Is the following assertion true?
Suppose $M$ is a smooth manifold of dimension $n$, and $S$ an embedded submanifold of dimension $k$. If there is an embedding $S\times \Bbb R^{n-k}\to M$ (with $S\times 0$ corresponding to $S$ in the obvious way), then the normal bundle of $S$ in $M$ is trivial.
I think this can be proved as follows, but I'm not sure about my argument: If $f$ is such an embedding, then the image of $f$ should be an open subset of $M$, so the normal bundle of $S$ in $M$ and the normal bundle of $S$ in $f(S\times \Bbb R^{n-k})$ are the same, but the latter is a trivial bundle.
Best Answer
The sub-assertion
is false as stated literally. They aren't the "same". But it contains a grain of truth which you can then use to turn into a proof.
The key thing to keep in mind is that whenever you are tempted to use the "s" word, you should instead ask yourself: What kind of isomorphism are we talking about here?
The true statement which you should have in place of the one above is
And now you can go on to prove it.