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I have problem with this:

Prove that $F^{-1}(0,1)$ and $S^2$ are diffeomorphic, where

$$F(x,y,s,t)=(x^2+y,x^2+y^2+s^2+t^2+y).$$

I have found that

$$F^{-1}(\{0,1\})=\{ (x,y,s,t) \in \mathbb{R}^4: y^2+s^2+t^2=1, \ x^2+y=0 \}$$

but I can't find any diffeomorphism.

Mhenni Benghorbal
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amoneth
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    How is $F^{-1}({(0,1)})={(x,s,t) \in \mathbb{R}^3 : x^4+s^2+t^2=1 }$? Shouldn't $F^{-1}({(0,1)})$ a subset of $\mathbb{R}^4$? – Ashok Nov 01 '13 at 12:09
  • Of course you are right! $F^{-1}({0,1})={ (x,y,s,t) \in \mathbb{R}^4: y^2+s^2+t^2=1, \ x^2+y=0 }$. – amoneth Nov 01 '13 at 12:11
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    You can try sending $(x,y,s,t) \to (y,s,t) \subset \mathbb{R}^3$. The equation $y^2 + s^2 + t^2 = 1$ defines $S^2$. – Elchanan Solomon Nov 01 '13 at 12:19
  • Thanks Isaac Solomon, but i think it's not injective. For example f(1,-1,0,0)=(-1,0,0)=f(-1,-1,0,0) – amoneth Nov 01 '13 at 17:26

1 Answers1

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The equation $x^2+y=0$ can hold only when $y\le 0$. So, ignoring $x$ we have only the Southern hemisphere in $y,s,t$ space. But for every $y<0$, there are two $x$ values, so we actually have two copies of Southern hemisphere that are joined at the equator $y=0$. This does look like a sphere, right?

Perhaps the picture becomes clearer if we replace $y$ with $y=-x^2$ and get $$F^{-1}(\{0,1\})=\{ (x,-x^2,s,t) \in \mathbb{R}^4: x^4+s^2+t^2=1 \}$$ The projection $(x,-x^2,s,t)\to (x,0,s,t)$ is a diffeomorphism, the image of which is an ellipsoid-looking surface $x^4+s^2+t^2=1$ in $\mathbb R^3$. This surface is (a) smooth; (b) homeomorphic to $S^2$.

An explicit diffeomorphism onto $S^2$ is given by radial projection: $$(x,s,t)\mapsto \frac{(x,s,t)}{\sqrt{x^2+s^2+t^2}}$$

user103254
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