kth root of an open set in the circle toplogical group

Question

My intuition tells me that in the topological group of the circle, if I take an open set U, then its kth root (where k is some natural number) in the circle is also an open set. In order to show it I was thinking about a function that takes x in the circle and sends it to x^k, and showing that it is open. Then its inverse, although not a function, should be the answer. But that also gave me difficulties. It looks rather natural, but I can't seem to see it clearly. Can someone tell me how to approach it? Your help much appreciated.

Answer 1

Actually you don't need that $z\mapsto z^k$ is open. You simply need that it's continuous.

More precisely, consider the map $f_k : z\mapsto z^k$, which is obviously a continuous function on $S^1$. Let $U\subset S^1$ be open, then you can consider the set $$U^{1/k} := \{x\in S^1 : x^k\in U\}$$ Then $U^{1/k}$ by definition is precisely $f_k^{-1}(U)$ and is open since $U$ is open and $f$ is continuous.

This $U^{1/k}$ is really the only sensible way to define the $k$th root of a set. Any other definition will require arbitrary choices, corresponding to a "branch cut", which means that you have to exclude a point from the circle (there is no continuous $k$th root function on the circle).