Let $f: M \longrightarrow \mathbb {R} $ be smooth and not singular. Prove that there is a $ X \in X (M) $ field so that it is $f$-related to $ \partial_t \in X(M)$
To show this fact, we must show equivalently that there is a field $X$ such that $X(f) = 1$.
My suggestion is to use the partition of unity.
The Existence Theorem of partition of unity guarantees me a subordinate unit partition, a family $$(\psi_{\alpha}) _{\alpha \in A}$$
such that $\sum _{\alpha \in A}(\psi_{\alpha}) = 1$
any suggestion?
Let $(U_i)_{i\in I}$ an open covering of $M$ by charts (identified to open subset of $\mathbb{R}^n$. Let $x\in M, i\in I$ such that $x\in U_i$. Since $f$ is regular, there exists $v_i\in T_{x}U_i$ such that $df_{x}(v_i)\neq 0$. Since the tangent space of $U_i$ is trivial, we can find a vector field $V_i$ defined on $U_i$. such that $V_i(x)=v_i$. The function defined on $U_i$ by $g_i(x)=df_x.V_i(x)$ is continuous, and does not vanish at $x_i$, this implies that there exists $W_{i,x}\subset U_i$ such that the restriction of $g_i$ to $W_{i,x}$ is $>0$. Write $X_i={1\over{g_i(x)}}V_i$ and take a partition of unity $\phi_i$ related to a subcover $(W'_i)$ of $W_{i,x})$ and write $X=\sum_i\phi_iX_i$.