Obstructions to lifting a map for the Hopf fibration

This is a bit of an elementary question, but

suppose $\pi: \mathbb{S}^3\to \mathbb{S}^2$ is the Hopf fibration, are there reasonably computable obstructions to when a map $f:M\to \mathbb{S}^2$ can be lifted to a map $\tilde{f}:M\to \mathbb{S}^3$?

If it matters everything can be taken in the smooth category. Also, I am most interested in the case that $M$ is an open subset of a oriented closed surface if this simplifies things at all.

My understanding is that if $M$ is a disk this there is no obstruction,

I apologize if this is trivial but it is not by area of expertise...

Solution 1:

The Hopf fibration is of the form

$$S^1 \to S^3 \to S^2$$

Since the Hopf fibration is an $S^1$-bundle, there is a classifying map

$$S^2 \to BS^1$$

where $BS^1 \simeq K(\mathbb Z,2) \simeq Gr^+_{\infty,2}$, i.e. it is the Grassmannian of oriented 2-dimensional vector bundles. So the Hopf fibration is the pull-back of the unit circle bundle of this classifying map, moreover, the map $S^2 \to Gr^+_{\infty,2}$ is the generator of $\pi_2 Gr^+_{\infty,2} \simeq \mathbb Z$.

A map $M \to S^2$ lifts to a map $M \to S^3$ if and only if the pull-back of the generator of $H^2 S^2 \simeq \mathbb Z$ in $H^2 M$ is zero. This is because the tautological circle bundle over $Gr^+_{\infty,2}$ is the Stiefel manifold $V_{\infty,2}$ which is contractible. So the only obstruction is the map $M \to BS^1$, which via the Serre interpretation of $H^2$ is the pull-back of the generator under the map $M \to S^2$.