# Does every disc bundle come from a vector bundle?

Kosinski in his book "Differential Manifolds" states:

"A closed tubular neighbourhood $$E$$ of a compact submanifold $$M$$, which is closed neighbourhood in $$N$$, can always bee realised as a closed disc subbundle of a tubular neighbourhood of $$M$$".

He proves this statement like that:

"At first we reparametrize interior of $$E$$ to make it a vector bundle and then consider the unit disc subbundle $$E'$$. $$E'$$ can be expanded by an isotopy to cover $$E$$ and since it is compact this isotopy can be extended to an isotopy of $$N$$. The resulting isotopy will expand the interior of $$E$$ to a tubular neighbourhood of $$M$$ containing $$E$$ as a closed disc subbundle."

So, my question is: why is this reparametrization always possible?

P. S.: Kosinski defines a closed tubular neighbourhood as a subset of $$N$$, that has a stucture of a disc bundle over $$M$$ with $$M$$ as a zero section. Without any words about structure group. Must there be some kind of restrictions on it?

Equip $N$ with a Riemannian metric, and prove that the normal exponential map to any compact submanifold is a diffeomorphism onto its image on some closed $\epsilon$-neighborhood of the the zero section of the normal bundle. This is similar to proving that the exponential map is a local diffeomorphism, and boils down to computing the differential of the normal exponential map, and showing it the identity along the zero section (with appropriate identifications). There is of course a Riemannian-free proof as well but in the above the vector bundle structure is uniquely determined by the metric, which makes the conclusion somewhat cleaner.
• I still don't undetstand why exponential on that neighbourhood will be fiberwise map into interior of $E$. – Igor Ernst Jan 6 '17 at 16:10