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Is there any local diffeomorphism from $\mathbb R^2$ onto $S^2$?

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2 Answers 2

Yes, there are smooth onto functions $f : \mathbb R^2 \to S^2$ whose derivative is everywhere rank $2$.

This does not contradict Paul's answer since what Paul's answer tells you is $f$ can't be a covering map.

I think perhaps a good way to think about this is peeling an orange without "flaking" it:

The orange peel you can think of as as being like a neighbourhood of an embedded arc in $S^2$.

So let's see if there's a nice formula that does the job.

Step 1: $\mathbb R^2$ is diffeomorphic to $\mathbb R \times (-1,1)$. The map is $(x,y) \longmapsto (x,\frac{y}{\sqrt{1+y^2}})$.

Step 2: Let $\gamma : \mathbb R \to S^2$ be any immersion of bounded curvature whose image is dense in $S^2$.

Step 3: The onto submersion $\mathbb R \times (-1,1) \to S^2$ is given by sending $(x,y)$ to a little push-off of $\gamma(x)$ where you push in the direction a 90-degree counter-clockwise direction to $\gamma'(x)$, some distance proportional to $y$. The proportionality constant will be something like the reciprocal of the maximum curvature of $\gamma$ -- this ensures the map from the geometric normal bundle to the curve $\gamma$ is a submersion.

So I hope that gives you the idea. I imagine it's not too hard to cook up an explicit formula for such a $\gamma$ but it's too late for me to think-up one, apparently. umm... or maybe... $\gamma(t) = (\cos t \cos(t/a), \cos t \sin(t/a), \sin t)$ would appear to get the job done provided the vector subspace of $\mathbb R$ generated by $\pi$ and $a$ is $2$-dimensional over $\mathbb Q$.

There are of course simpler, perhaps more explicit constructions. The idea would be to think of such an onto submersion $\mathbb R \times (-1,1) \to S^2$ as a describing a "brush stroke" where you are painting a sphere. The first parameter $x$ is time, and the 2nd $y$ is the parameter along the brush. So to construct an onto submersion, the game is to entirely paint the sphere in one stroke, where the only constraint is your direction of travel has to be independent of the line where the brush is contacting the sphere. Clearly you can do this, it's just a matter of writing it out explicitly as some function.

orange

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thanks, I'll think about this idea. –  fiorerb Jan 21 '12 at 11:39

First note that there actually are no covering maps $\mathbb{R}^2 \to S^2$. This is because $S^2$ is simply connected and hence is its own universal cover; if there were a covering map $\mathbb{R}^2 \to S^2$, then by the universal property of the universal cover there would be a covering map $S^2 \to \mathbb{R}^2$. But there can't even be a continuous surjection $S^2 \to \mathbb{R}^2$ because $S^2$ is compact and $\mathbb{R}^2$ is not.

Thus any local diffeomorphism $\mathbb{R}^2 \to S^2$ answers your question. For example, the inverse of the stereographic projection map does the job.

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I just noticed that you asked about a local diffeomorphism ONTO $S^2$. My argument shows that there is no surjective proper local homeomorphism $\mathbb{R}^2 \to S^2$ (such a map would necessarily be a covering map) but there is probably some non-proper example. –  Paul Siegel Jan 21 '12 at 0:10
    
Thanks @Paul I've seen your point, I don't have to worry about if it is a cover or not, but still I can't find a surjective local diffeomorphism. –  fiorerb Jan 21 '12 at 1:11
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The question was «Is there any local diffeomorphism from $R^2$ onto $S^2$?» and you answered «Thus any local diffeomorphism $R^2\to S^2$ answers your question».Amazing! :D –  Mariano Suárez-Alvarez Jan 21 '12 at 4:14
    
@Mariano - In Paul's defence, the question was edited. The previous question asked if there was any local diffeomorphism that was not a covering space –  Juan S Jan 21 '12 at 7:11
    
Ah! In any case, it is fun :) –  Mariano Suárez-Alvarez Jan 21 '12 at 7:30

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