The unit disc is the set of points in $R^{2}$ given by $D=$ $\left\{(x, y) \mid(x, y) \in R^{2}\right.$ and $\left.x^{2}+y^{2} \leqq 1\right\} .$ Its boundary in $R^{2}$ is the unit circle $S$. Let $A$ be the subset of the circular cylinder $S \times[0,1]$ given by $S \times\{1\}$. Prove that $S \times[0,1] / A$ is homeomorphic to the disc $D$. $(S \times[0,1] / A$ is the cone over $S$, see the next problem.)
Proof: Let $\phi: S \times[0,1] / A \to D$ be the map $\phi (\theta , r) = (\theta, 1-r)$ (from cylinderical coordinates (surface only) to polar coordinates). Now how do I go about proving that this is an homeomorphism (not the bijective part). I have trouble even figuring out what are the open sets in $S \times[0,1] / A $ are suppose to be.
This is from Bert Mendelson's Intro to Topology which is quite basic so I'm looking for an answer that is not as advanced. I feel as though there's suppose to be a basic solution given what the book has covered so far. Perhaps more basic than found here Is the $S^1$ cone $D^2$?.
Best Answer
I would not use polar coordinates, but would simply define $$\varphi : S \times [0,1] \to D, \phi(z,r) = (1-r) \cdot z .$$ Here "$\cdot$" denotes the scalar multiplication on $\mathbb R^2$. It is easy to see that $\varphi$ is surjective and that $\varphi(z,t) = \varphi(z',t')$ iff $(z,t) = (z',t')$ or $t = t' = 1$. Hence $\varphi$ identifies $A = S \times \{1\}$ to a point (in fact, this set is mapped to $0 \in D$). The "level set" $S \times \{r\}$, where $0 \le r < 1$, is mapped to circle with center $0$ and radius $1-r$. By construction $\varphi$ induces a continuous bijection $$\phi : S \times [0,1]/A \to D .$$
But $S \times [0,1]/A $ is compact (as a quotient space of the compact space $S \times [0,1]$) and $D$ is Hausdorff; therefore $\phi$ is a homeomorphism.