What is the space by identifying the antipodal points of a cylinder? Define the equivalence relation for a cylinder $S^1\times [-1,1]\subset \mathbb{R}^3$ as $(x,y,z) \sim (-x,-y,-z)$. We know that identifying antipodal points of $S^2$ or of the boundary of the disk $D^1$ gives projective plane $\mathbb{RP}^2$, I wonder if there a simple surface for identifying the antipodal points of a cylinder? Is it $\mathbb{RP}^2$?
 A: Let $S^1 = \{ z \in \mathbb{C} = \mathbb{R}^2 \mid \lvert z \rvert = 1 \}$. Define $q : S^1 \to S^1, q(z) = z^2$. Let $C(q)$ denote the mapping cylinder of $q$ which is the quotient space obtained from the disjoint union $S^1 \times [0,1] + S^1$ with respect to the equivalence relation generated by $(z,0) \sim q(z)$. Let $r : S^1 \times [0,1] + S^1 \to C(q)$ denote the quotient map. Note that $\rho = r \mid_{S^1 \times [0,1]}$ is surjective because $q$ is surjective.
Define $\phi : S^1 \times [-1,0] \to S^1 \times [0,1], \phi(z,t) = -(z,t)$. Then $\rho(z,0) = \rho(\phi(z,0))$ because $(z,0) \sim z^2 = (-z)^2 \sim (-z,0) = \phi(z,0)$. Next define
$$f : S^1 \times [-1,1] \to C(q), f(z,t) = 
\begin{cases}
\rho(z,t) & t \ge 0 \\
\rho(\phi(z,t)) & t \le 0
\end{cases}
$$
This is a continuous surjection (note that both parts of the defnition agree on $S^1 \times [0,1] \cap S^1 \times [-1,0] = S^1 \times \{ 0 \}$).
We have $f(z,t) = f(w,s)$ if and only if $(w,s) = \pm(z,t)$. The "if"-part is trivial. So let $f(z,t) = f(w,s)$. If $s = 0$, we must have $t = 0$ and $z^2 = w^2$, i.e. $w =  \pm z$. If $s \ne 0$, we must have $t = \epsilon s$ with $\epsilon = \pm 1$. But then clearly $w = \epsilon z$.
Hence $S^1 \times [-1,1]/\sim$ is homeomorphic to $C(q)$. The latter is homotopy equivalent to $S^1$.
Note that the space $C(q)$ is homeomorphic to $\mathbb{RP}^2 \setminus O$ where $O$ is an open disk in $\mathbb{RP}^2$.
A: Here is another answer. Let $S = S^1 \times [-1,1] \cup D^2 \times \{ -1, 1 \}$. There is a homeomorphism $h : S \to S^2$ such that $h \circ a = a \circ h$, where $a$ denotes the antipodal map both on $S$ and $S^2$. Hence $S/ \sim$, where $\xi \sim a(\xi)$, is homeomorphic to $\mathbb{RP}^2$. Let $p : S \to R = S/\sim$ denote the quotient map and $U = \mathring{D} \times  \{ -1, 1 \}$. Then $O = p(U)$ is an open disk in $R$ and $p$ restricts to a quotient map $p ' : S^1 \times[-1,1] = S \setminus U \to R \setminus O$.
