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given a vector bundle $V \rightarrow N$ over a manifold $N$ and let's assume $N \hookrightarrow M$ is embedded into a manifold $M$ is there a way to extend $V$ to a bundle over $M$, i.e. is there a bundle $\tilde V \rightarrow M$ such that $\tilde V|N=V$. I hardly believe that this is true in general, e.g. if I look at $S^2 \hookrightarrow \mathbb{R}^3$ and its tangent bundle it seems to me, that there is no way how to extend $TS^2$ continously but are there criterions. Of course, if the bundle is trivial the answer is yes...

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Since any vector bundle is the pullback of the tautological bundle over a Grassmannian, the problem boils down to the problem of extending a map from $V$ to a Grassmannian, to a map from $M$ to a Grassmanian. This is what homotopy theory is good for (think obstruction theory). However, as far as I know There are many necessary conditions, but few necessary and sufficient conditions. – Liviu Nicolaescu Jul 24 '13 at 13:47
Here is a counterexample. Take a closed submanifold $N$ of a contractible manifold $M$(eg ${\bf R}^n$). Suppose that $N$ carries a non-trivial vector bundle $V$. Then any extension of $V$ to $M$ would be trivial (see Atiyah, K-theory, Lemma 1.4.4), which is not possible. – Damian Rössler Jul 24 '13 at 14:12
Here is something in the positive direction. If $N$ is compact then there exists an open (tubular) neighborhood $W$ of $N$, together with a projection map $\pi:W\to N$. So $V$ can be extended to $\pi^*V$ on $W$. – Damian Rössler Jul 24 '13 at 14:16
The Moebius band is a line bundle over the circle. A line bundle over a disc is trivial, so the Moebius band (as a bundle) does not extend from the circle to the disc. – Ryan Budney Jul 24 '13 at 19:01
Generalizing @DamianRössler's remark: if the embedding $N\subseteq M$ is a deformation retract, i.e., it admits a homotopy inverse, then the extension is possible. E.g., you can extend $TS^2$ to $\mathbb{R}^3\smallsetminus\{0\}$. – Giovanni Moreno Aug 20 '13 at 7:44

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