Rank 2 vector bundle

Question

$E$ is a rank $2$ vector bundle.

Why is $E\simeq E^*\otimes \det E$?

Any generalization (arbitrary rank, $E$ non locally free etc.)?

Answer 1

More generally, any vector bundle $V$ (or representation of a group, etc.) of rank $d$ comes equipped with a natural nondegenerate pairing (the exterior product)

$$V \otimes \wedge^{d-1} V \to \wedge^d V$$

which gives an isomorphism

$$V \cong \wedge^{d-1} V^{\ast} \otimes \wedge^d V.$$

Now take $d = 2$.