How to show the quotient of dual space and the annihilator of its subspace is equal to dual of the subspace?

Question

Let $W$ be a subspace of a vector space $V$ over a field $F$. Let $i : W \to V $ denote the inclusion map.

Show that $ \pi: V^*\to W^*$given by $\pi(f) = f\circ i$ is a surjective linear map, with kernel equal to $W^0$. Hence show that $W^* = V^*/W^o$.

Answer 1

Spelled out differently, all you need to observe is that every linear functional on $W$ can be linearly extended to $V$. The statement then follows from the definition.