Let $\sim$ denote the equivalence relation on the cylinder $S^1 \times [-1, 1]$ defined by $(v,-1) \sim (v',-1)$ for all $v, v' \in S^1$, and $(v, 1) \sim (v', 1)$ for all $v, v' \in S^1$. Here $S^1$ is the unit circle $\{(x, y) \in \Bbb{R}^2 | x^2 + y^2 = 1\}$. Prove that the quotient space $S^1 \times [-1, 1]/\sim$ is homeomorphic to the unit sphere
$S^2 = \{(x, y, z) \in \Bbb{R}^3| x^2+y^2+z^2 = 1\}$. State carefully results about homeomorphisms and the quotient topology that you use in your proof.
Let $X = S^1 \times [-1,1]$ and $Y = S^2$, where $S^2$ is the sphere centered at the origin with radius $1$. Define $f: X \rightarrow Y$ by homeomorphically mapping each circle $S^1 \times \{t\}$ for $t \in (-1,1)$ to the correspoding circle in $S^2$ at $z = t$. Map $S^1 \times \{1\}$ and $S^1 \times \{-1\}$ to the top and bottom points on the sphere respectively.
Obviously, $f$ is both surjective and continuous. Since both $S^1$ and $[-1,1]$ are compact, so is there product $S^1 \times [-1,1]$. Since $S^2$ is a subspace of a Hausdorff space, it must also be Hausdorff. This tells us that $f$ must be closed, implying that $f$ is a quotient map.
Since $f$ is a quotient map, the map $\bar{f}: X/\sim \rightarrow Y$ (induced by $f$) is a homeomorphism as it is an injective quotient map.
I was wondering if there was a more explicit way of showing that $f$ is surjective and continuous. I just said it's "obvious" since it is obvious when you think about it, but is there a more explicit way of proving that (without getting too messy)?
Best Answer
Explicit means tedious in this case.
The quotient map $\pi:S^1 \times [-1,1] \to X$ is given by $\pi((s,c)) = \begin{cases} \{s\} \times \{c\}, & c \in (-1,1) \\S^1 \times \{c\}, & c \in \{\pm1 \} \end{cases}$.
Let $N=S^1 \times \{1\}$, $S=S^1 \times \{-1\}$.
A set $V$ is open in $X$ iff $\pi^{-1}(V)$ is open in $S^1 \times [-1,1]$ (with the product topology).
Suppose $p \in X$ and $p=\pi((s,c))$. If $p \notin \{N,S\}$, then the sets $\pi(B((s,c), \epsilon))$ form a neighbourhood base of $p$ for sufficiently small $\epsilon$ (so $\{N,S\} \cap \pi(B((s,c), \epsilon)) = \emptyset$). If $p=N$, then the sets $\pi(S^1 \times (1-\epsilon,1])$ form a neighbourhood base of $p$. Similarly for $S$.
Define $f:S^1 \times [-1,1] \to S^2$ by $f((a,b), t) = (\sqrt{1-t^2} a, \sqrt{1-t^2} b, t)$.
It is straightforward to see that $f$ is surjective. We see that $f((a,b),c) = (0,0,c)$ for $c \in \{-1,1\}$, and the restriction $f \mid_{S^1 \times (-1,1)} $ is injective. We note that if $\pi((s_1,c_1)) = \pi((s_2,c_2))$, then $f(s_1,c_1) = f(s_2,c_2)$, so we can unambiguously define $\tilde{f}:X \to S^2$ by $\tilde{f}(\pi((s,c))) = f(s,c)$. Note that $\tilde{f}$ is bijective. Let $V$ be open in $S^2$. Then since $\pi^{-1}(\tilde{f}^{-1}(V)) = f^{-1}(V)$, we see that $\tilde{f}^{-1}(V)$ is open, hence $\tilde{f}$ is continuous.
Let $n=(0,0,1), s=(0,0,-1)$ and define $\phi:S^2 \setminus \{ n,p \} \to S^1 \times (-1,1)$ by $\phi((x,y,z)) = (({x \over \sqrt{x^2+y^2}}, {y \over \sqrt{x^2+y^2}}), z)$. Note that $\phi$ is continuous and surjective, and since $f(\phi((x,y,z)) = (x,y,z)$ (since $x^2+y^2+z^2 = 1$), we see that $\phi$ is a bijection.
Now define $\tilde{\phi}: S^2 \to X$ by $\tilde{\phi}(r) = \begin{cases} \pi(\phi(r)), & r \in S^2 \setminus \{ n,p \} \\ N, & r=n \\ S, & r=s \end{cases}$. First suppose $V$ is open in $X$ and $V \cap \{N,S\} = \emptyset$, then $\tilde{\phi}^{-1}(V) = \phi^{-1}(\pi^{-1}(V))$, which is open. This shows that $\tilde{\phi}$ is continuous on $X \setminus \{N,S\}$.
Now suppose $V$ is open with $N \in V$. Hence there is some $\epsilon>0$ such that $N \in \pi(S^1 \times (1-\epsilon,1]) \subset V$. We have $\tilde{\phi}^{-1}(\pi(S^1 \times (1-\epsilon,1])) = \tilde{\phi}^{-1}(\pi(S^1 \times (1-\epsilon,1))) \cup \{n\} = \{ (x,y,z) \in S^2 | z > 1 -\epsilon \}$, which is open (in $S^2$). Similarly for $S$. Hence $\tilde{\phi}$ is continuous.
Finally, we note that $\tilde{f} \circ \tilde{\phi}$ is the identity on $S^2$, hence $\tilde{f}$, $\tilde{\phi}$ are inverses.