CW structures of identification spaces, fundamental groups and universal covers I am working on this exercise concerning the relation between some identification spaces and their CW structure and I am a bit confused.
First, the question is the following: For each identification space below, find a homeomorphic CW structure. Then for each CW complex, draw its 1-skeleton, give a presentation of the fundamental group, and find its universal cover.


Label as $a$ the edge with one arrow and as $b$ the edge with two arrows. 


*

*I thought the identification spaces as above were already the CW structure. If not, how can I guess this structure from the images?

*Now without having the CW structures, can I guess the fundamental group directly from the identification spces? I would say:
$$\pi_1(X)=<a,b|aba^{-1}b^{-1}>=\mathbb{Z}^2$$
$$\pi_1(Y)=<a,b|aa^{-1}b^2>=<a,b|b^2>=\mathbb{Z}\ast\mathbb{Z}_2$$
$$\pi_1(Z)=<a,b|aa^{-1}bb^{-1}>=<a,b|>=\mathbb{Z}$$
Is this correct?


*As for the universal cover, how can I get it just by looking at the identification spaces?


Remark: Maybe it would be easier if I recognized the spaces in question? I know $X$ is the torus, so I know its CW structure, its fundamental group and its universal cover. But I don't know what the other spaces are.
Many thanks!
 A: NO...your fundamental group computations of $Y$ and $Z$ are absolutely wrong (what you do in 2). The first calculation was right. 
The theorem which you are trying to use in order to get fundamental group is only applicable if $a$ and $b$ are loops. But in $Y$, $a$ is not at all a loop and in $Z$, neither $a$ nor $b$ are loops.
You can think this in a different way. $X$ is space torus (this is well known).
You can think of $Y$ as two triangles glued at a common edge, i.e. as in the picture 
Now the upper triangle is a copy of mobius strip, whereas the lower triangle is a copy of a cone (or $D^2$); they are identified in their common boundary circle. In other words, we are attaching a copy of the disc to the boundary of a mobius strip by a homeomorphism, so the resulting space is real projective space.
In similar fashion you can see that $Z$ is basically two copies of $D^2$ identified at their boundary by a homeomorphism, so the resulting space is a sphere. 
And now you can do the rest I guess.
