Condition for a submanifold to have a trivial normal bundle

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.

The sub-assertion

... the normal bundle of $$S$$ in $$M$$ and the normal bundle of $$S$$ in $$f(S \times \mathbb R^{n-k})$$ are the same...

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

... the derivative $$Df : T(S \times \mathbb R^{n-k}) \to TM$$ restricts to an isomorphism between the normal bundle of $$S \times 0$$ in $$S \times \mathbb R^{n-k}$$ and the normal bundle of $$f(S \times 0)$$ in $$M$$...

And now you can go on to prove it.

• Is this the tubular neighborhood theorem? Nov 12, 2020 at 14:06
• No, in this question the existence of a product tubular neighborhood is already being assumed. The tubular neighborhood theorem is about the proof of existence of tubular neighborhoods in general. Nov 12, 2020 at 14:59