Covering spaces of $S^1$

Question

Put $\tilde X=\lbrace (exp(2\pi if(t)),t)| t\in \mathbb{R} \rbrace$ where $f:\mathbb{R}\rightarrow \mathbb{R}$ is any continuous function and let $\pi_1$ be the projecction on the first coordinate. If we define $p:\tilde X\rightarrow S^1$ by $p=\pi_{1}|_{\tilde X}.$ Proof that if $f$ is strictly monotone then $p$ is a covering map.

Answer 1

You're right; $p$ doesn't need to be surjective with the assumptions you state. $f$ could be a constant function, in which case $\pi_1\left. \right|_\tilde{X}$ is not a covering map.

Edit: Whoops, forgot the monotonicity. Daniel Rust points out in the comments that $\frac{1}{2\pi}\tan^{-1}x$ is a monotone function where we don't get surjectivity.