# Is a regular point of a smooth function also a regular point for the restriction to a submanifold?

Suppose I have a smooth map $F \colon M \rightarrow N$ and an embedded submanifold $S \subset M$. Suppose that the differential of $F$ is non-singular at a point $z \in M$.

If $z$ also happens to lie in $S$, is it true that the differential of $F\vert_S$ (as a smooth map from $S$ to $N$) is non-singular at $z$?

$\textbf{Edit}$

So it seems this was a bad question to ask. What I'm trying to do is the following:

I have a smooth vector field $\widetilde{V} \colon \Bbb R^3 \rightarrow T\Bbb R^3$ which sends $$(x,y,z) \mapsto (-y,x,x^2+y^2+z^2-1).$$

Its restriction to $\Bbb S^2$ is then a smooth vector field $V$.

I want to somehow see that $dV_p$ is non-singular at the points $p =(0,0,\pm1)$.

• No, of course not. Take $S=z$. If $F^{-1}(F(z)) \subset S$, then it's true. (Why?) – user98602 Nov 5 '15 at 4:14
• Sorry I'm confused, what would be the tangent space $T_zS$ in this case? – Open Season Nov 5 '15 at 4:19
• It would be trivial in the case $S=z$. – user98602 Nov 5 '15 at 4:29