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

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.

1. 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.

2. 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.

• Right inverses of surjective maps (sections) are, by definition, injective, yes. Nov 5, 2019 at 10:18
• not by definition, but yes
– user661541
Nov 5, 2019 at 10:29

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.
• You said $f\colon\Bbb R^n\to\Bbb R^n$ and not $f\colon\Bbb R^n\to\Bbb R^{2n}$. So the undergraduate vector field is more of $f\colon\Bbb R^n\to\Bbb R^n$ which is different from the one in differential geometry which is more of $f\colon\Bbb R^n\to\Bbb R^{2n}$? Nov 6, 2019 at 0:31
• As I already said, the vector field in differential topology is of the form $\sigma(x)=(x,f(x))$, i.e., the graph of the calculus student's vector field (in order to keep track of the base point). Nov 6, 2019 at 0:40
• I still won't think of a section of the trivial bundle as a map $\Bbb R^n\to\Bbb R^{2n}$, because it is specifically the graph of a function. Of course, as such, it is a special mapping as you specify. Nov 18, 2019 at 16:45