Parallelisable three dimensional orientable manifolds and Stiefel's theorem On Steenrod's book "The topology of fibre bundles" by the end of page 203 and beginning of page 204, Steenrod claims that any orientable three dimensional manifold is parallelisable, and he cites an article by Stielfel: "Richtungsfelder und Fernparallelismus in n-dimensionalen Mannigfaltigkeiten".
I cannot read mathematics in German; thus my question is:

Does Stielfel's theorem about the parallelisability of orientable three manifolds applies to noncompact ones, or his proof relies on the assumption that the manifold is compact? 

If his proof depends on the manifold being compact,

Is it true that any noncompact and orientable three dimensional manifold is parallelisable?     

PVAL suggested a proof in his answer to this question: Elementary proof of the fact that any orientable 3-manifold is parallelizable . However, I should like to see a reference in the literature with more details than Steenrod's comments (in English, if it is the case that Stielfel's theorem holds for noncompact manifolds).
 A: Stiefel's article deals only with closed (= compact, without boundary) manifolds. The first paragraph of it says:

Die $n$-dimensionalen Mannigfaltigkeiten, welche in dieser Arbeit betrachtet werden, sollen geschlossen und stetig differenzierbar sein.

which roughly means:

The $n$-dimensional manifolds which will be considered in this work will be closed and continuously differentiable.

The theorem is nonetheless true for the other orientable $3$-manifolds as well. Actually, we have a stronger result.

Theorem (Whitehead, 1961). Let $M$ be a non-closed connected orientable $3$ manifold. Then there is an immersion $M  \to \mathbb
 R^3$.

The reference is Whitehead, The Immersion of an open 3-Manifold in Euclidean 3-space, Proc. London Math. Soc., 1961. The proof uses very different techniques than the standard one for the closed cases (the ones quoted in the question you mentioned). By puncturing a closed $3$-manifold, it is actually easy to deduce the closed case from the non-closed one.

Remarks on Kirby's proof. In Kirby's little The Topology of 4-Manifolds (which features a very small but quite brilliant chapter on 3-manifolds), one can find some remarks proving that the theorem for the closed case actually prove the general theorem. Let me summarise these remarks.
First, the "closed" theorem implies the "compact with boundary" (or "bounded" in a quite old but charming terminology) theorem: if $M$ is a compact manifold, its double $DM$ is a closed one, and it is oriented if $M$ is oriented. The triviality of $T_{DM}$ implies the triviality of its restriction $T_{M} = \left(T_{DM}
\right)_{|M}$.
Then, Kirby claims without proof that if a non-compact manifold $M$ has a non-trivial tangent bundle $T_M$, then the tangent bundle is already non-trivial in restriction to a "compact piece". Since every compact part of $M$ is included in a compact codimension-0 submanifold $N$, the previous step shows that such a situation is impossible.
I actually believe Kirby's claim, but I was unable to prove it. It turns out that more general statements are actually false, which surprised me a little. This proof and the subsequent question on the validity of what I've called Kirby's claim have already come up on MSE, but I haven't been able to find a definitive answer. Maybe it would be worth another question.
A: If you just want to deduce the noncompact case from the closed one, this requires little machinery.
Basically, you find first an exhaustion of $M$ with connected compact manifolds with boundary $M_k$. Then you inductively construct linearly independent vector fields $X,Y,Z$ on each $M_k$. Extending them from $M_k$ to $M_{k+1}$ is not trivial, and can be done e.g. taking an appropriate harmonic extension. See my answer to a very similar question here.
