Prove that two quotients of a bounded cylinder are not homeomorphic 
Let $C$ be the cylinder $\{(x,y,z)\mid x^2+y^2\leq 1, |z|\leq 1\}$. Consider two relations $R$ and $R'$ on $C$ which identify a point such that $x^2+y^2=1$ with $(-x,-y,z)$ and $(-x,-y,-z)$ respectively. Show that the two quotients $C/R$ and $C/R'$ are not homeomorphic.

I tried the three classic homeomorphism invariants: connectedness, compactness (both trivially verified) and the fundamental group. It seems that the retraction of $C$ onto the two-dimensional closed disk provides a retraction of both quotients onto $\mathbb P^2(\mathbb R)$(the disk with the well-known boundary identification).
I tried then to remove a point and check the three invariants, but it seems that connectedness is always verified and compactness never (is this all true?); and we can't take much advantage by this remarks.
Also, I have some difficulties in finding the fundamental group of the quotients without a point. This would be interesting also as a stand-alone problem.
Is there perhaps a more simple invariant? And more in general, what homeomorphism invariants are generally useful in these situations?
Thank you in advance.
 A: Let's write $C = D^2 \times [-1,+1]$.
Before flipping through a card catalogue of invariants, it is often helpful to just try to visualize the spaces in question. Sometimes that visualization will assist you in figuring out which invariants to apply. I use the term "visualize" quite broadly. Really what I mean is simply try to analyze each space as much as you can, describe it as clearly as you can, using whatever descriptive tools you have.
Each of the spaces $C/R$ and $C/R'$ is a manifold with boundary. Furthermore, as you have noticed, each space has a natural projection to $\mathbb{P}^2$. Also, each of these projection maps is a locally trivial fiber bundle whose fiber is homeomorphic to $[-1,+1]$.
In the case of $C/R$, the fiber bundle structure is trivial. The space $C/R$ is homeomorphic to $\mathbb{P}^2 \times [-1,+1]$. Also, its manifold boundary $\partial(C/R)$ is the image of $D^2 \times \{-1,+1\}$. That image has two components $\mathbb{P}^2 \times \{-1\}$, $\mathbb{P}^2 \times \{1\}$, each homeomorphic to $\mathbb{P}^2$. 
In the case of $C / R'$, the fiber bundle structure is nontrivial. Its boundary is also the image of $D^2 \times \{-1,+1\}$. However, in this case that image is homeomorphic to $S^2$.
Another way to think of the boundaries is that the projection maps $p :C/R \to \mathbb{P}^2$ and $p' :C/R' \to \mathbb{P^2}$ each restrict to double covering maps $\partial p :\partial(C/R) \to \mathbb{P}^2$ and $\partial p' :\partial(C/R') \to \mathbb{P}^2$. The first covering map $\partial p$ is the disconnected double cover of $\mathbb{P}^2$, and the second $\partial p'$ is the orientation double cover.
Okay, enough visualization (for me, that's the fun part). Now for invariants. Any homeomorphism between manifolds with boundary restricts to a homeomorphism between the boundaries, by invariance of domain. Hence, all homeomorphism invariants of the boundary are also homeomorphism invariants of the manifold as a whole. Since the boundary of $C/R$ is disconnected whereas the boundary of $C/R'$ is connected, it follows that $C/R$ and $C/R'$ are not homeomorphic.
