Finding the inverse of a continuous bijection from $(0,1] \to S^1$.
I would like to find the inverse of a function $f$ with domain and codomain as above, given by
$$
f(x) = \bigl( \cos(2 \pi x), \sin(2 \pi x) \bigr)
$$
Using the normal process of finding the inverse of a parametric equation, I calculated that
$$
t = \frac{\arccos(x)}{2\pi},
\quad
t = \frac{\arcsin(y)}{2\pi}
$$
But for $(x,y)=(0,-1) \in S^1$ these equations give me $t = \frac{1}{4}$ and $t = \frac{-1}{4}$ which is a contradiction, but is also not the map that I want since $f( \frac{1}{4} ) = (0,1)$.
Is there some way to modify the equation I have found to make it a true inverse from $S^1 \to(0,1]$? I realize it cannot be a continuous inverse since $f$ is not a homeomorphism.
Best Answer
The standard definitions of the inverse sine and cosine functions have domain $[-1, 1]$ and produce angles in $[-\pi/2, \pi/2]$ for sine and $[0, \pi]$ for cosine. You have to mess around a bit, patching together shifted versions of these functions to build your inverse function. But it's probably easier to use the inverse tangent function, which has the added advantage of not requiring your circle to have unit radius. Let's call the inverse $g: S^1 \to (0, 1]$.
We go quadrant by quadrant, building the function piecewise. Then we have to check that the definitions agree where they overlap on the axes, so that we have a well-defined function. (It will also be continuous everywhere except at $(1,0) \in S^1$, but I digress.)
For convenience, when $x \neq 0$, let $a(x,y) = \tfrac{1}{2\pi}\arctan \tfrac{y}{x}$, whose range is $\bigl( -\frac14, \frac14 \bigr)$. Then, $$ g(x,y) = \begin{cases} \hfil a(x,y), &x>0, y>0 \\ \hfil \tfrac14, &x=0, y>0 \\ \hfil \tfrac12 + a(x,y), &x<0 \\ \hfil \tfrac34, &x=0, y<0 \\ 1 + a(x,y), &x>0, y \leq 0 \\ \end{cases} $$ works. You need to check that the pieces fit together at $(0,1)$ and $(0,-1)$. Can you take it from here?