Is any homomorphism $\Bbb R\to\mathrm U(1)$ an exponential? Is any continuous homomorphism $f:\Bbb R\to\mathrm U(1)$ necessarily of the form $f(x)=\mathrm e^{\mathrm ikx}$, for $k\in\Bbb R$? It certainly seems plausible, but I can't nail down a proof. I don't know how to guarantee that an exponential appears without using some analytic argument, which I don't want to do. (So maybe it is false?)
 A: The result is true. More generally, any continuous homomorphism $f \colon \mathbb{R} \rightarrow G$ between $\mathbb{R}$ and a Lie group $G$ is of the form $f(t) = e^{itX}$ for $X \in \mathfrak{g}$ (and in particular, automatically smooth).
To see this in your case, note that any continuous map $f \colon \mathbb{R} \rightarrow U(1)$ can be written as $f(x) = e^{i\varphi(x)}$ for a continuous map $\varphi \colon \mathbb{R} \rightarrow \mathbb{R}$ with $\varphi(0) = 0$ (this follows for example from the theory of covering spaces). Now, the identity $f(x + y) = f(x)f(y)$ implies that $\varphi(x + y) = \varphi(x) + \varphi(y) + 2\pi c(x,y)$ for some $c \colon \mathbb{R}^2 \rightarrow \mathbb{Z}$. Since $c$ is evidently continuous and $c(0,0) = 0$, we must have $c(x,y) \equiv 0$ and so $\varphi(x + y) = \varphi(x) + \varphi(y)$. Now use the well-known result that a continuous additive map $\varphi \colon \mathbb{R} \rightarrow \mathbb{R}$ must be a linear map to deduce that $\varphi(x) = kx$ for some $k \in \mathbb{R}$.
A: Yes. The continuity shows we get a connected subset of $U(1)$, and since any neighborhood of the identity generates the whole group, if the homomorphism is trivial, it's $e^{i0x}$ and if it's not, then the map is onto, and we know the kernel is isomorphic to $\Bbb Z$ since all subgroups of $\Bbb R$ are either discrete or dense, and the latter case plus continuity would give a trivial homomorphism. But then let $T>0$ be the smallest positive number for which the map is zero, then we can see that it is periodic with period $T$, and so $e^{2\pi ix/T}\cdot f(x)$ is constant since homogeneity of the fibers over any point we know that it traverses the circle at constant speed.
