Lie group homomorphism from $\mathbb{R}\rightarrow S^1$

This is from an answer I posted to one of my own questions:

Let $f: \mathbb R \to S^1$ be a continuous group homomorphism. Then $f$ are of the form $e^{i \lambda x}$ for $\lambda \in \mathbb{R}$. To see this note that $(e^{ix}, \mathbb{R})$ is a covering space of $S^1$. Then by the unique lifting property we get that for a continuous homomorphism $f: \mathbb{R} \to S^1$ there is a unique continuous homomorphism $\alpha : \mathbb{R} \to \mathbb{R}$ such that $f = g \circ \alpha$ where $g (x) = e^{ix}$ is the covering map. By (i) we get that $f$ has to be of the form $x \mapsto e^{i\lambda x}$.

Here (i) is the following: If $\alpha : \mathbb{R} \to \mathbb{R}$ is a continuous homomorphism then $\alpha$ is of the form $x \mapsto \lambda x$ for some $\lambda \in \mathbb{R}$. This follows from the fact that $\alpha$ is a linear map and one dimensional matrices are multiplication by scalars.

I found an elementary proof which expands on the intuition given by user38268 years ago.

I'm posting this new answer because the already accepted answer given by Rudy the Reindeer requires knowledge of the theory of covering spaces and liftings in order to be understood and I think it's useful to have access to elementary proofs which don't relie on another theory, just in case. In fact, I'm not capable myself of following Rudy's answer, since I don't know that much stuff about covering spaces and liftings. So I post this proof in order to help future people in my same situation who get to this thread for help, as it was my case.

This proof was motivated by the hint given by Brian C. Hall in his book Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, chapter 1, exercise 18. This exercise asks the very same question of this thread.

Let $Φ:(\mathbb{R},+)\to (S^1,\cdot)$ be a continuous group homomorphism. We shall prove that there exists $a\in\mathbb{R}$ such that $Φ(x)=e^{iax}$ for all real $x$. The hint given by Hall is the following: Since $Φ$ is continuous and $Φ(0)=1$ (because group homormophisms preserve the identity) there exists some $ε>0$ such that, if $|x|\leq ε$, then $Φ(x)$ belongs to the (open) right half of the unit circle.

Using the customary continuous determination of the complex argument $\operatorname{arg}:\mathbb{C}\setminus\mathbb{R}_{\leq 0}\to(-π,π)$ we define the continuous function $θ:[0,ε]\to (-\frac{π}{2},\frac{π}{2})$ as $θ(x)=\operatorname{arg}(Φ(x))$. Thus $$\tag{1}\label{ec:phi} Φ(x)=e^{iθ(x)} $$ for every $x\in[0,ε]$.

The whole trick of the proof is that it will suffice to know the value of $Φ$ in $[0,ε]$ to deduce its value in all $\mathbb{R}$. And for what has been already presented, working with $Φ|_{[0,ε]}$ amounts to work with $θ$, the latter having a more tractable codomain than the former. Furthermore, since $θ$ is continuous, it will be sufficient to know what $θ$ evaluates in a dense subset of $[0,ε]$. This will be calculated thanks to $Φ$ being a homomorphism, which will transmit $θ$ some sort of additivity:

Let's see that $θ$ is additive under closed sums in $[0,ε]$. Let $x,y\in [0,ε]$ be some numbers such that $x+y\in [0,ε]$. Then $$ e^{iθ(x+y)}=Φ(x+y)=Φ(x)Φ(y)=e^{iθ(x)}e^{iθ(y)}=e^{i[θ(x)+θ(y)]}. $$ Since the exponential function is injective up to the sum of an integer multiple of $2π$, $$ θ(x+y)=θ(x)+θ(y)+2kπ $$ for some $k\in\mathbb{Z}$. And due to the triangle inequality, $$ 2|k|π=|θ(x+y)-θ(x)-θ(y)|\leq \frac{3π}{2}, $$ from which $k=0$ is deduced. This proves that $θ(x+y)=θ(x)+θ(y)$ for every $0\leq x,y \leq ε$ satisfying also $0\leq x+y\leq ε$. A similar property follows via induction for $n$ numbers contained in $[0,ε]$ for which its total sum remains inside this interval.

