Let $T^1 =\mathbb{S}^1 \times \mathbb{S}^1$ be the torus, $D^2$ the unit disk and $X = T^1 \cup D^2$. Find the fundamental group of $X$.
Try: I believe there should be a simpler argument applying Seifert Van Kampen's Theorem, but I was unable to find $U, V$ open such that $X = U \cup V$ in the first place, so I tried looking at the space $X$ as the adjunction space obtained by gluing the disk to the torus along the unit circle $\{ x^2 + y^2 = 0, z = 0 \}$. Define a map $f : \mathbb{S}^1 \to T^1$ by $$e^{2\pi i t} \mapsto (e^{2 \pi i t}, 1)$$ Then we get a commutative diagram ($i$ and $\pi_i, i =1,2$ denote the inclusion and canonical maps to the quotient space)
$\require{AMScd}$ \begin{CD} \mathbb{S}^1 @>{i}>> D^2\\ @V f\ VV @V \pi_1\ VV\\ \mathbb{S}^1 \times \mathbb{S}^1 @>{\pi_2}>> (\mathbb{S}^1 \times \mathbb{S}^2) \cup_f D^2 \end{CD} which induces a commutative diagram on the fundamental groups
\begin{CD} \pi_1(\mathbb{S}^1) @>{i_*}>> \pi_1(D^2)\\ @V f_*\ VV @V \pi_{1*}\ VV\\ \pi_1(\mathbb{S}^1 \times \mathbb{S}^1) @>{\pi_{2*}}>> \pi_1((\mathbb{S}^1 \times \mathbb{S}^2) \cup_f D^2) \end{CD} Now, since $\pi_{2*} \circ f_* = \pi_{1*} \circ i_*$, take $\gamma(t) = (\cos t, \sin t)$ on $\mathbb{S}^1$, then $$(\pi_{1*} \circ i_*)([\gamma]) = \pi_{1*}([i \circ \gamma]) = \pi_{1*}([\gamma]) = [\epsilon]$$ since $D^2$ is simply connected, where $\epsilon$ denotes the trivial loop. On the other hand $$(\pi_{2*} \circ f_*)([\gamma]) =\pi_{2*}([f \circ \gamma]) = \pi_{2*}([\sigma]) = [\epsilon]$$ and by the way we defined $f$, $[\sigma] = [f \circ \gamma]$ is a generator of the torus. So by attaching the disk to the torus we have "killed off" one generator of the torus, and thus the fundamental group of $X$ is $$\pi_1(X) = \{0 \}\times \mathbb{Z} \cong \mathbb{Z}$$
Is this proof okay? Thank you in advance.