\pi_1(U_1\cap U_2)$ is represented in $\pi_1(U_1)$ or $\pi_1(U_2)$ respectively. Because of $\pi_1(U_2)=\pi_1(e^2)=0$ only $\pi_1(U_1)=\pi_1(S^1\times S^1)$ is interesting. Let us take $\alpha$ to be the generator of the intersection. We have $\beta$ and $\gamma$ as generators of $\pi_1(U_1)=\Bbb{Z}\times \Bbb{Z}$. Now $\alpha$ is under $\iota_1$ in $U_1$ just homotopic to $\beta^2\gamma^3$.
You get $\pi_1(S^1\times S^1\cup e^2)=\langle \beta, \gamma|\beta^2\gamma^{3}=1\rangle$.