I need to show that the map:
exp: $\mathbb{R}\rightarrow \mathbb{S}^1,f(\theta)=e^{2\pi i\theta}$ is surjective.
I know that $f$ is surjective if $\forall y\in\mathbb{S}^1 \exists x\in \mathbb{R}$ such that $y=f(x)$, so with that in mind this is what I did:
Let $\theta=\frac{\log(y)}{2\pi i}$ then $e^{2\pi i\theta}=e^{2\pi i \frac{\log(y)}{2\pi i}}=e^{\log(y)}=y$
But $\theta=\frac{\log(y)}{2\pi i}\notin \mathbb{R}$ so I don't think what I did is right at all.
Could somebody please help me show that this map is surjective?
Since the derivative of $f$ does not vanish on $\mathbb{R}$, $f$ is open. Hence, $f(\mathbb{R})$ is open in $\mathbb{S}^1$. Since $f(\mathbb{R})$ is nonempty and $\mathbb{S}^1$ is connected, it suffices to show that $f(\mathbb{R})$ is closed is $\mathbb{S}^1$. Let $H:=f(\mathbb{R})$, notice that $H$ is a subgroup of $\mathbb{S}^1$. Therefore, let $(g_i)_{i\in I}$ be a set of representants of $(\mathbb{S}^1/H)\setminus\{H\}$, one has: $$H=\mathbb{S}^1\setminus\bigcup_{i\in I}g_iH.$$ Since each $x\mapsto {g_i}^{-1}x$ is continuous, $g_iH$ is open as the inverse image of $H$ by the previous map. Finally, $H$ is closed in $\mathbb{S}^1$.
Remarks.
xyzzyz gave in the comments an easier argument to show that $f(\mathbb{R})$ is closed.
The argument I developed gives right away that $f$ is surjective from $\mathbb{C}$ to $\mathbb{C}^*$ (knowing that $f$ is holomorphic).