Existence of a local transverse embedded submanifold for a flow Let $M$ be a smooth manifold and $\Phi$ the flow of a non-vanishing vector field. There always exists around every point $x$ at least locally an embedded submanifold $S_x$ that is transversal to $\Phi$ at $x$, right? How can one show this? I thought that somehow, since $\Phi$ foliates $M$, there is a foliated atlas, so that for any $x$ there is a chart $(U,\varphi_\perp,\varphi_\parallel)$ with $x\in U$ and $\{y: \varphi_\parallel(y)=\varphi_\parallel(x)\}$ is naturally transverse to $\Phi$.Would this go in the right direction?

You can suppose that $x$ is in the domain of a chart $U$ that you identify with an open subset of $R^n$ and $x=0$. Suppose that $X$ is the vector field which does not vanish, $X(x)$ is a vector of $R^n$, consider an hyperplane (which contains the origin) $H$ defined by $\alpha(u)=0$ which does not contain $X(x)$. The function defined on $U$ by $f(y) =\alpha(X(y))$ is continue. Since $f(x)\neq 0$, there exists an open interval $f(x)\in I$ and $0$ is not in $I$. $f^{-1}(I)=V$ is open since $f$ is continue. If, $y\in V\cap H, f(y)=\alpha(X(y))\neq 0 $.
