Why is the Tangent bundle orientable?

Question

Let $M$ be a smooth manifold. How do I show that the tangent bundle $TM$ of $M$ is orientable?

Answer 1

Start with an atlas for $M$ and construct the corresponding atlas on $TM$, each of whose charts is constructed from one on $M$. Check that the transition functions in the latter have Jacobian with positive determinant.

Answer 2

There is also a way to see it with fibre bundles and characteristic classes. It's possible the original poster is not familar, but other people might be and more importantly I need practice. It is based on two relatively basic facts (at least I'm mostly sure I've seen them before):

1) A smooth $n$-manifold $M$ is orientable iff the first Stiefel-Whitney class of its tangent bundle $\tau_M$ vanishes,

and

2) If $\xi$ is a smooth $k$-plane bundle with base space $M^n$, total space $E^{n+k}$ (both smooth manifolds) and projection $\pi:E\rightarrow M$, then $$\tau_E=\pi^*(\tau_M)\oplus\pi^*(\xi)$$

Then, if $TM$ is the total space of the $n$-plane bundle $\tau_M$ with projection map $\pi\colon TM\rightarrow M$, it is a smooth manifold with its own tangent bundle $\tau_{TM}$. Since $\pi$ is the projection map of $\tau_M$ we have $$\tau_{TM}=\pi^*(\tau_M)\oplus\pi^*(\tau_M)$$ so by the Whitney product formula $$\omega_1(\tau_{TM})=(\pi^*\omega_0)(\tau_M)\cup(\pi^*\omega_1)(\tau_M)+(\pi^*\omega_1)(\tau_M)\cup(\pi^*\omega_0)(\tau_M)=2\pi^*\omega_1(\tau_{M})=0\in H^1(TM;\mathbb{Z/2})$$

Hence the manifold $TM$ is orientable.

(But for all intents and purposes, writing down charts is the easiest way to go)