Show that a critical point of one function can't be a critical point of another function

Question

Let $S$ be compact surface (i.e. $2d-$manifold) embedded in $\mathbb{R}^3$ with system of coordinates $(x,y,z)$. Let $\pi$ be the projection $(x,y,z)\rightarrow(x,y)$ and let $p \in S$ be a critical point of $\pi$.

Show that $p$ is not a critical point of $z-ax-by$

This is from a badly-written past exam which does not specify what $a$ and $b$ are. I assume they are reals, and it might be that the assertion is to be proven "for almost all $(a,b) \in \mathbb{R}^2"$ (but I am not sure, maybe this works for all reals a,b unconditionally)

This is part of a larger exercise which I have solved except for this question on which I am completely stuck. I just can't see how being a critical point of $\pi$ allows to draw such conclusions.

If I understood correctly, if $\varphi:\mathbb{R}^3 \rightarrow S$ is chart of a neighborhood of the manifold $S$ (or reverse chart, I don't remember the name), then I basically have to prove that:

$(x,y,z)\rightarrow (\begin{matrix} \partial_x\varphi_1(x,y,z) & \partial_y\varphi_1(x,y,z) & \partial_z\varphi_1(x,y,z) \\ \partial_x\varphi_2(x,y,z) & \partial_y\varphi_2(x,y,z) & \partial_z\varphi_2(x,y,z) \end{matrix})$ has rank $\leq1$ (i.e. is not surjective) at $p$

implies

$(x,y,z)\rightarrow \partial_i\varphi_3(x,y,z)-a\partial_i\varphi_1(x,y,z)-b\partial_i\varphi_2(x,y,z)$

is non-zero at $p$ for at least one of $i \in${$x,y,z$} (this should suffice to show that the differential of $z-ax-by$ is surjective at $p$)

But for example the first map could have rank $0$ (i.e. all the partial derivatives are $0$ in the matrix) and from there I should be able to deduce that one of $\partial_x \varphi_3$, $\partial_y \varphi_3$, $\partial_z \varphi_3$ is non-zero: how?

Any help or hint is welcome

Answer 1

It works for all $a,b\in\mathbb{R}$, and in fact works for all smooth functions $a,b$ (I'll leave this last part for you).

Assuming $a,b\in\mathbb{R}$. Just note that $\psi=(x,y,z-ax-by)$ is just another (global) chart for $\mathbb{R}^3$ compatible with $(x,y,z)$. Remember $S$ being embedded implies the derivative of all charts of $\mathbb{R}^3$, in particular $d\psi$, must restricts to have full rank at every point of $S$, i.e. rank 2. Note that $d\psi=d\pi\oplus d(z-ax-by)$ and since $d\pi(p)\colon T_pS^2\to\mathbb{R}^2$ does not have rank 2, $d(z-ax-by)(p)$ must not be zero.