Isomorphism between dual of quotient and annihilator

Question

While doing exercises I found this particularly surprising result: $(V/W)^* \cong W^0$, where $W^0$ denotes the annihilator of $W$. I wanted to prove this result by giving an explicit linear map (the isomorphism) from $(V/W)^*$ to $W^0$, but I don't know how to define it. First of all, the elements of $(V/W)^*$ are linear functionals, and so are the ones in $W^0$, so how am I supposed to define this linear map? A linear functional mapping to another linear functional? If so, how can you then prove this map is invertible?

Answer 1

Recall how equivalence relations works: if $A$ is a set, and $\sim$ is an equivalence relation on $A$, then for every function $f$ whose domain is $A$ such that $$(\forall x,y \in A) \quad x \sim y \implies f(x) = f(y) \tag{$*$}$$ there exists a unique function $\overline f$ whose domain is $A/{\sim}$ so that $\overline f \circ \pi = f$, where $\pi : A \to A/{\sim}$ is the canonical projection.

In linear algebra everything works in the same way: if $V$ is a vector space, and $W$ subspace of $V$ (so we have an equivalence relation on $V$), then for every linear map $T$ whose domain is $V$ such that $W \subseteq \ker T$ (please check the equivalence between this and $(*)$) there exists a unique linear map $\overline T$ whose domain is $V/W$ so that $\overline T \circ \pi = T$, where $\pi : V \to V/W$ is the canonical projection.

In other words, for every vector space $Z$, the function $$\begin{split} \{\textrm{linear maps $S : V/W \to Z$}\} & \longrightarrow \{\textrm{linear maps $T : V \to Z$ with $W \subseteq \ker T$}\} \\[0.5mm] S & \longmapsto S \circ \pi \end{split} \tag{$\star$}$$ is a bijection (a function $g : X \to Y$ is a bijection if for every $y \in Y$ there exists a unique $x \in X$ so that $g(x)=y$).

Finally, in $(\star)$ when we take $Z$ to be the scalar field, we have the desired isomorphism $(V/W)^* \to W^0$ (just check the linearity).

Answer 2

Question: "First of all, the elements of (V/W)∗ are linear functionals, and so are the ones in W0, so how am I supposed to define this linear map? A linear functional mapping to another linear functional? If so, how can you then prove this map is invertible?"

answer: There is an exact sequence

$$0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$$

and a dual exact sequence

$$0 \rightarrow (V/W)^* \rightarrow V^* \rightarrow^{\phi} W^* \rightarrow 0$$

and by definition it follows $W^{0}:=ker(\phi) = (V/W)^*$.