This is from John Lee's Introduction to Smooth Manifolds. In the first sentence of the second paragraph of the proof, I cannot show how we can shrink $S$ so that $V$ is nowhere tangent to $S$.
So if $S$ is an embedded hypersurface, and $V_p \notin T_p S$, then we can find some smooth coordinates $(U, x^i)$ such that $\partial/ \partial x^1 \notin T_p S$, but $V^1(p) \neq 0$. By smoothness, we can find some neighborhood of $p$ such that $V^1(q) \neq 0$ for all $q$ in this neighborhood. But can I just replace $S$ by the intersection with this neighborhood and still have an embedded hypersurface of $M$?
Best Answer
This should work. If $U_0$ is a neighborhood of $p$ such that $V^1(q) \neq 0$ for all $q \in U_0$ then $U_0 \cap S$ is open in $S$ ($S$ has the subspace topology since it is embedded), so $U_0 \cap S$ is an open submanifold of $S$. So we have a composition of embeddings $U_0 \cap S \to S \to M$ which shows that $U_0 \cap S$ is embedded in $M$.