# Orientable manifold $M$ ,then $\partial M$ is orientable

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.

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).
• 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 – Tommaso Scognamiglio Nov 11 '18 at 18:05