I would like to show that the function $f(x) = (\textrm{cos}2 \pi x, \textrm{sin}2 \pi x)$ is a quotient map; I have already shown that it is surjective and continuous (the latter by invoking the universal property for functions into a product topology (I am considering $S^1$ as a subspace of $\mathbb{R} \times \mathbb{R}$)). However, I'm stuck trying to prove the final condition for $f$ to be a quotient map. Here's my work so far:
Let $U \subset S^1$. Suppose $f^{-1}[U]$ is open in $[0,1]$. Then we need to show that $U$ is open in $S^1$. Let $\bar{y} \in U$; either $\bar{y} = (1,0)$ or $\bar{y} \neq (1,0)$. If $\bar{y} = (1,0)$, then by using the openness of $f^{-1}[U]$, we can deduce that there exists $\epsilon, \epsilon' \in \mathbb{R}_{> 0}$ such that $[0,\epsilon) \cup (\epsilon',1] \subset f^{-1}[U]$ (since $f^{-1}[\{ (1,0) \}] = \{0, 1 \}$). Similarly, if $\bar{y} \neq (1,0)$, there exist $\delta, \delta'\in \mathbb{R}_{> 0}$ such that $(\delta,\delta') \subset f^{-1}[U]$. Of course these sets, $[0,\epsilon) \cup (\epsilon',1]$ and $(\delta,\delta')$ are open in $[0,1]$.
How do I (or can I) proceed from here to show that $U$ is open in $S^1$? On the other hand, I'd also love to hear of some more general methods for showing that the function is a quotient map (maybe certain theorems or something). Thanks in advance for the help!
Best Answer
Maybe here is a general method:
Note that $[0,1]$ is compact and $S^1$ is Hausdorff. So, actually, your function is a closed map. Therefore, your function is a quotient map.
Edit:
Let $f:X\to Y$ be a continuous function where $X$ is compact and $Y$ is Hausdorff. Then, $f$ is a closed map.
Proof:
Pick a closed set $U$ in $X$. Since $X$ is compact, $U$ is compact. Hence, $f(U)$ is compact. Since $Y$ is Hausdorff, $f(U)$ is closed.