Let $X$ be the union of a torus with an interval that meets the torus as shown. Use Van Kampen to find a presentation for the group.

Question

I need to come up with some kind of cell structure here right? How can I do this?

Answer 1

Alternatively, using a cell structure: note that your space is obtained from the standard cell structure on the torus (i.e., attach two arcs to a single point to get a wedge of two circles, then attach a disc by going around the first circle, then the second, then the first backwards, then the second backwards) by wedging another circle to the $1$-skeleton. The $\pi_1$ is therefore a quotient of $\pi_1(\bigvee_{i=1}^3 S^1) \cong \langle a,b,c \rangle$ by the normal subgroup generated by a loop through which the $2$-cell is attached. Hence, $\pi_1 = \langle a,b,c \mid aba^{-1}b^{-1}\rangle \cong \mathbb{Z}^2 * \mathbb{Z}$.

Answer 2

Write $U=$ torus,suppose that you attach the interval at $A$ and $B$, take a segment in the torus which connects $A$ and $B$, a contractible open subset $V'$ of the torus which contains that segment and set $V$ to be the union of the interval and $V'$. $U,V,U\cap V=V'$ are open and are arc connected you can apply Van Kampen and obtain that the $\pi_1(X)$ is the free product of $Z^2$ and $Z$ since $\pi_1(U\cap V)=1$.