Simple classification-proof of all continuous homomorphism from the reals (under +) to the unit circle group of complex numbers (under *) For every real number $r$, the function
$f(x)=e^{irx}$
is a continuous homomorphism.
If $g$ is any continuous homomorphism, show that there is a unique real number $p$ such that for all real numbers $x$,
$g(x) = e^{ipx}$
I am looking for a simple/direct proof, without recourse to Lie Groups, Hausdorff Spaces, etc.
 A: Let $p:\mathbb{R}\rightarrow S^1$ be the usual covering map, $p(t) = e^{it}$.
Claim:  For every continuous homomorphism $f:\mathbb{R}\rightarrow S^1$, there is a continuous homomorphism $\tilde{f}:\mathbb{R}\rightarrow \mathbb{R}$ for which $f = p\circ \tilde{f}$.
Proof:  Let $f:\mathbb{R}\rightarrow S^1$ be any continuous group homomorphism.  By the path lifting property of covering spaces, there is a unique lift $\tilde{f}:\mathbb{R}\rightarrow\mathbb{R}$ with $\tilde{f}(0) = 0$ and $p\circ \tilde{f} = f$.
Further, $\tilde{f}$ is a homomorphism.  To see this, note the map $\mathbb{R}^2 \ni (x,y)\rightarrow f(x+y)$ has two lifts $\tilde{f}(x+y)$ and $\tilde{f}(x)\tilde{f}(y)$.  Since both of these lifts map $0$ to $0$, they must agree everywhere.
$\square$
To summarize, for each continuous homomorphsim $f:\mathbb{R}\rightarrow S^1$, we can factor $f$ as $p\circ \tilde{f}$ where $p$ is the standard homomorphism and $\tilde{f}:\mathbb{R}\rightarrow \mathbb{R}$ is a continuous homomorphism of $\mathbb{R}$.
Thus, we have reduced the classification problem to classifying continuous homomorphisms from $\mathbb{R}\rightarrow \mathbb{R}$.
Claim:  Every continuous homomorphism $g:\mathbb{R}\rightarrow \mathbb{R}$ is of the form $g(t) = rt$ for some $r\in \mathbb{R}$.
Proof:  Set $r = g(1)$.  Since $g$ is a homomorphism, $g(n) = rn$ for each $n\in \mathbb{Z}$.  Further, for the rational number $\frac{1}{n}$, we have $r = g(1) = g\left(\underbrace{\frac{1}{n} + ... + \frac{1}{n}}_{n \text{times}}\right) = \underbrace{g\left(\frac{1}{n}\right) + ... + g\left(\frac{1}{n}\right)}_{n \text{times}} = n g\left(\frac{1}{n}\right)$, so $g\left(\frac{1}{n}\right) = r\frac{1}{n}$.
Thus, $g(t) = rt$ for any rational $t$.  Finally, by continuity of $g$, $g(t) = rt$ for all real numbers $t$.
$\square$
Combining these two claims, we have $f(t) = e^{irt}$, as claimed.
A: Without recourse to the trigonometric function or calculus, or even $\pi$ or $e$, you can show the following:
Proposition: There is a 1:1 correspondence between $\mathbb{R}$ and the continuous homomorphisms $f:\mathbb{R}\rightarrow S^1$
Math Machinery:
Minimal Group Theory
Intermediate value theorem
Proposition: Let $g$ be a monotone increasing function defined on an interval of $\mathbb{R}$. If the range of $g$ is an interval then $g$ in continuous.
Proposition: For any $w_o$ on the unit circle of the complex numbers, there are two solutions to the quadratic equation $z^2 = w_o$.
Sketch of Proof:
Proposition: Let $g:[0, 1] \rightarrow S^1$ be a continuous 1:1 function satisfying $g(0) = 1$ and $g(1) = -1$. Then the imaginary part Im(g) can't take on both positive and negative values.
Proof: Apply Intermediate Value Theorem
The continuous homomorphisms $f$ are constructed by repetitively taking square roots of -1 and focusing on the real coordinate of $f(x)$ (the 'cosine' function). In essence, the circle is divided into $2^n$ segments (square roots) and the homomorphisms are forced to map $2^n$ real numbers onto the vertices. As $n$ goes to infinity, these (unique) functions do indeed define the continuous homomorphisms that are being sought.
A: Here is a more detailed proof of uniqueness.
Let $z(\theta)$ be a continuous homomorphism of the real line onto the unit circle. Because the unit circle is compact and the real line is not, $z(\theta)$ cannot be injective, so there is a positive real $\theta$ satisfying $z(\theta)=1$. Let $2\pi$ be the inf of all such $\theta$. If $\pi=0$, then by continuity $z(\theta)=1$ for all $\theta$. Otherwise,  $2\pi>0$. Then $2\pi$ is the period of $z(\theta)$ in the sense $z(\theta+2\pi)=z(\theta)$, $z(\pm\pi)=-1$, and $z(\theta)$ is a bijection between $(-\pi,\pi)$ and the punctured unit circle $z\not=-1$.
Theorem: Let $\pi$ be a positive real and let $z(\theta)$ and $z'(\theta)$ be $2\pi$-periodic continuous homomorphisms of the real line onto the unit circle. Then $z(\theta)=z'(\theta)$ for all $\theta$ or $z(\theta)=z'(-\theta)$ for all $\theta$.
Proof:
Given $z=x+iy$ in the punctured unit circle $z\not=-1$, define
$$\sqrt{z}= \frac{(x+1)+iy}{\sqrt{2+2x}}.$$
Then $(\sqrt{z})^2=z$. Let $-\pi<\theta<\pi$. Since $z(\theta)$ is a homomorphism, $z(\theta/2)=\pm\sqrt{z(\theta)}$. Now use the intermediate value theorem to conclude $z(\theta/2)=\sqrt{z(\theta)}$.
Since $z(\pm\pi)=z'(\pm\pi)=-1$, $z(\pi/2)= z'(\pm\pi/2)$.
If $z(\pi/2)=z'(\pi/2)$, then $z(\theta)=z'(\theta)$ at all dyadic rationals in $(-\pi,\pi)$, hence by continuity $z(\theta)=z'(\theta)$.
Similarly, if $z(\pi/2)=z'(-\pi/2)$, then $z(\theta)=z'(-\theta)$.
