## finding an inclusion map to show submanifold

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:

++
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