How to prove a map is a local diffeomorphism Let $F:U \subset R^2 \rightarrow R^3$ by given by $F(u,v) = (u\sin{(\alpha)}\cos{(v)},u\sin{(\alpha)}\sin{(v)},u\cos{(\alpha}))$,  
$(u,v) \in U = \{(u,v) \in R^2 \space; u &gt; 0\}$ with $\alpha$  a constant. 
Show F is a local diffeomorphism of U onto a cone C with the vertex at the origin and $2\alpha$ as the angle of the vertex. 


From Do Carmo, problem 4.2.1


I was thinking of using the Inverse function theorem. $F$ is differentiable. We can compute the jacobian I suppose and show it's non-zero. I'm not sure how to construct this so it's a map onto a cone with the above properties.

Recall that the Inverse function theorem says if we have a function from $R^n \rightarrow R^n$ that is continuously differentiable on some open set containing a point $p$, and that the determinate of $Jf(a) \ne 0$ then there is some open set $W$ containing $f(a)$ such that $f:V \rightarrow W$ has a continuous inverse $f^{-1}:W \rightarrow V$ which is differentiable for all $y \in W$.    Basically it says $f$ is a diffeomorphism from $V$ to $W$ given it satisfies the above hypothesis.

Now above it wants us to prove it's a local diffeomorphism, not globally a diffeomorphism. That would just mean given any point $p$ in the domain of F, there exists an open set $U \subset \mathrm{dom}(F)$ containing $p$ such that $f:U \rightarrow f(U)$ is a diffeomorphism.
1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)\in U$, so that $F$ is an immersion, but certainly not an embedding: see  point 3) below.
The image $F(U)$ is exactly the subset $z\gt0, x^2+y^2=z^2\tan ^2(\alpha)$ of $\mathbb R^3$.        

2) This subset is indeed the intersection $C\subset \mathbb R^3$ of the upper half-space $z\gt 0$ with the cone of half-angle $\alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is   a locally closed submanifold of $\mathbb R^3$.     

3) The coinduced morphism of differential manifolds $F_0:U\to F(U)=C$ is the universal covering map of $C$, and its  restriction to any  vertical line $\{c\}\times \mathbb R\subset U \;(c\gt 0)$ is the universal  covering map of the  horizontal circle $C\cap\{z=c\cos \alpha\} $ of $C$ .  

An opinion
All in all, this exercise  could have been written in a slightly more explicit way ...
You have to do two things:


Show that the image of $F$ is the required cone
Show that $F$ is a local diffeomorphism


As for point 1, you can see that $F(u,v)=u (\sin \alpha \cos v , \sin \alpha \sin v, \cos \alpha)$, so it is clearly a cone (if you fix the variable $v$, and let $u \in \mathbb{R}$ you get a straight line through $0$, while if you fix the variable $u$ you get circles). To see that the angle of the vertex is $2 \alpha$, you can use the fact that
$$F(u,v) \cdot (0,0,1) = ||F(u,v)|| \ ||(0,0,1)|| \cos( \mbox{angle between $F(u,v)$ and $(0,0,1)$ } ) $$
where $\cdot$ denotes the scalar product. LHS gives you $u \cos \alpha$, while RHS gives you $u \cos( \mbox{angle between $F(u,v)$ and $(0,0,1)$ } )$, so (supposing that $\alpha \in \left( 0,\pi \right)$)
$$\mbox{angle between $F(u,v)$ and $(0,0,1)$ } = \alpha$$

As for point 2, compute the Jacobian
$$J(u,v) = \left( \begin{matrix} 
\sin \alpha \cos v &amp; \sin \alpha \sin v &amp; \cos \alpha \\
-u \sin \alpha \sin v &amp; u\sin \alpha \cos v &amp; 0
\end{matrix} \right)
$$
Since $u &gt;0$ (this hypothesis is important otherwise you would have the second row vanishing), you can see that this Jacobian has always rank $2$, because
$$
\left| \begin{matrix} 
\sin \alpha \cos v &amp; \sin \alpha \sin v  \\
-u \sin \alpha \sin v &amp; u\sin \alpha \cos v
\end{matrix} \right| = u \sin \alpha \neq 0
$$
($J(u,v)$ has an invertible $2 \times 2$ submatrix, so its rank is $\geq 2$). Since $2$ is the maximal rank $J(u,v)$ can achieve, "some version of the inverse function theorem" tells you that $F$ is a local diffeomorphism between $\mathbb{R}^2$ and the cone.

Clearly, all of this makes sense if $\sin \alpha \neq 0$, otherwise your map would be $F(u,v)=(0,0,u \cos \alpha)$, which gives you a line.