# Existence of boundary cylindrical neighborhood for a compact manifold

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. 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:

• What you want to show here follows from the continuity of the injectivity radius, which (in a compact manifold) implies a lower bound for it. See here for some thoughts on this and for some references, too: mathoverflow.net/questions/283467/… – Thomas Oct 20 '18 at 15:46

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.