# Prove that $\operatorname{range}\pi' = U^0$

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 $\operatorname{range}\pi'$.

• Do you know the universal property of the quotient space? Mar 5, 2017 at 16:25
• No. Universal properties have something to do with tensors, right? I haven't gotten to that. Mar 5, 2017 at 16:27
• Hi - I asked this just to check before I troubled you. math.stackexchange.com/questions/2173571/…. In your inset line beginning $f(u)$ I think it should reflect $\pi(u)=0$ so then you can say $\phi(0)$. Regards,
– user12802
Mar 5, 2017 at 22:32
• @TheBirdistheWord There is no difference between $U$, $u+U$ for $u\in U$, and $0\in V/U$. I decided to use the definition of $\pi$ "naively" (mostly so my audience could easily follow what I was doing), but certainly you're right that I could have skipped from $(\varphi\circ \pi)(u)$ straight to $=0$. There's nothing theoretically wrong with that line, though. Mar 5, 2017 at 23:00

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 $V/U$.
• I was reviewing this eloquent answer. Maybe you would please clarify something for me. At the end you say to check that $\varphi$ is linear. But do you not already assume that when you say $f(v_1-v_2)=0$ and thus $f(v_1)=f(v_2)$? Thanks. With regards,
• @Andrew I only use the fact that $f$ is linear, which is true by hypothesis, since $f\in U^0\subset V'$. Apr 2, 2017 at 12:17