I am trying to calculate the fundamental group of the torus with a line segment that touches the torus only at its endpoints. Let $X$ denote this space. For details and images, see here.
One sees that $X \simeq T \vee S^1$, where $T$ is the torus, and I define $U_1,U_2$ exactly as in the article. It is then easy to see that $U_1,U_2$ are open, path-connected and that $U_1 \cup U_2 = X$. Then, since $U_1 \simeq S^1$ and $U_2 \simeq T$, we get $\pi_1(U_1)\cong \mathbb{Z} \cong \langle \alpha \mid \emptyset \rangle$ and $\pi_1(U_2) \cong \mathbb{Z} \times \mathbb{Z}$. Now, the author writes that $\mathbb{Z} \times \mathbb{Z} \cong \langle \beta,\gamma \mid \emptyset \rangle$. Is this a typo? Shouldn't it be $\mathbb{Z} \times \mathbb{Z} \cong \langle \beta,\gamma \mid \beta\gamma\beta^{-1}\gamma^{-1} \rangle$?
Then, the author goes on to show that $\pi_1(X) \cong \langle \alpha,\beta,\gamma \mid \emptyset \rangle$, since $\pi_1(U_1 \cap U_2) \cong \{1\}$. Is this the correct answer to this question? Assuming the author made a mistake, I myself get the following: We have inclusion maps
\begin{align}
&\iota_1 : \{1\} \to \mathbb{Z}, &&1 \mapsto \alpha\\
&\iota_2 : \{1\} \to \mathbb{Z}\times\mathbb{Z}, &&1 \mapsto \beta,
\end{align}
where we chose $\beta$ w.l.o.g., since we could have chosen $\gamma$. Then, we get
$$
\pi_1(X) \cong \langle \alpha,\beta,\gamma\mid \beta\gamma\beta^{-1}\gamma^{-1}, \iota_1(1)=\iota_2(1) \rangle \cong \langle \beta,\gamma \mid \beta\gamma\beta^{-1}\gamma^{-1}\rangle \cong \mathbb{Z} \times \mathbb{Z},
$$
and I'm unsure whether or not this answer is correct, since we already have that $\pi_1(T) \cong \mathbb{Z}\times \mathbb{Z}$. Am I missing something?
Best Answer
$\mathbb{Z} \times \mathbb{Z}$ is indeed $\langle \beta,\gamma \mid \beta\gamma\beta^{-1}\gamma^{-1} \rangle$, you are right. Now, there are no group maps $\iota_1$ and $\iota_2$ as you described. $\{1\}$ is the zero group $0$, from which there is always only one map: any homomorphism has to take $1$ in the codomain to $1$ in the domain. Thus $\alpha$ and $\beta$ don't kill each other as in your, otherwise correct, computation.