[proofplan]
A continuous representation $\rho: S^1 \to \mathbb{C}^\times$ has compact image, which forces $\rho(S^1) \subseteq S^1$: any value $|\rho(z)| \neq 1$ would generate an unbounded sequence $|\rho(z)^n| = |\rho(z)|^n$, contradicting compactness. Pulling back through the covering $\mathbb{R} \to S^1$, $x \mapsto e^{ix}$, we obtain a continuous homomorphism $\mathbb{R} \to S^1$, which by the classification theorem has the form $x \mapsto e^{icx}$. The periodicity constraint $\rho(e^{2\pi i}) = \rho(1) = 1$ forces $c \in \mathbb{Z}$, so $\rho(z) = z^c$ for some integer $c$.
[/proofplan]
[step:Use compactness of $\rho(S^1)$ to prove $|\rho(z)| = 1$ for all $z \in S^1$]
The image $\rho(S^1) \subseteq \mathbb{C}^\times$ is the continuous image of a compact space, hence compact in $\mathbb{C}^\times$. In particular, $\rho(S^1)$ is bounded: there exists $M > 0$ with $|\rho(w)| \leq M$ for every $w \in S^1$.
Suppose for contradiction that there exists $z \in S^1$ with $|\rho(z)| > 1$. Since $\rho$ is a homomorphism, $\rho(z^n) = \rho(z)^n$ for every $n \in \mathbb{Z}_{>0}$, so $|\rho(z^n)| = |\rho(z)|^n$. As $|\rho(z)| > 1$, $|\rho(z)|^n \to \infty$, contradicting the bound $|\rho(z^n)| \leq M$ for all $n$.
Suppose for contradiction that there exists $z \in S^1$ with $|\rho(z)| < 1$. Then $|\rho(z^{-1})| = |\rho(z)|^{-1} > 1$, contradicting the previous case applied to $z^{-1} \in S^1$.
Hence $|\rho(z)| = 1$ for every $z \in S^1$, i.e., $\rho(S^1) \subseteq S^1$. We may therefore consider $\rho$ as a continuous group homomorphism $S^1 \to S^1$.
[/step]
[step:Lift $\rho$ along the covering $\mathbb{R} \to S^1$ to obtain a continuous homomorphism $\mathbb{R} \to S^1$]
Define
\begin{align*}
\tilde{\rho}: \mathbb{R} &\to S^1, \\
x &\mapsto \rho(e^{ix}).
\end{align*}
This is continuous as the composition $\rho \circ \varepsilon$ where $\varepsilon: \mathbb{R} \to S^1$, $x \mapsto e^{ix}$. It is a homomorphism from $(\mathbb{R}, +)$ to $(S^1, \cdot)$:
\begin{align*}
\tilde{\rho}(x + y) = \rho(e^{i(x + y)}) = \rho(e^{ix} \cdot e^{iy}) = \rho(e^{ix}) \cdot \rho(e^{iy}) = \tilde{\rho}(x) \cdot \tilde{\rho}(y),
\end{align*}
where the third equality uses that $\rho$ is a homomorphism.
By the [Classification of Continuous Homomorphisms $\mathbb{R} \to S^1$](/theorems/2470), there exists $c \in \mathbb{R}$ with $\tilde{\rho}(x) = e^{i c x}$ for every $x \in \mathbb{R}$.
[/step]
[step:Use $2\pi$-periodicity to force $c \in \mathbb{Z}$]
Evaluating at $x = 2\pi$,
\begin{align*}
e^{i c \cdot 2\pi} = \tilde{\rho}(2\pi) = \rho(e^{2\pi i}) = \rho(1) = 1,
\end{align*}
where the last equality uses that $\rho$ is a homomorphism: $\rho(1) = \rho(1 \cdot 1) = \rho(1)^2$, hence $\rho(1) = 1$ (the equation $w = w^2$ in $\mathbb{C}^\times$ forces $w = 1$).
The equation $e^{2\pi i c} = 1$ in $S^1$ is equivalent to $2\pi c \in 2\pi \mathbb{Z}$, i.e., $c \in \mathbb{Z}$. Set $n := c \in \mathbb{Z}$.
[/step]
[step:Conclude $\rho(z) = z^n$ for every $z \in S^1$]
Every $z \in S^1$ has the form $z = e^{ix}$ for some $x \in \mathbb{R}$. By Step 2 and the choice $n = c$,
\begin{align*}
\rho(z) = \rho(e^{ix}) = \tilde{\rho}(x) = e^{i n x} = (e^{ix})^n = z^n.
\end{align*}
Hence $\rho = \rho_n$ where $\rho_n: z \mapsto z^n$, with $n \in \mathbb{Z}$.
[guided]
The argument has three logical movements: a *boundedness* movement (using compactness of $S^1$ to bound the image), a *lifting* movement (passing through the covering $\mathbb{R} \to S^1$ to convert a homomorphism on $S^1$ into a homomorphism on $\mathbb{R}$), and a *quantisation* movement (using $2\pi$-periodicity to force the parameter $c$ to be an integer).
Why does compactness force $|\rho(z)| = 1$? The orbit $\{\rho(z)^n : n \in \mathbb{Z}\}$ lies inside the compact set $\rho(S^1)$ and so must be bounded. But $|\rho(z)^n| = |\rho(z)|^n$ tends to $\infty$ if $|\rho(z)| > 1$ or to $0$ if $|\rho(z)| < 1$, with $0$ also outside $\mathbb{C}^\times$. The only escape is $|\rho(z)| = 1$. This is the same compactness mechanism that, in a more sophisticated form, yields the unitarisability statement for compact group representations.
Why does the lifting through $\varepsilon: \mathbb{R} \to S^1$ help? The classification of continuous homomorphisms $\mathbb{R} \to S^1$ already gives the form $e^{icx}$ on the connected non-compact group $\mathbb{R}$. We do not have an analogous classification on $S^1$ directly — the topology of $S^1$ obstructs this — but we can pull a homomorphism on $S^1$ back through the universal cover to reduce to the classification on $\mathbb{R}$. The price is that the parameter $c$ obtained on $\mathbb{R}$ must respect the kernel of the covering, $2\pi\mathbb{Z}$, which translates into the integrality $c \in \mathbb{Z}$.
What goes wrong without continuity? The proof uses continuity of $\rho$ in two places. First, to pass to a continuous lift $\tilde{\rho} = \rho \circ \varepsilon$ and apply the classification on $\mathbb{R}$. Second, the boundedness step in the previous paragraph relied on compactness of $\rho(S^1)$, which in turn relied on continuity of $\rho$. A non-continuous group homomorphism $S^1 \to \mathbb{C}^\times$ may fail to land in $S^1$: existence of such pathological maps (using the axiom of choice) shows that continuity is essential.
The classification produces a $\mathbb{Z}$-indexed family $\{\rho_n\}_{n \in \mathbb{Z}}$ of one-dimensional representations, and these are pairwise non-isomorphic (their characters $z \mapsto z^n$ are distinct functions on $S^1$). The set $\widehat{S^1}$ of irreducible characters is therefore canonically identified with $\mathbb{Z}$ — a manifestation of Pontryagin duality $\widehat{S^1} \cong \mathbb{Z}$.
[/guided]
[/step]
