Let $C:=\{ (x,y,z) : z^2=x^2+y^2\}$, $M=S^1 \times \mathbb{R}$, $A=S^1 \times \{ 0\}$, $q:M \rightarrow M/A$ the quotient map ($A$ is identified with a point). I want to see that $M/A \simeq C$.
In one direction things are clear. The function $\psi: M \rightarrow C$, $\psi(\theta, w) = (w \cos \theta, w \sin \theta, w)$ is continuous and constant on fibers, so it passes to the quotient to a continuous $\tilde\psi$.
The inverse of $\tilde\psi$ is $\phi: C \rightarrow M/A$,$\phi(x, y, z)=\begin{cases} A, (x, y, z) = (0, 0, 0) \\ \{ (arg(x,y), z) \} , \textrm{ otherwise}\end{cases}$.
Although it seems obvious, I don't know how to show that $\phi$ is continuous. It obviously returns open sets not containing the point $A$ to open sets, but is there a sleek argument why if $A \in U \subseteq M/A$ with $U$ open, then $\phi^{-1}(U)$ is open?
Best Answer
A direct way is to use a classical property of compact spaces: assume $X, Y$ are topological spaces where $X$ is compact, $y_0 \in Y$ and $G \subseteq X \times Y$ is an open set containing $X \times \{ y_0 \}$. Then there is an open neighborhood $V \subseteq Y$ of $y_0$ such that $X \times V \subseteq G$.
Proof: for each $x \in X$ pick open sets $U_x \subseteq X, V_x \subseteq Y$ such that $\left< x, y_0 \right> \in U_x \times V_x \subseteq G$. Then $\{ U_x : x \in X \}$ is an open cover of $X$, so we can find $x_1, \ldots, x_n \in X$ such that $X = U_{x_1} \cup \ldots \cup U_{x_n}$. Taking $V = V_{x_1} \cap \ldots \cap V_{x_n}$ we easily check that $y_0 \in V$ and $X \times V \subseteq G$, as required. $\blacksquare$
It follows that if $G \subseteq S^1 \times \mathbb{R}$ is an open set containing $S^1 \times \{ 0 \}$, then it contains $S^1 \times (-\varepsilon, \varepsilon)$ for some $\varepsilon > 0$. The rest is a matter of technical details.
A more elegant way would be as follows: first prove that if $f : X \to Y$ is any bijection between topological spaces, $X = F_1 \cup \ldots \cup F_n$ for some closed subsets $F_1, \ldots, F_n \subseteq X$ where each $f[F_i] \subseteq Y$ is closed and $f \restriction F_i : F_i \to f[F_i]$ is a homeomorphism, then $f$ itself is a homeomorphism. Now write $M/A = F_1 \cup F_2$ where
$$F_1 = S^1 \times [-1, 1] / A \qquad \qquad F_2 = S^1 \times \big( \mathbb{R} \setminus (-1, 1) \big) / A$$
are closed subsets of $M/A$ and consider your bijection $\overline{\psi} : M/A \to C$. Note that $\overline{\psi}$ is a homeomorphism between $F_1$ and its image because it is a continuous bijection, $F_1$ is compact (as a quotient of a compact space*) and $C$ is obviously Hausdorff. Also it is clearly a homeomorphism between $F_2$ and its image as $A$ is disjoint from $F_2$. Hence $\overline{\psi}$ is a homeomorphism.
(*) Actually there is a hidden fact to prove here: that the subspace topology induced from $M/A$ coincides with the quotient topology induced from $S^1 \times [-1, 1]$.