Why $p(t)=e^{2\pi it}$ is open?

Question

I try to show that $p: \Bbb R\rightarrow \Bbb S^1\subset \mathbb{R}^2$,$p(t)=e^{2\pi it}$ is identifaction map.

• I proved $p$ is the continuous and surjective map.

$\textbf{Question:}$: I can't see how to prove $p$ is an open map?