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!
Best Answer
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.