# Tangent bundle of a noncompact surface

Let $\Sigma$ be a connected noncompact orientable surface. I'm not assuming that $\Sigma$ is of finite type or anything -- for instance, I'm allowing $\Sigma$ to be the $2$-sphere minus a Cantor set. I'm pretty sure that the tangent bundle of $\Sigma$ is trivial. Here are two pieces of evidence.

1. This is true if $\Sigma$ can be embedded in a closed surface. Proof : it is easy to see that a closed surface minus a point has a trivial tangent bundle, so $\Sigma$ must have one too.

2. If $U\Sigma$ is the unit tangent bundle, then we have the standard short exact sequence $$1 \longrightarrow \mathbb{Z} \longrightarrow \pi_1(U\Sigma) \longrightarrow \pi_1(\Sigma) \longrightarrow 1,$$ where the $\mathbb{Z}$ is the loop around the fiber. Since $\Sigma$ is noncompact, the group $\pi_1(\Sigma)$ is free, so this short exact sequence splits (just like you would have if the tangent bundle was trivial).

Does anyone know how to prove this in general? Thanks!

In general, the tangent bundle of a smooth $n$-manifold $M$ is classified by a (homotopy class of) map $\phi:M\rightarrow BO(n)$. The manifold $M$ is orientable iff there is a lift of this map to $BSO(n)$.
For $n=2$, and $M$ orientable, we see see that the tangent bundle to $M$ is classified by a map $\phi:M\rightarrow BSO(2) = BS^1 = \mathbb{C}P^\infty$. But homotopy classes of maps from $M$ into $\mathbb{C}P^\infty$ is canonically isomorphic to $H^2(M)$, and in this case this group is trivial since $M$ is not compact. It follows that there is a unique (up to isomorphism) rank 2 orientable vector bundle on $M$. Since both $TM$ and the trivial bundle are orientable, they must be isomorphic.