Notation:
- Let $V$ be an $\Bbb F$-vector space with $U$ a subspace thereof.
- Let $\pi: V \to V/U$ be given by $\pi(v) = v+U$.
- Let $U^0\subseteq V'$ denote the annihilator of $U$, which is a subspace of the dual space of $V$.
- Let $\pi': (V/U)'\to V'$ be given by $\pi'(\varphi) = \varphi\circ \pi$.
Prove that $\operatorname{range}\pi' = U^0$.
Here's what I've got:
Let $f\in \operatorname{range}\pi'$. Then there exists $\varphi \in (V/U)'$ such that $f=\varphi \circ \pi$. Let $u\in U\subseteq V$. Then $$f(u) = (\varphi \circ \pi)(u) = \varphi(u+U) = \varphi(U) = 0.$$ Thus $f\in U^0$. Hence $\operatorname{range}\pi' \subseteq U^0$.
But then I've basically got nothing for the reverse inclusion. I've been using dimensional arguments for these types of problems, but this is one of the rare exercises in this section that doesn't make the assumption that $V$ or $U$ is finite dimensional. So I know I have to start with "let $g\in U^0$", but I'm not seeing how to show that is in $\operatorname{range}\pi'$.
Let $f\in U^0$, so that $f(u)=0$ for all $u\in U$. You want to find some $\varphi : V/U\to \Bbb F$ such that $\varphi \circ \pi =f$, or equivalently, such that $$\varphi(v+U)=f(v)$$ for all $v\in V$. So the simple thing to do is : just define $\varphi$ by the above formula.
But of course you need to make sure that it really is well-defined! In other words, you need to make sure that $\varphi(v+U)$ does not depend on $v$. But notice that $v_1+U=v_2+U$ if and only if $v_1-v_2\in U$. If this is the case, then $f(v_1-v_2)=0$, and thus $f(v_1)=f(v_2)$; thus $\varphi$ is well-defined. All that remains to check that $\varphi$ is linear, and it is immediate given the vector space structure on $V/U$.