Continuous Bijective Map not Bicontinuous The classic example of a continuous bijective function that does not have a continuous inverse is the function $g:[0,2\pi) \rightarrow S^1$ given by
\begin{equation}
g(t)=(cos(t),sin(t)),
\end{equation}
where $S^1$ is the unit circle. I've always intuitively accepted this, because as we traverse the positive x-axis on the unit circle, the inverse function "jumps" between $0$ and $2\pi$. 
But thinking about this in terms of pre-images of open sets confuses me. The characterization I am thinking of is: a function $g$ is continuous if and only if the pre-image of every open set in the codomain is an open set in the domain. 
Since we want to show the inverse is not continuous, I want to show that given some open set $U$ in $[0,2\pi)$, its image $g(U)$ is not open. But I can't seem to find some open set $U$ for which this holds. Is there something I'm missing?
Thanks for the help!
 A: Take $U=[0,\pi)$ for instance: it is open in $[0,2\pi)$ but its image by the inverse map is not open due to the image of $0$ having only half of a neighbourhood (see below).
What you might find confusing is that $U$ is indeed open in $[0,2\pi)$. But in $[0,2\pi)$, there is nothing to the left of $0$ that would be missing in $U$ preventing it from being open.

A: Any open set in $S^1$ that contains $(1,0)$ must contain a set of the form $V=\{ (\cos t, \sin t) | |t| < \delta  \}$ for some $\delta >0$.
Then $g^{-1}(V)$ must contain a set of the form $[0,\epsilon) \cup (2 \pi -\epsilon, 2 \pi)$ for some $\epsilon>0$.
Addendum:
Let $h=g^{-1}$ to simplify notation, and let $p=(1,0) \in S^1$ to avoid confusion with an open interval. Note that $h(p) = 0$. To show that $h$
is continuous at $p$, we need to show that for any open $V \subset [0,2 \pi)$ containing $0$, there is some open $U \subset S^1$ containing $p$ such that
$h(U) \subset V$.
Now pick $V=[0,1)$, which is open in $[0,2\pi)$. Now observe that for any open $U\subset S^1$ that contains $p$, $h(U)$ contains some set of the form $(2 \pi-\epsilon , 2 \pi)$ for some $\epsilon>0$, and hence there is no open $U$
such that $h(U) \subset V$ and so $h$ is not continuous at $p$.
Note that $h$ is continuous everywhere else.
