V = U⊕W then Prove that (V/W)* is isomorphic to W^0

Question

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 $(V/W)^*$ is isomorphic to $W^0$.

note: (V/W)* is the dual space of (V/W).

W^0 is the annihilator of W.

W^0={f in V*|f(v)=0, for all v in W}.

V* is the dual space of V**.

for (a), I have considered somethings: Let ϕ∈Ann(W), then it is a form on V (i.e. ϕ∈V∗) such that W is in its kernel.

Answer 1

Consider the natural projection $\pi : V\to V/W$. This gives an induced map $$ \pi^{\ast} : (V/W)^{\ast} \to V^{\ast} $$ given by $$ \pi^{\ast}(f)(x) = f(\pi(x)) $$ Since $\pi$ is surjective, $\pi^{\ast}$ is injective (check this). Now what is the range of $\pi^{\ast}$?

For any $w\in W$, clearly, $\pi^{\ast}(f)(w) = f(\pi(w)) = f(0+W) = 0$. Hence, $\pi^{\ast}(f) \in W^0$.

Conversely, if $g\in W^0$, define $\hat{g} : V/W \to k$ by $\hat{g}(x+W) := g(x)$. Check that this is a well-defined linear functional. By construction, it follows that $$ \pi^{\ast}(\hat{g}) = g $$ Hence, $\pi^{\ast}$ maps $(V/W)^{\ast}$ bijectively onto $W^0$.