I'm studying harmonic analysis, and I'm trying to understand the fact that $$\phi: \mathbb{R}\to \widehat{\mathbb{R}}: s \mapsto (t \mapsto \exp(2\pi i st))$$ is an isomorphism of topological groups (here we consider $\mathbb{R}$ as a topological group for the usual topology and the usual addition $+$).
So far, I have managed to show that $\phi$ is a group isomorphism (surjectivity was the harder part). It remains to show that $\phi$ is a homeomorphism, i.e. that both $\phi$ and $\phi^{-1}$ are continuous.
To show continuity of $\phi$, we need to show that if $\{s_\lambda\}$ is a net that converges to $0$ in $\mathbb{R}$, then $$\exp(2\pi i s_\lambda t) \to 1$$ uniformly in $t\in K$ where $K\subseteq \mathbb{R}$ is compact. I think I can prove this using the fact that continuous functions are uniformly continuous on compact subsets.
Conversely, I also struggle to show that if $\exp(2\pi i s_\lambda t) \to 1$ uniformly on compact subsets, then $s_\lambda \to 0$. I guess we need some form of complex logarithm for this?
Thanks in advance for any hints/suggestions!
Assume that $\phi(s_\lambda) \to 1$ uniformly on compact subsets. Assume to the contrary that $s_\lambda \not\to 0$. Then there is a subnet $\{s_\mu\}$ and $\epsilon > 0$ such that $|s_\mu| \ge \epsilon$ for all $\mu$. By passing to a further subnet, we may assume that $s_\mu$ has the same sign for all $\mu$. By possibly replacing $\{s_\lambda\}$ by $\{-s_\lambda\}$, we may assume that $s_\mu \ge 0$ for all $\mu$, so that $s_\mu \ge \epsilon$ for all $\mu$.
By assumption, there is an index $\mu$ such that $$\sup_{t \in [0,\epsilon^{-1}]}|\exp(2\pi is_\mu t)-1| < 1.$$ However, $s_\mu [0, \epsilon^{-1}] = [0, s_\mu \epsilon^{-1}] \supseteq [0,1]$ so that $$2=\sup_{v \in [0,1]} |\exp(2\pi iv)-1| \le \sup_{t \in [0,\epsilon^{-1}]}| \exp(2\pi i s_\mu t)-1| < 1$$ a contradiction. Hence, $s_\lambda \to 0$, and this is enough to show that the group homomorphism $\phi^{-1}$ is continuous.