Let $A:V\to W$ be a linear transformation and $A^*:W^*\to V^*$ the corresponding pullback such that $A^*(g)(v)=g(A(v))$, where $g\in W^*$. There's a theorem in a book I'm reading (Semi-Riemannian Geometry: The Mathematical Language of General Relativity by Newman:
Prove that $\text{im}(A^*)=\text{ker}(A)^0$ for a linear transformation $A$
If $U$ is a subspace of $V$ then $U^0$ is the "annihilator set of $U$": $U^0=\{f\in V^*:f(u)=0\ \forall u\in U\}$.
The proof in the book (reproduced at the end of this question) uses a result that I don't want to use because of reasons mentioned in this question: Trouble in equating a linear transformation $A$ to $A^{**}$
I'm looking at an alternative approach: Let $f\in\text{im}(A^*)$ and $v\in\text{ker}(A)$. Then there exists $g\in W^*$ such that $f=A^*(g)$. So $f(v)=A^*(g)(v)=g(A(v))=g(0)=0$. Thus $f\in\text{ker}(A)^0$.
Conversely let $f(v)=0$ for all kernel vectors $v$ for some $f\in V^*$. We have to show that there exists $g\in W^*$ s.t. $A^*(g)=f$. This is where I'm having trouble - I may be missing something really easy or obvious.
I could use some help in proving the $\text{ker}(A)^0\subseteq\text{im}(A^*)$ part.
Proof in book: Note that it was proved in a previous part that $\text{ker}(A^*)=\text{im}(A)^0$
$$\text{ker}(A)=\text{ker}(A^{**})=\text{im}(A^*)^0\\\implies\text{ker}(A)^0=\text{im}(A^*)^{00}=\text{im}(A^*)$$
This proof uses another fact that $U^{00}=U$ for a subspace $U$ of $V$, which again seems erroneous because strictly speaking they are NOT equal, they are just naturally isomorphic. $U^{00}$ is a subspace of $V^{**}$, not $V$. Hence the reason why I don't want to use the book proof.
