We know that given a manifold $M$ that is connected and compact, there exist a real function with a finite number of critical points, and with at least two.
Now if we consider a point $x\in M$, is it true that there exists a function $f:M\setminus \{x\} \rightarrow \mathbb{R} $ with no critical points?
If we take M to be a sphere of dimension n, I think it holds, because if we remove a point, what is left is diffeomorphic to $\mathbb{R}^n$ so that we compose this diffeomorphism with a function $\mathbb{R}^n \rightarrow \mathbb{R}$ with no critical points, like the sum of the components?
Is it true in the general case? I think not, but is there a necessary and sufficient condition for this to hold?
Edit: The same way, we get that $M\setminus \{x\}$ having an atlas with only one chart is a sufficient condition, since this is what happen for the sphere (or maybe : $M$ having a chart with domain $M\setminus \{x\}$ for any $x$ could be a better statement ?). I heard that this was a necessary condition, but I have no idea whether it is true, or why.
Hint: Given any finite subset $F$ of a connected smooth $m$-manifold $M$, there exists a compact subset $D\subset M$ containing $F$ and diffeomorphic to the closed disk $D^m$, see here for the case of 2-point subsets. Next, prove that $M-D$ is diffeomorphic to $M-\{x\}$, for any $x\in M$. Now, take $F$ to be the critical set of a smooth function $f: M\to {\mathbb R}$ and consider the restriction $f|_{M-D}$.