Mean Value Theorem Like Statement About Manifolds Let $S$ be a connected $m$-dimensional embedded subamnifold in $\mathbf R^m\times \mathbf R^n$. Suppose that $S$ intersects $\{\mathbf 0\}\times \mathbf R^n$ at two different points. 

Conjecture: Then there is a point $\mathbf p$ on $S$ at which $S$ is not transverse to $\mathbf R^n$, that is, $T_{\mathbf p}S\cap \mathbf R^n\neq \{\mathbf 0\}$.

(One may assume that $S$ admits a global chart because right now that is the most important case for me.)
The statement is true if $m=n=1$, for then the situation is almost like that in Rolle's Theorem.
I came up with the above "conjecture" while trying to prove the implicit function theorem.
Can somebody help?
Thanks.
 A: Such a statement is true for compact manifolds even without your assumption. Think of $\Bbb R^{m+n}$ as $\Bbb R\times \Bbb R^{m+n-1} $ and let $f$ denote the smooth embedding of $S$ into $\Bbb R^{m+n}$. Then,  $f(x)=(g(x),h(x))$ for $g:S \to \Bbb R$ and $h: S\to \Bbb R^{m+n-1}$. As $S$ is compact, $g$ has a maximum value and hence a critical point $p$.  At $p$, we have $T_pS\subset 0\times \Bbb R^{m+n-1}$ of dimension $m$ and therefore intersects $0 \times \Bbb R^n$ non-trivially. 
To construct a non-compact counter-example take $f:S^2\rightarrow \Bbb R \times \Bbb R^3$, $f(x)=(g(x),h(x))$ where $g$ is the standard height function after $S^2$ is rotated by $\pi/4$ (in any direction) and $h$ is the standard embedding $S^2 \to \Bbb R^3$. $S$ intersects $0 \times \Bbb R^2$ at 2 points transversely. There are only 4 points where $(T_p S)\cap (0 \times \Bbb R^2)$ is non-trivial (the two critical points of $g$ and the points corresponding to the west and east pole with respect to $h$), so one can choose a smooth open 2-disk in $S$ missing those 4 points containing $S\cap (0 \times \Bbb R^2)$. Such an open 2-disk is the desired submanifold.