We'll compute now the value of $θ(\frac{k}{n}ε)$, where $k,n\in\mathbb{N}$ and $k\leq n$, in terms of the value of $θ(ε)$. Using the additivity for $θ$ we just showed,

$$ θ(ε)=θ\Bigg(\sum_{j=1}^n \frac{ε}{n}\Bigg)=\sum_{j=1}^n θ\left(\frac{ε}{n}\right)=nθ\left(\frac{ε}{n}\right). $$

Hence $θ(\frac{ε}{n})=\frac{1}{n}θ(ε)$. Now, if $k\in\mathbb{N}$ is such that $k\leq n$ and using again this additivity,

$$ θ\left(k\frac{ε}{n}\right)=θ\left(\sum_{j=1}^k\frac{ε}{n}\right)=\sum_{j=1}^kθ\left(\frac{ε}{n}\right)=kθ\left(\frac{ε}{n}\right) $$

and so $$\tag{2}\label{ec:A} θ(\tfrac{k}{n}ε)=\tfrac{k}{n}θ(ε). $$

Define now $$ A=\{\tfrac{k}{n}: k,n\in\mathbb{N}, k\leq n\}. $$ It is not difficult to see that $\overline{A}=[0,1]$. Given $t\in [0,1]$ and $δ>0$, taking $n$ such that $\frac{1}{n}<δ$, we can always find $0\leq k\leq n$ satisfying $|\frac{k}{n}-t|<δ$. Since $t\in [0,1]\mapsto tε\in [0,ε]$ is a homeomorphism, this gives us $\overline{Aε}=[0,ε]$.

Provided that we now the value of $θ$ on $Aε$ in terms of $θ(ε)$, \eqref{ec:A}, continuity gives us its value on $\overline{Aε}=[0,ε]$. Indeed, if $tε\in[0,ε]$ and $\{a_nε\}\subset Aε$ is some sequence with $a_nε\to tε$, then

$$ θ(tε)=\lim_n θ(a_n ε) =\lim_n a_n θ(ε)=tθ(ε). $$

This proves that $θ(x)=x\frac{θ(ε)}{ε}$ for every $x\in[0,ε]$.

Finally, we show that the knowledge of the function $θ$ on $[0,ε]$ gives us the value of $Φ$ on $\mathbb{R}$. Let $x\in\mathbb{R}$. If $x$ is non-negative, then take $n$ big enough so $0\leq\frac{x}{n}\leq ε$. Expressing $x=\sum_{j=1}^n \frac{x}{n}$, using that $Φ$ is a homomorphism and \eqref{ec:phi} gives $Φ(x)=e^{i\frac{θ(ε)}{ε}x}$. If $x$ is negative, then, since group homomorphisms preserves symmetric elements,

$$ Φ(x)=Φ(-(-x))=Φ(-x)^{-1}=(e^{i\frac{θ(ε)}{ε}(-x)})^{-1}=e^{i\frac{θ(ε)}{ε}x}. $$

Here is a proof that does not use as much advanced material. A Lie group homomorphism is a continuous group homomorphism between any two Lie groups. Any continuous group homomorphism $f : \Bbb{R} \rightarrow S^1$ is of the form $t \mapsto e^{i\theta(t)}$ for some function $\theta(t)$. Now observe that because $f$ is a group homomorphism we get that for all rational numbers $q$,

$$f(q) = e^{i\theta(1)q}.$$

Because $\Bbb{Q}$ is dense in $\Bbb{R}$, it follows that for all $t \in \Bbb{R}$ we have $f(t) = e^{i\theta(1) t}$. Since $\theta(1)$ can be any real number we conclude that any continuous group homomorphism from $\Bbb{R}$ to $S^1$ is of the form $e^{iat}$ for $a$ some real number.