Hi there

I've got the following exercise to solve and I don't believe my ideas are that bad but it doesn't seem to be enough yet:

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With an injective curve \gamma : ]0.1[ \rightarrow \mathbb{R}^2 that does not cross the axis z and a non-zero derivative \forall values consider the surface \Omega parametrised by the rotation of \gamma around the axis z.

Now show that \Omega is a submanifold of \mathbb{R}^3 with dimension 2. State an atlas.
++ (sorry when translating isnt perfect)

Concerning the lecture when \gamma(t)=(\gamma_1(t),\gamma_2(t)) then p(s,t)=(\gamma_1(t)*cos(s),\gamma_2(t)*sin(s), \gamma_2 (t) ) is a parametrisation of \Omega with t\in ]0,1[ and s\in [0,2*\pi) am I right?

My idea was taking the inverse of this p:

i : \Omega \rightarrow ]0,1[ \times [0,2\pi) \times \{0\} , (x,y,z) \mapsto (\gamma_2^{-1}(z),sin^{-1}(\frac{y}{z}),0)

\gamma is injective and surely surejective to the image of itself so invertable. So \gamma_2^{-1} is well-defined. But why is z never 0 ???


Now I want to argument that i is an inclusion map that embeds \Omega into \mathbb{R}^2 \times \{0\} . Is this the right way? Is this what one calls an "inclusion map"?

Then I'll have to show that the jacobian of i is injective everywhere on \Omega. This is equivalent to showing that the jacobian is invertible for all points on the surface is this correct?


What do you suggest?

Regards
Huberscher