Fundamental group of two discs connected by 3 bands I am working on a problem that asks me to 

(a) Find the fundamental group of the below space (on the left), called $\mathcal{X}$.
  (b) Glue a disc to the boundary cirle of the space, and find the fundamental group of this new space, called $\mathcal{Y}$.

I wanted to check and see if my solution is heading down the right path.
My thoughts:

(a) It seems as though the space is homotopy-equivalent to the wedge sum of two Moebius strips, $\mathcal{M} \vee \mathcal{M}$, so $\pi_1(\mathcal{X}) \cong \pi_1(\mathcal{M} \vee \mathcal{M}) \cong \mathbb{Z}\ast \mathbb{Z}$
(b) Since $\mathbb{R}P^2 - B_2 \sim \mathcal{M}$, then attached a disc to the boundary circle of our space should result in $\mathbb{R}P^2 \vee \mathbb{R}P^2$, which has fundamental group $\pi_1(\mathbb{R}P^2\vee \mathbb{R}P^2) \cong \mathbb{Z}/2\mathbb{Z}\ast \mathbb{Z}/2\mathbb{Z}$...?
 A: (a) is fine, but not (b). I'm not sure how you got $\mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/2\mathbb{Z}$ out of that, but what you have to do is add a relator which corresponds to the boundary of the surface. The boundary can be expressed most clearly in your middle picture. Suppose that in the top Mobius band the fundamental group is generated by the closed curve $a$, and in the bottom Mobius band the generator is the closed curve $b$ (up to path homotopy). You must, of course make sure that $a$ and $b$ have the same base point, which you can take to be the midpoint of the dotted line in the middle picture.
With more detail we have
$$\pi_1(\mathcal{X}) \cong \mathbb{Z}\ast \mathbb{Z} \cong \langle a \rangle \ast \langle b \rangle 
$$
The key fact is that the curve which goes around the boundary is freely homotopic to $a^2 b^2$. So the group you want has presentation
$$\langle a,b \,\bigm|\, a^2 b^2 = \text{Id}\rangle
$$ 
I don't know how you got $\mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/2\mathbb{Z}$ out of this, but that's not a valid conclusion.
