Working with the homological definition of orientation I'm trying to understand the homological definition of orientation given by a continuous choice of generators for local homology groups. Unfortunately, I am already unsure about the proof $\mathbb R ^n$ is orientable.
I tried to start by picking an orientation $\mu_0$ at zero and the defining the other local orientations $\mu_x$ to be the images of $\mu_0$ along the homology maps induced by translation $0\mapsto x$. However, I am not certain I understood the consistency condition well. I thought perhaps the composite $\mathrm H(\mathbb R^n\mid B)\to \mathrm H(\mathbb R^n \mid 0)\to \mathrm H(\mathbb R ^n\mid x)$ is equal to the excision isomorphism $\mathrm H(\mathbb R^n\mid B)\to \mathrm H(\mathbb R ^n\mid x)$ for any open ball $B$ containing $0,x$, and that this is enough, but I feel I am missing something.
What is the correct approach?
 A: The inclusion $(\mathbb{R}^n, \mathbb{R}^n - B) \to (\mathbb{R}^n, \mathbb{R}^n - \{0\})$ induces an isomorphism $H_n(\mathbb{R}^n | B) \to H_n(\mathbb{R}^n | 0)$. So for any neighbourhood $B$ of $0$, we can consistently choose an orientation for all points inside $B$. So locally, we have a well defined notion of orientation. You can propagate this (formally using a connectedness argument) to show that $\mathbb{R^n}$ is orientable. 
You might find the last paragraph on page 233 of Hatcher helpful.
EDIT: I've tried to make the exposition a little more explicit. Let us suppose that we have a chosen orientation $\mu_0 \in H_n(\mathbb{R}^n|0) \cong \mathbb{Z}$. Now, for any open ball $B$ around $0$, we have the inclusion $i^0:(\mathbb{R}^n, \mathbb{R}^n - B) \to (\mathbb{R}^n, \mathbb{R}^n - \{0\})$. This induces an isomorphism $i_\ast^0:H_n(M|B) \to H_n(M|0)$. So let us denote by $\mu_B$ the element of $H_n(M|B)$ such that $i_\ast^0(\mu_B) = \mu_0$. We will show that this $\mu_B$ helps choose generators for local homology at all the other points in $B$ and hence lets us make a consistent choice. Suppose we choose another point $y \neq 0 \in B$. What does it mean to choose a generator of $H_n(M|y)$ consistently? This means that the chosen generator $\mu_y = i_\ast^y(\mu_B)$. Thus, a choice of local generator is consistent in an open set $B$ if for every $x \in B$, $\mu_x = i_\ast^x(\mu_B)$. You can think of this as a local orientation being compatible with $\mu_B$. With this in mind, we look back at our definition of orientable.

A manifold is $R$-orientable if for all open balls $B$ in coordinate charts of $M$, there exists $\mu_B$, a generator of $H_n(M|B)$ such that $\mu_x \in H_n(M|x) = i_\ast^x(\mu_B)$ where $i_\ast$ is the map induced by inclusion. 

You might wonder if this is equivalent to Hatcher's and it is. The idea boils down to each local orientation is given by mapping the generator of $H_n(M|B)$ and seeing where it goes.
Now, for $\mathbb{R}^n$, for each point $x \neq 0$, we can consider the open ball of radius $||x||+1$. Then, given $\mu_0$, I get a $\mu_B$, and I let $\mu_x = i_\ast^x(\mu_B)$.
