$M$ is $k$ dimensional manifold connected, compact. Let $f:M \rightarrow \mathbb{R}$ smooth function. Prove there exists $x_0\in M$ that $f(x_0)=0$

Question

Let $M$ a $k$ dimensional manifold, connected and compact. Let $f: M \rightarrow \mathbb{R}$ a smooth function. Suppose there exists $x_1,x_2\in M$ such that $f(x_1)<0<f(x_2)$. Prove there exists $x_0\in M$ such that $f(x_0)=0$.

I know that since $f$ is smooth, then if $M=\cup_{\alpha}r_{\alpha}(V_{\alpha})$ where $(r_{\alpha},V_{\alpha})$ is a collection of maps, then to any ${\alpha}$, $f\circ r_{\alpha}$ is smooth.
Also, if $x_1,x_2\in r_{{\alpha}_{i}}$ then because $f\circ r_{\alpha}$ is continuous we get what we need from the intermediate value theorem, but I'm not sure what to do if $x_1\in r_{{\alpha}_{i}}$ and $x_2\in r_{{\alpha}_{j}}$ where $i\neq j$.

Would appreciate your help:)

Answer 1

Take a path $\gamma:[0, 1] \to M$ from $x_1$ to $x_2,$ prove that your function is continuous on $\gamma.$

Answer 2

$M$ is path connected, so consider a path connecting $x_1$ and $x_2$. Apply IVT to the pullback of the function to the interval defining the path.

Answer 3

An alternative approach (which does not use IVT explicitly, although the method of proof here is the same as that used to prove IVT assuming the connectedness of the closed interval $\left[0,1\right]$ has been already established): If the conclusion is false, then prove that $M = f^{-1}\left(-\infty,0\right)\cup f^{-1}\left(0,\infty\right)$ is a separation of $M$ (i.e., an expression of $M$ as a union of two disjoint nonempty open subsets).