equivalent definitions of orientation

Question

I know two definitions of an orientation of a smooth n-manifold $M$:

1) A continuous pointwise orientation for $M$.

2) A continuous choice of generators for the groups $H_n(M,M-\{x\})=\mathbb{Z}$.

Why are these two definitions equivalent? In other words, why is a choice of basis of $\mathbb{R}^n$ equivalent to a choice of generator of $H_n(\mathbb{R}^n,\mathbb{R}^n-\{0\})=\mathbb{Z}$?

See comments for precise definitions.

Thanks!

Answer 1

Observe that in (1) there is no difference between using tangent and cotangent bundle, and in (2) one can use $H^n$ instead of $H_n$.

Now, the equivalence becomes especially clear if in (2) one uses de Rham cohomology (instead of, say, singular).

Indeed, (1) is just existence of a (non-vanishing) section $\omega$ for $\Lambda^{top} T^*M$. So $\omega$ is a differential form, and for any $x\in U$ one can take a function $f_U$ that is 1 near $x$ and 0 outside of $U$ — and $\omega\cdot f_U$ is a generator of $H^n_{dR,c}(U)=H^n(M,M-\{x\})$. And using partitions of unit it's not hard to go in the opposite direction (i.e. reconstruct $\omega$ from local orientations).

Answer 2

Recall that an element of $H_n(M,M-\{x\})$ is an equivalence class of singular $n$-chains, where the boundary of any chain in the class lies entirely in $M-\{x\}$. In particular, any generator of $H_n(M,M-\{x\})$ has a representative consisting of a single singular $n$-simplex $\sigma\colon \Delta^n\to M$, whose boundary lies in $M-\{x\}$. Moreover, the map $\sigma$ can be chosen to be a differentiable embedding. (Think of $\sigma$ as an oriented simplex in $M$ that contains $x$.)

Now, the domain $\Delta^n$ of $\sigma$ is the standard $n$-simplex, which has a canonical orientation as a subspace of $\mathbb{R^n}$. Since $\sigma$ is differentiable, we can push this orientation forward via the derivative of $\sigma$ onto the image of $\sigma$ in $M$. This gives a pointwise orientation on a neighborhood of $x$.