# Any smooth mapping from $\mathbb{R}^n$ into $S^1$ is of the form $e^{if(x)}$?

Let $$F\colon \mathbb{R}^n \to S^1$$ be a smooth mapping.

Then, I strongly suspect that there must be a smooth function $$f\colon \mathbb{R}^n \to \mathbb{R}$$ such that $$$$F(x) = \exp \big( if(x) \bigr)$$$$

Moreover, $$f$$ must be unique up to integer multiple of $$2\pi$$.

Could anyone please tell me how to prove this or correct me if I am wrong?

• I think it is about this :notice $F_{*}(\pi_1(\mathbb{R}^n,\cdot)) = \{0\}\subseteq p_{*}(\pi_1(\mathbb{R},\cdot))$ where $p$ is the exponential function, which is a covering. When this condition is met, from nice spaces as those manifolds, $F$ lifts to a map into $\mathbb{R}$. Moreover, the lift has degeneracy upto each $2\pi k$ which is degeneracy of $p$ for a fixed base point Commented Jul 10 at 12:32

This is true. Recall that the map $$\phi\colon \mathbb{R} \to S^1$$, $$\phi(t) = \exp(it)$$ is a smooth covering map. By the smooth covering correspondence and the fact that $$\mathbb{R}^n$$ is contractible, any smooth map $$f\colon \mathbb{R}^n \to S^1$$ lifts to the covering, i.e. there's a smooth map $$\tilde{f}\colon \mathbb{R}^n \to \mathbb{R}$$ such that $$f = \phi \circ \tilde{f}$$, which is precisely what you were asking for. The uniqueness condition then follows from the fact that $$\phi(t) = \phi(t')$$ holds exactly if $$t = t' + 2n \pi$$ for some $$n \in \mathbb{Z}$$.
• (+1, but being a bit too picky:) The uniqueness condition then follows from the fact that $\phi(t)=\phi(t')$ holds exactly if $t=t′+2n\pi$ for some $n \in \Bbb Z$ and from the continuity of $f$ :) Commented Jul 10 at 12:39