Suppose that I have a smooth function $f:M\to S^1$ for a connected manifold $M$. I know that $p:\mathbb{R}\to S^1$ defined by $p(t) = e^{i t}$ is a covering map and that $\mathbb{R}$ is the universal cover of $S^1$. I am interested in what conditions I can put on $M$ such that if $f:M\to S^1$ is smooth, then there exists a smooth function $\tilde{f}:M\to \mathbb{R}$ so that $f = p\circ \tilde{f}$.
In this answer: Requirement for a given function to be smooth it seems to be implied that if $M = \mathbb{R}$, then the result should hold. I am thinking that it might suffice to assume that $M$ is simply connected so that it does not have a universal cover other than itself.
Another user made this post: When is it possible to lift a function to its covering space? which gives an equivalent condition for when we can have this lift property, but my Algebra/Category Theory is quite weak so I am unsure what conditions on $M$ will satisfy the condition. I am thinking that if $M$ is simply connected, then $\pi_1(M) = \{0\}$ so the condition should be satisfied.
My question could also be phrased as, if $f:M\to \mathbb{C}$ has constant magnitude so that $|f| = C > 0$, then does there exist a smooth function $g:M\to\mathbb{R}$ such that $f = e^{ig}$. A little bit of direction would be greatly appreciated.
There exist lifting of a map $X \to Y$ to a covering $\tilde Y\to Y$ if an only if the image of $\pi_1(X, x_0)$ is inside the image of $\pi_1(\tilde Y, \tilde y_0)$.
So in our case, we would need the image of $\pi_1(M)$ trivial. It surely is so if $M$ is simply connected.