Is this a smooth immersion of $\Bbb R^2$ to $\Bbb R^3$? This has to be a smooth immersion according to John Lee's book Introduction to smooth manifolds, page 78. I compute the rank of the Jacobian matrix of the map X in an arbitrary point $(u_o,v_o)$ but after the calculations, I just showed that $\text{rank}(J(X))=1$ or $2$. It has to be $2$! I really need help. It is self-study - manifolds for about 3 years. Any help is appreciated! 

 A: The Jacobian matrix is
$$
2\pi\cdot\begin{bmatrix}
-\sin(2\pi u)\cos(2\pi v) & -\sin(2\pi u)\sin(2\pi v)&\cos(2\pi u)\\
-(2 + \cos(2\pi u))\sin(2\pi v) & (2+\cos(2\pi u))\cos(2\pi v) & 0
\end{bmatrix}
$$
(or maybe it's the transpose; I can never remember).
We split into cases, depending on whether $\cos(2\pi u)= 0$.
If $\cos(2\pi u) = 0$, then $\sin(2\pi u) = \pm 1$. This reduces the matrix to
$$
2\pi\cdot\begin{bmatrix}
\pm\cos(2\pi v) & \pm\sin(2\pi v)&0\\
-2\sin(2\pi v) & 2\cos(2\pi v) & 0
\end{bmatrix}
$$
(where the $\pm$ stands for the same sign in both places). The determinant of the left $2\times 2$ submatrix is
$$
(4\pi^2)\cdot(\pm2\cos^2(2\pi v)\pm 2\sin^2(2\pi v)) = 4\pi^2\cdot (\pm2)\neq 0
$$
which means that the first two columns are linearly independent, and the matrix has rank 2.
If $\cos(2\pi u)\neq 0$, then for the rank to be $1$, the second component of all the  columns must be $0$. But $(2+\cos(2\pi u))\neq 0$, and $\sin(2\pi v)$ and $\cos(2\pi v)$ can't both be zero. So the rank must be 2.
