Are vector fields injective? (For higher maths: Are sections of a bundle injective?)

Question

For vector fields: 3 reasons why I think vector fields are injective

1. (This is from the undergraduate point of view) For a vector field $X: \mathbb R^n \to \mathbb R^{2n}$ (my undergraduate calculus textbook doesn't give a range for "vector fields", but we know from differential geometry that $\mathbb R^n$ is parallelisable tangent bundle, i.e. $T\mathbb R^n$ is isomorphic as a tangent bundle to $\mathbb R^n \times \mathbb R^n$) , I really don't see a way, geometrically, we have the same $2n$-vector $X_p = X_q$ for different $n$-points $p, q \in \mathbb R^n$, since $X_p$ is supposed to emanate from $p$.

The next two use differential geometry: Let $M$ be a smooth manifold.

2. Let $M$ have tangent bundle $\pi: TM \to M$. Consider a vector field, which need not be smooth, $X: M \to TM$, we have $\pi \circ X$ as the identity of $M$, which is bijective and thus injective. By this, $X$ is injective.

3. Alternatively, I think we don't need any (explicit) notion of tangent bundle. Let $X$ be a vector field on $M$ (a map whose domain is $M$ and such that for each $p$ in $M$, $X_p \in T_pM$). Let $p,q \in M$. We have $X_p \in T_pM$ and $X_q \in T_qM$. We don't really have that $T_qM$ and $T_pM$ intersect anywhere as sets even though as $\mathbb R-$vector spaces they are isomorphic to $\mathbb R^n$, where $n$ is the dimension of $M$. Saying that $X_p$ and $X_q$ are equal implies $T_pM$ and $T_qM$ intersect somewhere (namely at $X_p$), which is nonsensical unless $T_pM = T_qM$, which is also nonsensical unless $p=q$.

For vector bundles in general: Are sections (need not be smooth) injective? I think indeed so generalising from the case of tangent bundles.

Thanks in advance!

Answer 1

If you think of a vector field on $\Bbb R^n$ as a map $f\colon\Bbb R^n\to\Bbb R^n$, of course it need not be injective (the $0$-vector field is the easiest counterexample). If, however, you think of a vector field as a section of $T\Bbb R^n$, then, as you already proved, it's injective. We can see this directly because we're then looking at the map $\sigma\colon\Bbb R^n\to\Bbb R^{2n}$ given by $\sigma(x)=(x,f(x))$, and the identity map in the first coordinate makes the mapping injective.