Existence of a Tubular neighborhood of a hypersurface

Question

Suppose $H$ is a co-dimension 1 embedded submanifold of $M$. Let $X$ be a vector field on $M$ such that $\forall x \in H$, $T_xM = T_xH \oplus X_x$. Now I want to show that there exixts an open set $U$ such that $N \subset U$, $U$ diffeo to $H \times (-\epsilon, \epsilon)$.

The proof is supposed to go as follows:

Let $\phi(x,t): M \times (-a,a) \to M$ be an integral flow of $X$, such that $\phi(x,0) = id_M$, $\phi(\cdot,t)$ is a diffeo for each $t$, and $\frac{\partial}{\partial t} \phi = X$. Then by the inverse function theorem, $\phi$ is a diffeo on a neighborhood of $U \times V$ of each $(x,0)$.

How does the inverse function theorem work? I understand that the construction is trying to make that the differential of the map land in TH and X separately. But there is no guarantee that the first part is in TH. Why can the local diffeo be elevated to a neighborhood containing the entire H?

Answer 1

Here's a sketch of the argument that when $H$ is compact, we can find $\epsilon>0$ so that $\phi$ is injective on $X\times (-\epsilon,\epsilon)$. Suppose not. Then for each $n\in\Bbb N$ we have $x_n,x'_n\in H$ and $|t_n|,|t'_n|<1/n$ so that $\phi(x_n,t_n)=\phi(x'_n,t'_n)$. By compactness of $H$, we can find convergent subsequences $x_{n_k}\to x$ and $x'_{n_k}\to x'$. Then from $\phi(x_{n_k},t_{n_k})=\phi(x'_{n_k},t'_{n_k})$ we infer that $\phi(x,0) = \phi(x',0)$, and so $x=x'$. But, by the inverse function theorem, $\phi$ is a bijection on a neighborhood of $(x,0)$. This contradiction completes the proof.