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.

  • $\begingroup$ I still don't undetstand why exponential on that neighbourhood will be fiberwise map into interior of $E$. $\endgroup$ – Igor Ernst Jan 6 '17 at 16:10
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    $\begingroup$ Oh, I see that I misunderstood your question. I was explaining the existence of a closed tubular neighborhood. After looking at the definition of a closed tubular neighborhood in Kosinski's book (p.46) I agree that it looks sloppy. He never defines a disk bundle (as far as I can see), but I suspect he means a linear disk bundle. $\endgroup$ – Igor Belegradek Jan 6 '17 at 19:06

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