orientability of a vector bundle plus trivial bundle

Question

I'm trying to understand the following: if a vector bundle $E$ over $X$ is not orientable, then neither is $E \oplus \underline{\mathbb{R}}$, where $\underline{\mathbb{R}} = X \times \mathbb{R}$ is the trivial line bundle.

I have not been able to come up with a sketch for a proof, but to start I have been trying to prove: if $V$ is a vector space of dimension $k$, then there is a canonical isomorphism between $\Lambda^k(V)$ and $\Lambda^{k+1}(V \oplus \mathbb{R})$.

I know that both of them will be 1 dimensional, but I'm not sure how to provide the isomorphism. Also, where to go next in the overall proof.

Answer 1

As you observed, there is a natural isomorphism $\Lambda^{1+\dim V}(V\oplus\mathbb R)\cong\Lambda^{\dim V}V$ for every vector space $V$. From this you can get that there is an isomorphism of vector bundles $\Lambda^{1+\dim V}(V\oplus\underline{\mathbb R})\cong\Lambda^{\dim V}V$ for every vector bundle $V$ on any space $X$, with $\underline{\mathbb R}$ denoting the trivial 1-dimensional vector bundle.

What you want about orientability follows from this.

Answer 2

If $E\oplus \mathbb R$ had an orientation, this orientation would give rise to an orientation of $E$. There are a lot of reasons for this, can you think of one?

Reasons could be: Looking at gluing maps (cocycles) with values in $SO(n+1)$ (take basis corresponding to the splitting); looking at orientation reversing curves; continous explicit choice of basis; first Stiefel Whitney class.