Consider a space $X$ constructed by removing a disk $D$ from the torus $T$ and gluing a 2-cell via an attaching map to the boundary of this disk of degree 2. What is the fundamental group of the space $X$?
I've thought about van-Kampen's theorem but get a little confused right at the start. Let $U$ be a neighborhood of $T \setminus D$ in $X$ and let $V$ be a neighborhood of the attached 2-cell in $X$. This seems like the "usual" decomposition for van-Kampen's theorem (i.e. make $U, V$ each one of the separate "pieces").
However, I don't know what do to figure out what $U, V$, and $U \cap V$ are homotopy equivalent to. For instance, for $U \cap V$, if I retract "from the attached 2-cell" side, then it looks like I get two copies of the circle, since it's attached with a degree 2 map; but if I retract "from the torus" side, then it looks like I just get one circle.
Similarly, because the attaching map is not the identity, I don't know how $U$ and $V$ retract to a well-known space. If we let $c$ be the boundary disk of the attached 2-cell, I believe the attaching map sends $c \mapsto c^2$ i.e. the boundary of the disk cut out on the torus should be $c^2$ (not just $c$), and I am confused how to continue. Perhaps I should consider a different partition for van-Kampen? Is there a way to make it work with the $U, V$ I chose?
Edit after reading comments, with an attempted solution: Letting $U, V$ be as above, we have the homotopy equivalences mentioned in the comments (so $\pi_1(U \cap V) \mathbb{Z}, \pi_1(U) = \mathbb{Z} * \mathbb{Z}, \pi_1(V) = 0$). Therefore $\pi_1(X)$ is $\mathbb{Z} * \mathbb{Z}/ N$ for some almalgamation subgroup $N$. This is found by the inclusion of the generating loop of $\pi_1(U \cap V)$ into $\pi_1(U)$ (since $\pi_1(V)$ is trivial). Since the $U \cap V$ is a circle, which can be deformed to the boundary of the attached 2-cell which is then wrapped twice around the cut-out disk in the torus, and then this is homotopy equivalent to tracing out the boundary of the torus (i.e. in the standard unit-square representation), which is $(aba^{-1}b^{-1})^2 = [a, b]^2$. Thus, $$ \pi_1(X) = \langle a, b | [a, b]^2 \rangle. $$