Consider the $2$-torus $T$ obtained by revolving about the $z$-axis the circle $(x-2)^2+z^2=1$. I want to find the critical points of the map $f:T\to\Bbb{R}$ defined by $f(x,y,z)=x$.
The equation of $T$ is given by $(r-2)^2+z^2=1$ where $r=\sqrt{x^2+y^2}$. We need to compose $f$ with a diffeomorphism (parametrization) to see the critical points.
One parametrization would be $\phi:\Bbb{R}^2\to T$ defined by $\phi(y,z)=\left(\sqrt{1-(\sqrt{z^2+y^2}-2)^2},y,z\right)$. $\phi$ parametrizes some regions of $T$.
Now look at the map $f\circ\phi:\Bbb{R}^2\to\Bbb{R}, (y,z)\to\sqrt{1-(\sqrt{z^2+y^2}-2)^2}$. And $$d(f\circ\phi)=\left(\frac{-z(\sqrt{z^2+y^2}-2)^2}{...},\frac{-y(\sqrt{z^2+y^2}-2)^2}{...}\right)$$
So this gradient loses rank when both coordinates are $0$. This occurs when either $z=y=0$ or $\sqrt{z^2+y^2}=2$. But I can see only one critical point here, intuitively there are four. How do I find the rest of them?