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For an orientable compact manifold $M$ with boundary, is its boudary orientable as well?

If $M$ is smooth, the conclusion obviously holds. For the general manifold, my plan is to use long exact sequence $H_n(M,\partial M)\rightarrow H_{n-1}(\partial M) $ to show the latter homology group is $\mathbb Z$, but I stuck there.

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closed as off topic by Ryan Budney, Mark Grant, Chris Gerig, Anton Petrunin, Fernando Muro Jul 31 '12 at 11:30

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My guess is that whatever argument you have in mind which makes it obvious in the smooth case will have a counterpart in the nonsmooth case. – Paul Siegel Jul 31 '12 at 5:08
The tools in the topological category are a little heavier though. You basically need to know the boundary has a collar - a product neighbourhood. – Ryan Budney Jul 31 '12 at 5:34
I can recommend several sources, G. Bredon Topology and Geometry, or Dold's lecture on algebraic topology. – Liviu Nicolaescu Jul 31 '12 at 8:48
Let $N^{n-1}$ be a boundary component of oriented $n$-manifold $M$, and let $x\in N$. Complete a basis $B$ for $T_x N$ to a basis $B_M$ for $T_x M$ by appending a vector $v$ (possible because the boundary has a collar, as Ryan Budney pointed out). Vector $v$ points either into or out of $M$ (again, because the boundary has a collar). Say $B$ is positive if $B\cup v$ is positive whenever $v$ points inwards, with respect to the orientation of $M$. And voilá, you've oriented $N$. – Daniel Moskovich Jul 31 '12 at 11:33
I'm pretty sure you don't need global collars in either the smooth or topological cases. All you need is a covering of $\partial M$ by open subsets of $M$ which are $n$-dimensional ``half-balls'' and which intersect $\partial M$ in $n-1$-dimensional balls. – Lee Mosher Jul 31 '12 at 13:56