Why $F: [0,1) \rightarrow S^1$ with $f(x) = (\cos2\pi x, \sin2\pi x)$ continuous function?

Question

I am teaching myself Topology using Munkres book and have problem with the example of continuous function $F\colon [0,1) \rightarrow S^1$. Assume that the topology $[0, 1)$ is order topology whose basis are $\{[0, a) \mid a < 1\}$ and $\{(a, b) \mid 0<a<b<1\}$ while the basis of $S^1$ is dictionary order.

A point $\{(x,y) \mid x \notin \{0,1\}, y > 0\}$ has the exact point $(x,-y)$ lies before it with nothing between. Therefore, the open set $\{((x_1,-y_1),(x_2,y_2)) \in S^1 \mid y_1, y_2 > 0\}$ can be written as $[(x_1,y_1),(x_2,-y_2)]$. What is the open set in $[0,1)$ that is the preimage of such set in $S^1$ for a function $f(x) = (\cos2\pi x, \sin2\pi x)$ to make it continuous.

I understand why $f^{-1}$ is not continuous but can't figure out why $f$ is continuous.

Answer 1

Since you're not allowed to use that for $f= f^1 \times f^2$, $f$ is continuous if and only if $f^1$ and $f^2$ are continuous:

An open subset of $S^1$ is going to be a disjoint union of open arcs. The preimage of the disjoint union of open arcs under $F$ is going to be a disjoint union of open intervals in $\mathbb R$. Finally, if $A \subseteq B \subseteq C$ and $A$ is open in $C$, then $A$ is open in $B$. Here, $C=\mathbb R$, $B = [0,1)$ and $A$ is the disjoint union of open intervals

Answer 2

You reasoning is quite correct: F is not continuous.

As you have noted, the set $[(x_1,y_1),(x_2,-y_2)]$ is open but its preimage is not.

In fact it is clopen - both open and closed. $[0,1)$ has no clopen subsets except itself and the empty set, so we can deduce that no map from $[0,1)$ onto $S^1$ with this topology can be continuous. (Look up the topological property "connectedness" if you haven't already met it. A continuous image of a connected space is always connected; $[0,1)$ is connected but $S^1$ is not.)