In the case of orientable closed $3$-manifolds, we have 4 ingredients that ensure parallelizability:


1a) Closed smooth $n$-manifolds have homotopy type of a $CW$-complex


1b) Closed smooth $n$-manifolds are Poincare duality spaces.


2) First Stiefel-Whitney class is $0$, from orientability


3) $BSU(2)$ is $3$-connected


1b) gives us Poincare duality and all of the familiar machinery of Steenrod squares and Wu's classes, which, combined with 2), provides for the vanishing of higher classes.


This, in turn, allows us to construct a map to $BSU(2)$. Throwing in 1a), we have that this map is null-homotopic QED.


For open 3-manifolds, I don't see why the computations with characteristic classes should still be valid (as 1b doesn't hold), but one can still show that the map to $BSU(2)$ is null-homotopic, provided it exists.


I am aware of Whitehead's paper <https://www.sciencedirect.com/science/article/pii/B9780080098722500260>, but I haven't read it, and I don't know if he uses obstruction theory, or something else.


Can the classical proof using obstruction theory still be salvaged in open case?


EDIT: Per Andy Putman's comment, I've replaced the false statement in 1b).



Are open orientable 3-manifolds parallelizable via obstruction theory?
I think you are saying: for a closed $3$-manifold, vanishing $w\_1$ implies vanishing $w\_2$ by Wu's relations, but is this still true if the manifold is not closed? The answer is yes.


For a compact manifold with boundary, you can form the double (union of two copies of $M$ along $\partial M$). This will again be orientable, therefore parallelizable, therefore so is $M$.


For a noncompact manifold without boundary, you can prove the vanishing of $w\_2$ by observing that $w\_2$ restricts to zero on each compact set (because of the previous paragraph). This forces $w\_2$ to be zero.


(This implication -- a cohomology class must be zero if it restricts to zero on every compact set -- is valid for homology with field coefficients, but not for integral cohomology.)


