Let $M$ be a compact riemannian manifold with boundary $\partial M\neq \varnothing$. I would like to show that there is some neighborhood $U$ of $\partial M$ which is diffeomorphic to $[0,a)\times \partial M$, for some (and thus for all) $a>0$. I would like also this to be true for a differentiable manifold (without having to consider a riemannian metric but since any differentiable manifold admits a riemannian metric, this is not a problem).
What I've tried: first, note that $\partial M$ is also compact, since it's closed in $M$. Also, since $M$ is compact, the injectivity radius $inj(M)>0$. Start with any real number $0<b<inj(M)$. Then define
$$\begin{array}{cccc}F:&[0,b]\times\partial M&\to&M\\ &(t,p)&\mapsto&\exp_p(t\nu(p)) \end{array},$$
where $\nu(p)$ is the inward unit vector normal to $\partial M$ through $p$.
Then I verified that $d F_{(0,p)}$ is an isomorphism, for all $p\in \partial M$ and then tried to use the inverse function theorem.
Using the compactness of $\partial M$ this gave me a finite collection of open sets $[0,a_i)\times U_i$, $U_i\subset \partial M$, such that $U_i$ covers $\partial M$ and the restrictions on any of these sets is a diffeomorphism.
Then I chose $a<\min\{a_i\}$.
I wished $F$ restricted to $[0,a)\times \partial M$ to be a diffeomorphism on its image. But this seems not to work...
For example, how do I avoid such "transversal" intersections? (the injectivity radius condition avoid this?)
Or, if $a$ is too big it seems that something like this can happen:
Since you have that the map $F$ restricted to $[0,a) \times \partial M$ is a local diffeomorphism, it suffices to see that it is injective order to conclude what you want.
Suppose there is no such $a>0$ that makes it injective. It follows that there exists sequences $x_n,y_n \in \partial M$, together with sequences $a_n, b_n$ which go to $0$, for which $F(a_n,x_n)=F(b_n,y_n)$ and $(a_n,x_n) \neq (b_n,y_n)$. Since $\partial M$ is compact, passing to subsequences if necessary, we have that $x_n \to x$ and $y_n \to y$.
Since $F(a_n,x_n) \to F(0,x)=x$ and $F(b_n,y_n) \to F(0,y)=y$, it follows that $x=y$. But since $F$ is locally a diffeomorphism (hence, injective) near $x$, we have a contradiction, because for $n$ sufficiently big $(a_n,x_n),(b_n,y_n)$ are inside a neighbourhood of $(x,0)$ on which $F$ was supposed to be one-to-one, but $F(a_n,x_n)=F(b_n,y_n)$ by construction.