Continuous function $ f：\operatorname{SO}(3) \to \operatorname{SU}(2)$

Question

Given the usual surjective homomorphism $ Φ:\operatorname{SU}(2)\to \operatorname{SO}(3)$ that maps a quaternion to a rotation matrix,

Does there exist a continuous function $f:\operatorname{SO}(3)\to \operatorname{SU}(2)$ such that $ \Phi \circ f = \operatorname{Id}$ on $\operatorname{SO}(3)$?

If yes, what would be this map be?

Thank you

Answer 1

No. A map $f$ as in your question would induce an embedding of fundamental groups $f_* :\pi_1(\text{SO(3)}) \to \pi_1(\text{SU}(2))$.

If you look at Fundamental group of $SO(3)$, you will see that $\pi_1(\text{SU}(2)) = 0$, while $\pi_1(\text{SO(3)}) = \mathbb{Z}_2$. Therefore $f$ cannot exist.

Answer 2

Here is another proof. It is known that $\Phi$ is a covering projection with two sheets. The map $f$ would be a section of $\Phi$, but this would imply that $\Phi$ is a homeomorphism. See If a covering map has a section, is it a $1$-fold cover?.