Let $f(x,y,z)=z$ be the height function on a Torus in $\mathbb{R}^3$.
Computing the representation of the differential of $f$ with respect to these coordinates, we get $f_{*,p}=\left(\begin{array}{c c c}0& 0& 1\end{array}\right)$, which has rank $1$ (and this rank is coordinate independent). Therefore, every point in the Torus should be a regular point. However, Loring Tu's book states that there are 4 critical points... Where did I go wrong?
Addendum for future reference: In page 91, the proposition 8.11 is written with respect to the charts of the Torus. The Torus is of dimension 2, see wikipedia link, so our charts with domain the Torus are not $x,y,z$ but $\phi(x,y,z)=(\theta,\varphi)$, such that $\theta,\varphi$ parametrize the Torus as in the wikipedia link. (We must switch the z with x in the wiki's parametrization to get the vertical Torus in Tu's book, i.e. $z=(R+r\cos(\theta))\cos(\varphi)$ ). We must think with respect to the parametrization to apply proposition 8.11.
Let $\mathbb{T}$ be the torus. Your matrix $\begin{pmatrix}0&0&1\end{pmatrix}$ is the differential of the map $\pi:\mathbb{R}^3\to \mathbb{R}$ given by $(x,y,z)\mapsto z$ (which is linear hence equals its own differential). Your function $f:\mathbb{T}\to \mathbb{R}$ is the composition of the inclusion $\iota:\mathbb{T}\to\mathbb{R}^3$ and $\pi.$ Hence $df_p=\pi\circ d\iota_p:T_p\mathbb{T}\to \mathbb{R}.$ So $df_p$ fail to be surjective when $d\iota_p(T_p\mathbb{R})$ is contained in the kernel of $\pi,$ which is the span of $(1,0,0)$ and $(0,1,0),$ i.e. the $xy$ plane. So $p$ is critical when its tangent plane is horizontal.