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.
 A: First, it suffices to consider the case when $M$ and $\partial M$ are both connected. (If $M$ is connected and $\partial M$ contains several components $D_i$, consider manifolds $M_i=(M - \partial M) \cup D_i$ for each $i$.) As you noted, $\partial M$ admits a collar $C\cong \partial M\times [0,1)$, where the homeomorphism sends $C\cap \partial M$ to $\partial M\times \{0\}$. Suppose that $\partial M$ is non-orientable. I will prove that it will imply that $C-\partial M\cong \partial M\times (0,1)$ is non-orientable either. But that would contradict orientability of $M$ in the topological sense. For the following, see Example 6.25 (page 289) in Dold's book "Lectures in Algebraic Topology." 
Consider a connected $n$-dimensional topological manifold $N$. Then:  


*

*$H^n_c(N; {\mathbb Z})\cong {\mathbb Z}$ if and only if $N$ is orientable.

*$H^n_c(N; {\mathbb Z})\cong {\mathbb Z}_2$ if and only if $N$ is non-orientable. 
Hence, with rational coefficients (which I will assume from now on), $H^n_c(N)=0$ if and only if $N$ is non-orientable. 
Thus, if our $n$-dimensional manifold $M$ has non-orientable boundary, $H^{n-1}_c(\partial M)=0$. By Kunneth formula:
$$
H^n_{c}(\partial M\times (0,1))\cong H^{n-1}_c(\partial M)\otimes H^1_c((0,1))=0\otimes {\mathbb Q}=0.
$$
Thus, if $\partial M$ is non-orientable, then $\partial M\times (0,1)$ is non-orientable, hence, $C-\partial M$ is non-orientable, hence, $M- \partial M$ is non-orientable.  
A: If $b\in\partial M$ is a point in the boundary, there exists an open neighbourhood of $b$ homeomorphic to the half-space $\mathbb{R}^{n-1}\times\mathbb{R}_{\ge 0}$. Take any point $x\in\mathbb{R}^{n-1}\times\mathbb{R}_{>0}$ in the interior, we have $x\in M\setminus\partial M$ and thus the local orientation of $M\setminus\partial M$ at $x$ induces a local orientation of $\partial M$ at $b$. The fact that $M\setminus\partial M$ is orientable implies that this is well defined and coherent along $\partial M$.
