Homology of Identifications of 2-torus While working through old algebraic topology quals, I found these questions and wasn't sure if I had the right idea.
Let $X$ be a closed genus–2 surface. Let $\alpha$ be a non-separating circle in $X$ and $\beta$ be a separating circle that is disjoint from $\alpha$ and not null–homotopic. Let $Y$ be the space obtained from $X$ by identifying $\alpha$ and $\beta$ by some choice of homeomorphism. Let $Z$ be the space obtained from gluing together two copies of $X$ along a homeomorphism of $\alpha$ and $\beta$. Compute the fundamental groups and integral homology of these spaces.
I've tried drawing the CW-complex structure of these spaces:
CW-Complex for X,Y,Z
My questions are:

*

*Is this the right CW-complex structure?

*How can I read the fundamental group/homology off of these? I'm confused in $Y$ if there is still only one 2-cell with the way $\alpha$ is drawn, and how I would read the fundamental group in either case with the way $\alpha$ now cuts the CW complex.

Edit: Following Kevin's answer, I have two possible CW complexes? I'm unsure whether the separating circle in the larger copy of X gets identified with the non-separating circle $\alpha$.
Option 1
Option 2
Edit 2: Option 1 it is!
 A: The images of $X$ and $Y$ seem fine to me.
For your second question(s), $Y$ certainly continas two $2$-cells, where $e_1^2$ is glued via the attaching map $zyz^{-1}y^{-1}\alpha^{-1}$ and $e_2^2$ is glued via $\alpha x\alpha x^{-1}\alpha^{-1}$. So you can obtain a group presentation $$\pi_1(Y)=\langle x,y,z,\alpha\mid zyz^{-1}y^{-1}\alpha^{-1}, \alpha x\alpha x^{-1}\alpha^{-1}\rangle$$
This result follows from Proposition 1.26 of Hatcher.
The image of $Z$ is confusing for some reason:

*

*The question asks you to glue $\alpha,\beta$ of two copies of $X$ but not other identified edges, so you should really change the letters of the labelling  scheme $yz...$ of one copy. Otherwise it's like $z$'s of the two copies are also identified...

Once these corrections are done, the CW structure is again clear, and you can compute the fundamental group by the same technique.
Hints for computing homology groups:

*

*The cell structure yields a cellular chain complex $$0\hookrightarrow\Bbb{Z}^m\overset{\partial}{\to}\Bbb{Z}^n\twoheadrightarrow0$$ where $m,n\in\Bbb{N}$.

*$\partial(e_i^2)$ is a linear combination of $e_j^1$'s, which form the 'boundary'. This is exactly what the attaching map tells you.

You may try solving the problem yourself before telling me to add the full answer.
