So let $M$ be a smooth closed orientable real manifold such that $M$ is parallelizable, i.e., the tangent space $TM$ of $M$ is trivial. From the triviality of $TM$ we get that the Stiefel-Whitney and Pontryagin classes are trivial and therefore the Stiefel-Whitney and Pontryagin numbers are all equal to $0$. We have the following amazing theorem of C.T.C. Wall:


**(Wall's theorem)** A closed orientable real manifold $M$ is the boundary of a compact oriented manifold (with boundary) iff its Stiefel-Whitney and Pontryagin numbers are trivial.


**Q**: Using the stronger additional assumption that $M$ is parallelizable, is it possible to give a simple proof that $M$ is the boundary of an oriented manifold? 



A simple proof that parallelizable oriented closed manifolds are oriented boundaries?
I sketch the proof of Buoncristiano and Hacon:


Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big $q$ (*i.e.*, one may chose a $\mathbb Z/2$-equivariant embedding $N\hookrightarrow S^q$). The boundary of $N$ is $M \times S^{m-1}$, and the involution on the boundary can be induced from that on $S^{m-1}$.


So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $\mathbb{RP}^q$, and the boundary of $N'$ is mapped into $\mathbb{RP}^{m-1} \subset \mathbb{RP}^q$. Take an $\mathbb{RP}^{q-m+1}$ in $\mathbb{RP}^q$ that intersects $\mathbb{RP}^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $\mathbb{RP}^{m-1}$ (this can be supposed) then $f^{-1}(\mathbb{RP}^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.


