Let $(\mathbb{R}, +, \tau_{\mathbb{R}})$ and $(\mathbb{S}^1, \cdot, \tau_{\mathbb{S}^1})$ be a topological groups, where $\tau_{\mathbb{R}}$ is the usual topology of real numbers and $\tau_{\mathbb{S}^1}$ is the subspace topology induced by $\mathbb{C}^*$ with the open ball topology.
I need proof that the homomorphism $f : \mathbb{R} \longrightarrow \mathbb{S}^1$, given by $f(x) = \cos(2\pi x) + i\sin(2\pi x) $ is continuous but I still didn't get it. I thought about using a result that says that the homomorphism will be continuous if and only if it is continuous on the neutral element, but I didn't succeed.
Do you suggest any Theorem of complex analysis or some composition of functions that will help me to get around this problem?
See the function $f$ as a function from $\Bbb R$ into $\Bbb C$. Then $f=\exp\circ g$, where $g\colon\Bbb R\longrightarrow\Bbb C$ is the function defined by $g(x)=2\pi ix$. Then $f$ is continuous, since it is obtained from two continuous functions by composition.
Since $f$ is a continuous function from $\Bbb R$ into $\Bbb C$, since the range of $f$ is a subset of $\Bbb S^1$, and since the topology that you are taking into account in $\Bbb S^1$ is the subspace topology, $f$ is a continuous function from $\Bbb R$ into $\Bbb S^1$.