Let $V$ be a vector space (not necessary being finite dimensional) and let $U,W$ be subspaces of $V$ such that $V = U\oplus W$.
Prove that $V^\ast/(W^0)$ is isomorphic to $W^\ast$.
Notation and Definitions:
$W^0$ is the annihilator of $W$
$W^0=\{f \in V^\ast\mid \text{f(v)=0, for all v in W}\}$
$V^\ast$ is the dual space of $V$
Before doing this proof, I have a question: when we construct a mapping from $V^\ast$ to $W^\ast$, do we need $W^0$ to be the kernel of that mapping?
Consider the following linear mapping, $$\phi:V^*/W^o \rightarrow W^*$$ $$[f]\longmapsto f|_W$$ Lets see if it's well defined, consider $f,g\in V^*$ such that $[f]=[g]$, this means $f-g\in W^o$. Now given any $w\in W$ we have that: \begin{align*} (f-g)(w)&=0\\ \Rightarrow f(w)&=g(w)\\ \Rightarrow f|_W&=g|_W\\ \end{align*}
Next we are going to prove that $\phi$ is inyective, if $f$ is such that $\phi(f)=0\in W^*$ we have that $$\forall w\in W:\quad f|_W(w)=0$$ \begin{align*} \Rightarrow f&\in W^o\\ \Rightarrow [f]&=0 \end{align*}
Finally lets see if $\phi$ is onto. Take any $g\in W^*$, we define $f\in V^*$ such that $f(v=u+w)=g(w)$ where $u+w$ is the unique decomposition of $v$ given by $V=U\oplus W$. $f$ is clearly linear and $f|W=g$.