Let $M^{m}$ be a connected smooth manifold and let $\Sigma^{m-1} \subset M$ be a smooth hypersurface that seperates M into two connected components. Prove that there exists a smooth vector field $X$ on M, such that $X(p) \not \in T_p(\Sigma^{m-1})\: \forall p \in \Sigma^{m-1}$.
My ideas so far:
I chose countably many slice-charts $\{(U_i,\phi_i)\}_{i\in I}$ that cover M, i.e. such that $\phi_i(U_i \cap \Sigma^{m-1})=\mathbb{R}^{m-1}\times\{0\}$.
That means $\phi_i(U_i \setminus \Sigma^{m-1}) = \mathbb{R}^m_+ \cup \mathbb{R}^m_-. $
Therefore $U_i$ itself consists of two connected components $C_i^+, C_i^-$.
I then chose $X = \pm \frac{\partial}{\partial{x^{m}}}\cdot \psi(x)$. Plus in $C_i^+$ and minus in $C_i^-$ where $\psi$ is a smooth bump function supported in $M\setminus \Sigma$. But then I didn't know how to expand this beyond one chart. Maybe using a partition of unity ?
Thanks in advance for any help
Thank you @Didier I want to try to solve the problem now:\
First I cover $\Sigma$ with countably many slice-charts $\{(U_i,\phi_i)\}_{i \in I}$ such that $\phi_i(U_i \cap \Sigma) = \mathbb{R}^{m-1} \times \{0\} \:\forall i \in I$.
That means $\phi_i(U_i \setminus \Sigma^{m-1}) = \mathbb{R}^m_+ \cup \mathbb{R}^m_-. $
Therefore $U_i$ itself consists of two connected components $C_i^+, C_i^-$. With $C_i^+$ being mapped to $\mathbb{R}^m_+$ and $C_i^-$ to $\mathbb{R}^{m}_-$. Further all components on one side of the hyperplane should get a + and all components on the other side a minus sign. This can be achieved by inverting the last component $\phi_i^m$ if necessary.
I define the vector-field $X_i = \frac{\partial}{\partial{x^m}}$ locally. (Pointing from $C_i^-$ to $C_i^+$).
I then take a partition of unity $(\rho_i)_{i \in I}$ subordinate to the given cover of $\Sigma$. Then $X:=\sum_{i \in I}\rho_i\cdot X_i$ is clearly a smooth global vector field. I check that it is nowhere tangent to $\Sigma$. Let $p \in \Sigma$. Since the supports of the $\rho_i$ are locally-finite, there exists an open neighbourhood U of p such that only finitely many $\rho_1,...,\rho_n$ are nonzero in U. Say now the chart $\phi_1$ contains p. Since $\phi_{1*}$ is an isomorphism of tangent-spaces, it suffices to show that the m-th component of $\phi_{1*}(X)$ is nonzero. $\phi_{1*}(X) = \sum_{i=1}^{n}\rho_i(p)\cdot \phi_{1*}(X_i)$ where all $\rho_i(p) \geq 0$. I can write $\phi_{1*}(X_i) = \phi_{1*}(\phi_{i*}^{-1}\cdot \phi_{i*}(X_i)) = \phi_{1*}\cdot\phi_{i*}^{-1}(e_m)$. Now it should follow that the m-th component of $\phi_{1*}\cdot\phi_{i*}^{-1}(e_m) \: $is > $0$ with the chosen orientations.(And that should finish the proof). But I am struggling to show this rigorously.
Thanks in advance for any help.