Can a smooth manifold be embedded into its tangent bundle? Given a smooth $n$-dimensional manifold $M$, one can always find an immersion into its tangent bundle $TM$ by looking at the zero-section, i.e. the map that sends $p\in M$ to $(p,0)\in TM$. One can see this is an immersion for example by writing everything in local coordinates and checking that this is locally an inclusion.
Can it be proved that this is also an embedding? (i.e. a homeomorphism onto its image). It seems intuitively clear to me that this should be true because "the tangent bundle has a copy of $M$ inside" but I can't find a way to prove or disprove it. Could you help me?

(EDIT: This is false! leaving it up cause there's a useful comment.) Let $\phi: X \to Y$ be an injective map of topological spaces which you wish to check is a homeomorphism onto its image. You can check this locally: if $\{U_i\}$ is a cover of $X$ such that $\phi|_{U_i}$ is a homeomorphism onto its image for all $i$, then $\phi$ is a homeomorphism onto its image.
This fact works out great in your case, where $X$ is a manifold, since you can take $\{U_i\}$ to be a coordinate chart, and you are reduced to proving that $\mathbb{R}^n$ is homeomorphic to its image in its (trivial) tangent bundle $\mathbb{R}^n \times \mathbb{R}^n$, which is true.
