# Global section of a smooth fiber bundle is an embedding

Given a smooth fiber bundle E over a manifold $$M$$, is any global section $$s: M \to$$ E is an embedding?

Here I always mean a locally trivial bundle. I think the above assertion is correct. If so, does this really imply that a smooth vector field over $$M$$ is actually the same as an embedding of $$M$$ in the tangent bundle T$$M$$?

• Yes, if by section you mean smooth section. Commented Sep 23, 2021 at 13:16

Is any global section $$s \colon M \to E$$ an embedding?
The answer is yes, provided that the section is smooth. To see this, check that if $$s\colon M \to E$$ is a smooth section, then $$s(M)$$ is indeed a smooth submanifold and that the restriction of the projection $$\pi\colon E \to M$$ to $$s(M) \subset E$$ is a smooth inverse.
If so, does this really imply that a smooth vector field over M is actually the same as an embedding of $$M$$ in the tangent bundle $$TM$$?
The answer is no. It is possible that there exist plenty of embeddings $$f\colon M \to TM$$ such that $$\pi \circ f \neq \mathrm{id}_{M}$$. In other words, nothing forces any embedding $$M \to TM$$ to respect base-points.
For instance, $$T\mathbb{S}^1 = \mathbb{S}^1\times \mathbb{R}$$, and $$f(e^{i\theta}) = (e^{i(\theta +\pi)},0)$$ is not a vector field while it is an embedding of the circle into its tangent bundle.
• Sir, according to the definition of smooth embedding mentioned in Lee, Introduction to smooth manifolds, 2nd edition, it is a smooth immersion which is also a topological embedding i.e a homeomorphism onto its image. So, according to it, given a smooth map $f \colon M \rightarrow N$ (not necessrily smooth fibre bundle), any smooth section $s$ of $f$ should be an embedding, as the section $s$ is always an injective immersion. Commented Jul 9, 2023 at 13:23