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Let $M$ a topological manifold of dimension $n$ with boundary $\partial M$.

We define $M$ to be orientable if $M- \partial M$ is orientable. Here when I say orientable, I mean there is a locally coherent choice $\mu_x$ of generators of $H_n(M,M-x) \cong \mathbb{Z}$,so that every point $x \in M$ has a $U \cong \mathbb{R}^n \ni x$ such that for every $y \in B(0,1) \subset U$, we have $\mu_y$ comes from the isomorphism $$H_n(M,M-x) \cong H_n(M,M-B(0,1) \cong \mathbb{Z})$$

There is an exercise on Hatcher that says that this implies that $\partial M$ is orientable as an $n-1$ manifold without boundary.

I tried to use the fact that $\partial M$ has a collar neighbourhood to "induce " the orientation on $M$ to an orientation on $\partial M$ but I did not really achieve anything.

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Let $n$ be the unit inward normal to $\partial M$. If you choose a basís $e_1, \ldots e_{n-1}$ of the tangent space in some point you can, because $M$ is oriented, decide whether $e_1, \ldots e_{n-1}, n$ is a positively or negatively oriented basis.

This observation allows you to define an orientation on $\partial M$. (E.g. by defining that $e_1, \ldots e_{n-1}$ has positive orientation iff $e_1, \ldots e_{n-1}, n$ has).

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  • $\begingroup$ I'm assuming $M$ is just a topological space(with no tangent space etc.) or,at least,this is the definition I had of orientation of a manifold,but in my question was really unclear. I'm gonna edit $\endgroup$ – Tommaso Scognamiglio Nov 11 '18 at 18:05

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