Let $M$ be smooth manifold and $S^s \hookrightarrow M^m$ a smooth embedding. Denote the normal bundle of $S$ in $M$ via $N_{S/M}$.
A tubular neighbourhood (of $S$ in $M$) is a disc bundle over $S$ with smoothly varying radius; it is embedded in the normal bundle $N_{S/M}$ induced by an embedding $S \hookrightarrow M$. While the tubular neighbourhood is only locally resembling the normal bundle (in the embedded setting), abstractly it is isomorphic to $N_{S/M}$. Indeed,
- if $S$ is compact, the radius is fixed and we can use the same diffeomorphism from the disc to $\mathbb{R}^{m-s}$ in each fibre.
- for non-compact $S$, a more detailed argument is needed, I assume, but the statement remains true, nonetheless.
So, tubular neighbourhoods are exactly normal bundles of $S$ in $M$ (over varying embeddings)? Now, since every vector bundle $E$ over $S$ can be realized as $E \cong N_{S/E}$, does this mean:
Tubular neighbourhoods = Normal bundles = Vector bundles over $S$?
This seems wrong, but I keep confusing myself.
Edit: With "this equals that equals that", I mean the following: Let
- $\tau(S,M)$ be the set of tubular neighbourhoods of $S$ in $M$
- $\mathcal{N}(S,M)$ the set of normal bundles of $S$ in $M$, one for each embedding of $S$,
- $\operatorname{Vect}(S)$, the space of vector bundles over $S$
Are there bijections between $\tau(S,M)$, $\mathcal{N}(S,M)$ and $\operatorname{Vect}(S)$?
Edit 2: I guess $\tau(S,M) \to \mathcal{N}(S,M)$ is a bijection, but the bijection to the space of vector bundles over $S$ is false the reason being that $M$ may not embedded in $E$ although $S$ does. So, $N_{S/E} \not \cong N_{S/M}$. For example, let $S = [0,1]$ and $E = \mathbb{R} \times [0,1]$ and $M$ be the Mobius band.