Integral curves of a vector field not tangent to an embedded submanifold $S$

Question

I got stuck on a problem. Let $M$ be a smooth $n-$dimensional manifold and let $S$ be a compact embedded submanifold. Suppose $V$ is nowhere tangent to $S.$ Prove that there exists $\epsilon>0$ such that the flow of $V$ restrict to a smooth embedding
$$\Phi:(-\epsilon,\epsilon)\times S \to M.$$ Firstly, by the compactness of $S,$ we find a positive $\epsilon>0$ for which all the integral curves with initial conditions in $S$ are defined on $(-\epsilon,\epsilon).$ The differential of $\Phi$ if I am not mistaken should be of this form: $$(\frac{\partial}{\partial t}, di_{S}),$$ where $i_{S}$ is the inclusion of $S$ in $M,$ practically by definition this should have maximum rank.
The only part I'm not sure is on how to prove that $\Phi$ is open.
Since $S$ is embedded for every point $y \in S$ there is a chart $(U,\varphi)$ in $M$ in which $U\cap S$ is diffeomorphic under $\varphi$ to a set of the form
$$\{ x \in \mathbb{R}^{N} : (x^1,\dots,x^{k},\,0, \dots,\,0) \}.$$ Since $V$ is nowhere tangent to $S$ it's not restrictive to suppose the differential of $\varphi$ sends $V$ to the constant vector field $e_{k+1}.$ So the integral curves with initial conditions on $S$ should all be of the form
$$\gamma(t)=(x^1,\dots,x^{k},t,\dots,0).$$ So the set $\Phi(S\times (-\epsilon,\epsilon))$ is diffeomorphic to the following relative open set of $\mathbb{R}^n$
$$ \{(x^1,\dots,x^{k},t,\dots,0): ((x^1,\dots,x^{k}) \in \mathbb{R}^{k+1}, \, t \in (-\epsilon,\epsilon) \},$$ so the map $\Phi$ is open on is image and we are done.
Is this reasoning correct? Where should I be more precise?

Answer 1

You are missing a bit more.

Let $F: M\times {\mathbb R}\to M$ be the (partially defined) flow of the vector field $V$. By compactness of $S$, there exists $\epsilon>0$ such that you have a well-defined map $$ \Phi: S\times (-\epsilon,\epsilon)\to M, \Phi(x,t)=F(x,t), x\in S. $$ As you correctly noted, the map $\Phi$ is an immersion at each point of $S\times \{0\}$. Hence, taking $\epsilon>0$ small enough, you get that $\Phi$ is an immersion. [Indeed, immersion means injectivity of the derivative. But if the derivative is injective at one point, then it is also injective at all nearby points. Compactness of $S$ then implies the existence of $\epsilon>0$ such that $\Phi$ defined above is an immersion.]

Thus, for each $x\in S$ there exists $\epsilon_x\in (0,\epsilon)$ an a neighborhood $U_x$ of $x$ in $S$ such that the restriction $$ \Phi: U_x\times (-\epsilon_x, \epsilon_x)\to M $$ is 1-1. Since $\Phi$ restrict to the (identity) embedding on $S\times \{0\}$, for each pair $(x,y)\in S\times S$ there exists $\epsilon_{xy}\in (0,\epsilon)$ and a product neighborhood $U_{x,y}\times U_{y,x}$ of $(x,y)$ in $S\times S$ such that the restriction $$ \Phi: (U_{x,y}\cup U_{y,x})\times (-\epsilon_{xy}, \epsilon_{xy})\to M $$ is 1-1. [Indeed, the fact that $\Phi$ is an immersion implies that for each $x\in S$ there exists a product neighborhood $U_x\times (-\epsilon_x, \epsilon_y)$ of $(x,0)$ such that $\Phi$ restricted to this neighborhood is an embedding. Now, if $x=y$ then we can take $U_{x,y}= U_{y,x}=U_x$ and $\epsilon_{xy}=\epsilon_x$. Suppose that $x\ne y$. Then, after shrinking $U_x, U_y$ if necessary, we can find disjoint neighborhoods $V_x, V_y$ of $x, y$ in $M$ and $\epsilon_{xy}>0$ such that $$ \Phi(U_x\times (-\epsilon_{xy}, \epsilon_{xy}))\subset V_x,$$ $$ \Phi(U_y\times (-\epsilon_{xy}, \epsilon_{xy}))\subset V_y.$$ This is a direct consequence of the fact that $\Phi$ restricted to $S$ is the identity map. Lastly, injectivity of $\Phi$ on each $U_x\times (-\epsilon_{xy},\epsilon_{xy})$, $U_y\times (-\epsilon_{xy}, \epsilon_{xy})$ implies the claimed injectivity of $\Phi$ on the union of these two open subsets. Hence, we can take $$ U_{xy}=U_x, U_{y,x}=U_y, $$ thereby proving the claim.]

By compactness of $S$, we can find a finite subcover of $S\times S$ by these product neighborhoods. Hence, there exists $\delta\in (0,\epsilon)$ such that $$ \Phi: S\times (-\delta,\delta)\to M $$ is 1-1. Since $S\times [-\delta/2,\delta/2]$ is compact and $M$ is Hausdorff, the injective continuous map $$ \Phi: S\times [-\delta/2,\delta/2]\to M $$ is a homeomorphism to its image. Thus, the restriction $$ \Phi: S\times (-\delta/2,\delta/2)\to M $$ is an immersion and a homeomorphism to its image, hence, an (smooth) embedding. qed