Global section of a smooth fiber bundle is an embedding

Question

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$M$?

Thank you in advance!

Answer 1

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.