How to prove that $X_1 := \{(x, y, z) ∈ \mathbb{R}^3 : x^2 + y^2 = 1\} / (S^1 × \{0\} )$ is homeomorphic to the union $X_2$ of two tangent spheres minus two points?
What I know:
Let $C$ be cylinder $:=\{(x, y, z) ∈ \mathbb{R}^3 : x^2 + y^2 = 1\}$,
$ (S^1 \times \{0\})$ collapses the circle in the $xy$-plane to a point which leaves a double cone connected by a single point which is given by the equation $x^2+y^2=z^2$. So, firstly, we need to find an identification turning the cylinder into the infinite cone.
Could I please get a hint on how to do that?
Looking at the coordinate function for z, would it be the homeomorphism between 0 to (0,+infinity)
f_z: {0} $\to$ (0,+ infinity)
z $\to$ 1/z
Consider the two spheres $N$ and $S$, each with radius 1, $N$ with center $(0, 0, 1)$ and $S$ with center $(0, 0, -1)$. Now look at the following map from $C$ to $N \cup S$:
For $(x, y, z) \in C$ with $z > 0$, define $h(x, y, z)$ to be the intersection of the segment from $(x, y, z)$ to $(0, 0, 0)$ with $N$; for $z < 0$, define it to be the intersection of the segment with $S$. (In brief: consider $h$ as "the inverse of stereographic projection from the origin of $A \cup B$ to $C$").
For $(x, y, 0) \in C$, define $h(x, y, z) = (0, 0, 0)$.
I'm going to assume you can prove that $h$ is continuous, and that its image misses only the north pole of $N$ and the south pole of $S$. Let's say that $N$, without its north pole, is $N_0$ and similarly for $S_0$. (Note that $N_0 \cup S_0 = X_2$.) So $h: C \to S_0 \cup N_0$ is a surjection.
Now all that's left is to see that $h$ passes to the quotient. Well, $h$, restricted to the $z = 0$ circle of $C$, is the constant map to the origin. So we get a map $\bar{h}: X_1 \to S_0 \cup N_0$ defined by $$ \bar{h}(\pi(p)) = h(p) $$ where $\pi :C \to X_1$ is the projection onto the quotient.
$\bar{h}$ is the required homeomorphism.