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Suppose there exists such a map $f:\mathbb{R}^2\to\mathbb{S}^2$. Basic (topological) cover space theory tells us that this map cannot be a covering map. If we abandon the condition that $f$ must be a local diffeomorphism, then clearly such a map exists – we can simply use the map $$f(\theta,\phi) = (\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi).$$

However, I suspect (but cannot prove) that if we impose the condition of being a local diffeomorphism, there does not exist any such map. Algebraic topology seems to be of no direct use here (since I don't see how to use smoothness), and I don't know of any theorems in differential geometry that would be very useful in proving the nonexistence of such a map.

Dominic Wynter
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    It seems that this question has already been addressed here http://math.stackexchange.com/questions/100884/local-diffeomorphism-from-mathbb-r2-onto-s2?rq=1. I may decide therefore to retract this question. – Dominic Wynter Mar 14 '17 at 17:12
  • Yes, this is an exact duplicate. – Moishe Kohan Mar 17 '17 at 00:25

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