Show that the boundary of an \(R\)-orientable manifold is also \(R\)-orientable.
Solution
Let \(M^n\) be \(R\)-orientable. A collar identifies a neighborhood of the boundary with \[ \partial M\times[0,\varepsilon). \] For \(x\in\partial M\), choose a small half-ball neighborhood \(B_+^n\) in \(M\). The local fundamental class of \(M\) determines a generator of \[ H_n(B_+^n,B_+^n-\{x\};R) \] in the relative form appropriate to a boundary point. The connecting homomorphism for the local pair sends this generator to a generator of \[ H_{n-1}(\partial M,\partial M-\{x\};R). \] Use this image as the local orientation of \(\partial M\) at \(x\).
Naturality of the connecting homomorphism shows that these boundary local classes vary locally constantly whenever the local orientation classes of \(M\) do. Hence they define an \(R\)-orientation of \(\partial M\).
Equivalently, in oriented collar charts this is the usual rule that an oriented basis of the boundary is chosen so that the outward normal followed by that basis gives the orientation of \(M\). Thus \[ \boxed{\partial M\text{ is }R\text{-orientable}.} \]