Dimensionality/Combinatorial Argument for Parallelizability of 1-Manifolds It just occurred to me that all one-dimensional manifolds are parallelizable. We clearly have examples of many, many (many) manifolds that are not, but since any one-dimensional manifold can be written as a countable disjoint union of $\mathbb{S}^1$ or $\mathbb{R}$, and both of these spaces are parallelizable, then any one-dimensional manifold is as well.
I was wondering if there's anything in the classical literature or modern research literature that gives a proof of this simply based on dimensionality.  Is there a particular reason 1-manifolds all have this property even though it doesn't extend to higher dimensions?  Is there a particular property of manifolds this is indicative of?
 A: An idea is that you can put a Riemannian metric on $M$ and flow a unit vector through a geodesic, thus finding a global section of $TM$.
To make this rigorous, we need to prove the fact that any geodesic will be surjective on a (connected) $1$-manifold. To see this, define the equivalence relation $x \sim y$ if they belong to the trace of a same geodesic. It is easy to see that every equivalence class is open (this is due to the fact that the manifold has dimension $1$!). Therefore, there must be only one.
However, we are not done, since what we have is a map $I \to TM$. If the geodesic is also injective, then we are done. If it isn't, we can pick the moment it starts not being and see that it must come back and make a smooth loop (just pick the $\inf$ of the $t$ for which $\gamma(t) \neq \gamma(s)$ for all $s < t$ and use local charts to see that it must come back the right way). Since it will come back the right way, the map $M \to TM$ is well defined.
However, this is essentially a proof of classification of $1$-manifolds.
A: To complete Aloizio's answer, I'd like to point out that parallelizability quickly implies a full classification of 1-manifolds. Suppose $M$ is parallelizable. Then in each tangent space there is a unique vector $v$ corersponding to $1 \in \Bbb R$; construct the vector field $V$ that corresponds to 1 at each point. Flowing a point along this vector field gets you a map $\Bbb R \to M$; a standard argument shows it's surjective (the set of points in the image is open and closed). It's also a local diffeomorphism. If it's injective, this identifies the image with $\Bbb R$. Finish the proof in the case it's not injective!
