Is there a continuous function $\theta: S^1 \to \mathbb{R}$ such that $e^{i \theta(x)} = x$ for each $x \in S^1$?

Question

I have stumbled accross this question while working out the details of why Example 1 of this document https://web.ma.utexas.edu/users/vandyke/notes/256a_notes/lecture21.pdf is not a sheaf. It seems to me that such a function cannot exist, for then it would be possible to define an angle function. But I couldn't find a rigorous proof of it.

Answer 1

For such a function to exist, the function must be one-to-one. That is $\theta(x)=\theta(y)$ only when $x=y.$

More generally, there is no one-to-one continuous function $\theta:S^1\to\mathbb R.$

The basic idea:

There are two directions to go from $1$ to $-1$ on the circle. $\theta(x)$ along each direction must pass through $\frac{\theta(1)+\theta(-1)}2.$ But that means their are at least two solutions to $\theta(x)=\frac{\theta(1)+\theta(-1)}2,$ so no $\theta$ can be one-to-one.

In more detail.

Assume a one-to-one $\theta$ exists.

Now, let $f_1,f_2:[0,\pi]\to S^1$ be defined as:

$$f_1(x)=e^{ix},f_2(x)=e^{-ix}$$

Then for $i=1,2,$ $f_i(0)=1, f_i(\pi)=-1.$

Then, for $i=1,2,$ define $g_i(x)=\theta(f_i(x)),$ which are continuous functions $[0,\pi]\to\mathbb R,$ with $g_i(0)=\theta(1),$ $g_i(\pi)=\theta(-1).$

Since $\theta$ is one-to-one, $\theta(1)\neq \theta(-1).$

In particular, by the intermediate value theorem, there is an $x_1\in(0,\pi)$ such that $g_1(x_1)=\frac{\theta(1)+\theta(-1)}2.$

Similarly, there is an $x_2\in(0,\pi)$ such that $g_2( x_2)=\frac{\theta(1)+\theta(-1)}2.$

So this gives us $$\theta(f_1(x_1))=g_1(x_1)=g_2(x_2)=\theta(f_2(x_2))$$

Since $\theta$ is one-to-one, then $f_1(x_1)=f_2(x_2).$ But this isn’t possible for $x_1,x_2\in(0,\pi),$ because then the imaginary part of $f_1(x_1)$ is positive, while for $f_2(x_2)$ it is negative.

Answer 2

If the composition $$ S^1 \stackrel{\theta}{\to} \mathbb{R} \stackrel{\exp(i\cdot\_\ )}{\longrightarrow} S^1 $$ were the identity map, then it would induce the identity map on $\pi_1$, but $\pi_1(\mathbb{R})$ is trivial whereas $\pi_1(S^1)$ is not.