Let $V$ be a finite-dimensional vector space and let $W$ be a subspace of $V$. I shall use the following notation: $V^*$ will be the dual space of $V$, $Wº$ will be the annihilator of $W$, that is the set of linear forms that annihilate every vector in $W$. Finally, $W^*$ will be the dual of the subspace $W$.
I am asked to prove the relationship:
$$ W^* \oplus Wº = V^* $$
Now, my approach initially would be to take a basis of $W$ and extend it to a basis of $V$. Then the corresponding linear forms in $V^*$ of the former set of vectors will correspond to the dual of $W$ and the rest will be a basis for the annihilator.
However, I was told I should be using an exact sequence of the following type:
$$ 0 \rightarrow W \xrightarrow{i} V \xrightarrow{\pi} \frac{V}{W} \rightarrow 0$$
Here, $i$ denotes the natural inclusion, $\pi$ denotes the quotient map to the quotient space $V$ by $W$. Now, since this sequence is exact, the corresponding sequence in duals will also be exact, that is:
$$ 0 \rightarrow (\frac{V}{W})^* \xrightarrow{\pi^*} V^* \xrightarrow{i^*} W^* \rightarrow 0 $$
Now since this sequence is also an exact sequence, we have that $i^*$ is surjective, $\pi^*$ is injective and:
$$ \text{Im}(\pi^*) = \text{Ker}(i^*)$$ Now, how could one use this relationship to prove that the annihilator of $W$ is supplementary to its dual? This question seemed rather odd since I have never seen an explicit definition of the dual of a subspace, but have rather been told that it defined by the above exact sequence and is not a subspace of the dual, but instead is equal to some quotient that can be seen, again, with the sequence above. So, how does one define the dual of a space with this sequence, and with that, prove the relation above?