Regular and critical values of a map from the torus to the 2-sphere Let the 2-sphere be denoted by $S^2$ and $T^2$ be the torus realized as a regular submanifold of $\mathbb{R^3}$ by taking a circle of radius 1 at $(0,2,0)$ in the $yz$-plane and rotating it about the $x$-axis. Let $\gamma:\mathbb{R^3}\to S^2$ be $\gamma(x,y,z)=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$. Let $g:T^2\to S^2$ be the restriction of $\gamma$ to $T^2$. I want to find the regular and critical values of $\gamma$. I guess this can be done by parametrizing the torus but it seems very tedious. I was wondering if there is any other method. If $\pi:\mathbb{R^3}\to \mathbb{R^2}$ is the projection $\pi(x,y,z)=(y,z)$ and $p:T^2\to\mathbb{R^2}$ the restriction of $\pi$ to $T^2$. Then $\gamma=g\circ p \circ \pi$. Or equivalently, $$g=\gamma \circ \pi^{-1}\circ p^{-1}$$. Now we can find the differential of $g$ and it's kernel will give all the critical points and their image will be all the critical values. I am not too sure how I can take the differential of the compositions and the inversions. Any help is appreciated. Thank you!
 A: In regards to what I wrote above, although "tedious", you should just parametrize $T^2$ in terms of the rectangular coordinates $x,y,z$ or curved-coordinates $u,v$. This actually isn't too hard.
$\bullet$  Consider a circle of $1$ in the $yz$-plane with center $(0,2,0)$
$\bullet$ If you rotate any point on this circle about the $z$ axis then it's distance from the origin doesn't change.
$\bullet$ Hence if $(x,y,z) \in T^2$ then $\left(\sqrt{x^2+y^2} - 2\right)^2 + z^2 = 1^2 =1$. In general we would have $\left(\sqrt{x^2+y^2} - a\right)^2 + z^2 = r^2$. See if you can figure out which triangle are these lengths corresponding to.
$\bullet$ You can now define a function $f: D \to \mathbb{R}^3$ in which $f$ gives local coordinates for $p \in T^2 \subset \mathbb{R}^3$ i.e $f(u,v) = (u,v,z(u,v))$ where we solve for $z$ above. This is just an example of getting coordinates arising from the graph of a smooth function. In this case we can take $g = \gamma \circ f$. 
This is probably the best way to go about the problem. Although we are taking these coordinates from the graph of a function, you can safely use $z = \sqrt{1-\left(\sqrt{x^2+y^2}-2\right)^2}$ (or the one with $-$) since the only difference is a sign. This is the $z$ coordinate in the above description of $f$ and so your restriction will look like,
$$(\gamma \circ f)(u,v)= \gamma(u,v,z(u,v))$$
$\bullet$ Another route you can go is to uncover the two-parameter parametrization of the torus. You wish to rotate a circle of radius $1$ with center $(0,2,0)$ about the $z$-axis. Just apply the rotation matrix.
$$\\$$
$$\underbrace{\begin{pmatrix} \cos v & - \sin v & 0 \\ \sin v & \ \ \cos v & 0 \\ 0 &\ \  0 & 1 \end{pmatrix}}_{\textrm{rotation about}\  z-axis} \underbrace{\begin{pmatrix} 0 \\ R+r \cos u \\ r \sin u \end{pmatrix}}_{\textrm{parametrization of circle we wish to rotate}} = \begin{pmatrix}  (R + r \cos u)(-\sin v) \\ (R + r \cos u)(\cos v) \\ r \sin u \end{pmatrix}$$
$\bullet$ Letting $R = 2$ and $r = 1$, we get the desired parametrization,
$$ \sigma(u,v) = (-(2+ \cos u) \sin v, (2+ \cos u) \cos v, \sin u)$$
Hence you can take $g=\gamma \circ \sigma$.
