Does there exist a continuous function $g:S^1 \to S^1$ such that $(g(z))^2=z , \forall z \in S^1$?

Question

Let $S^1:=\{z \in \mathbb C:|z|=1\}$ ; does there exist a continuous function $g:S^1 \to S^1$ such that $(g(z))^2=z , \forall z \in S^1$ ?

Answer 1

There is no such function.

Here's an elementary argument. Parametrize $S^1$ as $e^{it}$ where $0 \le t < 2\pi$. Then $$ g(e^{it}) = e^{it/2} \cdot s(t) $$ where $s(t) \in \{ \pm 1 \}$ for every $t$. If $s$ is not constant, then $g$ will be discontinuous at some $0 < t < 2\pi$. If $s$ is constant, things go wrong at $t = 0$/$t=2\pi$ (which both correspond to $z=1$).

Answer 2

If you are familiar with degrees, one can argue as follows. The identity selfmap $z\mapsto z$ of the circle has degree $1$. The squaring map $z\mapsto z^2$ has degree $2$. The function $(g(z))^2$ is the composition of $g$ and $z\mapsto z^2$. Under composition the degree multiplies, i.e., the degree of $g$ would have to be $\frac{1}{2}$ in order for the composed map to have degree $1$. But since the degree is always an integer, there couldn't be such a map $g$.

Answer 3

No.

(i). For $t\in (-\pi, \pi)$ let $h(e^{i t})=e^{i t/2},$ If $g(e^{ i t})^2=e^{i t}$ then $g(e^{i t})=\pm h(e^{i t}). $ Note that the point $-1$ is excluded from the domain of $h.$

Let $g(z)^2=z,$ with $g(e^{i t})$ continuous for $t\in (-\pi,\pi).$

(ii). Suppose that $g(e^{i t_0})=h(e^{i t_0})$ for some $t_0\in (-\pi,\pi).$ The continuity of $g(z)$ at $z=e^{i t_0}$ implies that for some $r,s$ \in $(0,\pi-t_0)$ we have $$\forall t\in (t_0-s,t_0+r)\;(\;g(e^{i t})=h(e^{i t})\;).$$ (iii). Consider the union $U$ of the set of all open intervals $J\subset (-\pi,\pi)$ such that $t_0\in J$ and such that $\forall t\in J\;(\;g(e^{i t})=h(e^{i t})\;).$ Let $t'_0=\sup U$ and $t''_0=\inf U$... We have $ U \ne \phi...$ If $t'_0\ne \pi$ we can apply the argument of (ii) to find $t'_0+r\in (t'_0,\pi) $ such that $[t'_0,t'_0+r)\subset U, $ contradicting the def'n of $t'_0. $ So $t'_0=\pi. $

Similarly we show that $t''_0=-\pi. $

(iv). Hence $U=(-\pi,\pi). $ So $g(e^{i t})=h(e^{i t})=e^{i t/2}$ for all $t\in (-\pi,\pi).$ But this makes $g(z)$ discontinuous at $z=-1.$

(v). Similarly if $g(e^{i t_0})=-h(e^{i t_0})$ for some $t_0\in (-\pi,\pi)$ we also find that $g(z)$ is discontinuous at $z=-1.$