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For $a,b,c >0$ let,

$E(a,b,c) = \{ (x,y,z) \in \mathbb {R^3} : \frac {x^2}{a^2} +\frac {y^2}{b^2} + \frac {z^2}{c^2} =1 \}$

I am asked to Show that $E(a, b, c)$ is diffeomorphic to the 2-dimensional sphere $S^{2} ≡ E(1, 1, 1)$

But i can' show that this is diffeomorphic to a 2-dimensional sphere cause i dont what a 2-dimensional sphere even looks like.

Could someone show me a 2-dimensional sphere or perhaps (even better) show my how to put the unit sphere from $\mathbb {R^3}$ into $\mathbb {R^2}$ assuming thats even what this means.

its also likely i got the tags wrong.

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    The unit sphere in $\mathbb R^3$ _is_ the $2$-dimensional sphere. – 2017-01-24
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    The unit 2-dimensional sphere is $E(1,1,1) = \{ (x,y,z) \in \mathbb R^3 : x^2 + y^2 + z^2 = 1 \}$ just use the map $\phi:E(a,b,c) \to E(1,1,1)$ defined $$\phi(x,y,z) = (x/a,y/b,z/c), \,\,\,\, (x,y,z) \in E(a,b,c).$$ – 2017-01-24
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    doesn't it have 3 dimensions? – 2017-01-24
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    It is a 2-dimensional surface: imagine you were a tiny bug standing on the unit sphere, what would the ground look like? It would look like the plane $\mathbb R^2$. Alternatively, using polar coordinates, it can be parameterized by two variables $(\theta, \phi)$ since the radius is fixed. – 2017-01-24
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    hmm is there a 3 dimentional surface? in $\mathbb {R^3}$ – 2017-01-24
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    Usually "surface" is by definition a "2-dimensional manifold". So you are probably asking for a 3-dimensional manifold. However the answer is yes, e.g. any open ball. – 2017-01-24

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