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Let $(M, \partial M)$ be an orientable $n$-dimensional topological manifold with boundary. Suppose that $(N, \partial N)$ is an $n$-dimensional topological manifold with boundary and $N \subset M$.

If $N- \partial N$ is an open subset of $M$ then I believe that $N$ is also orientable since the orientation of $M$ (which can be viewed compatible choice of local orientations at each point $x \in M - \partial M$) should induce an orientation of $N$.

If $N - \partial N$ is not an open subset of $M$ then can I say if $N$ is orientable or not?

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  • $\begingroup$ If $\partial M$ is isotopic to $\partial N$, certainly. $\endgroup$
    – Loki Clock
    Mar 14 '13 at 21:14
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Intuitively, yes, because orientability is a local property in the first place. So I'm assuming the tangent bundle of N is a subbundle of the tangent bundle of M. You can have nonorientable subspaces of an orientable manifold because orientability is defined relative to the dimension of the space.

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