I have the map $f:[0, 2 \pi) \rightarrow S^1, f(t) = e^{it}$. I'm trying to show that the inverse map $g: S^1 \rightarrow [0, 2 \pi), \space \space \space g=f^{-1}$ is not continuous by finding an open set $V$ in $[0,2 \pi)$ such that $g^{-1} (V)$ in not open.
I realise there is a simpler solution than the one I'm about to present (involving noticing that $g^{-1} (V) = f(V) $ and working from that), but I'm interested to see if my thinking is at all correct.
First I go about explicitly finding $g(t)$. By going from $f(t) = e^{it}$ to $t = e^{ig(t)}$ and rearranging, I get $$ g(t) = -i\text{log}|t| +\text{Arg}(t)+2 \pi k \space \space \space (k \in \Bbb Z).$$ where I define the argument to be such that $ 0 \leq \text{Arg}(t) < 2 \pi$. Then for a small $ \epsilon > 0$, let $V=[0, \epsilon)$, open in $[0, 2 \pi)$.
Note that in our domain $S^1$, $t$ is such that $|t|=1$. Therefore we get that $$ g(t) = \text{Arg}(t)+2 \pi k.$$ From now on I will set $k=0$ because I don't think I need any extra multiples of $2 \pi$ in the solution. Let's say that for some $\gamma , \delta$ related to $\epsilon$, we have that under $g$: $$t=1 \mapsto 0 $$ $$t= \delta + \gamma i \mapsto \epsilon. $$
Note that $|t|=1$ must hold. Then we have that
$$ g^{-1} (V) = \{\delta' + \gamma' i \in \Bbb C : \delta' , \gamma' \in \Bbb R, \space \gamma' \in [0, \gamma), \space \delta' \in (\delta, 1] \} $$
is not open in $S^1$ due to $[0, \gamma), (\delta, 1]$ not being open in $\Bbb R$ (can I say this?). $\square$
I feel there are some errors in my proof attempt (most likely in finding the inverse, and also defining the pre-image), but I was wondering if this could be done along the same lines I have taken/using the same idea (i.e. explicitly defining what $g=f^{-1}$ is and working from there in roughly the same vein) if my working is incorrect? Many thanks for any help.
Edited: to include $\delta$ to ensure that $|t|=1$ holds; to try to fix the proof with regards to mercio's comments.