Dual map and solving linear equation 
Let $f:V\to W$ be a linear transformation from a vector space $V$ to a vector space $W$.  Suppose that $b\in W$ satisfies $\phi(b)=0$ for all $\phi\in\ker(f^*)$. Show that there exists $x\in V$ such that $f(x)=b$.

I have to confess that this is part of a homework problem that was handed to me before christmas. I spent the last two weeks think on this problem and I am desperate.. I learned that if $f$ is injective, $f^*$ is surjective and I expect this to be the dual statement to this but when I try to mimic the proof, I fail. Any help would be nice..
 A: This answer does not assume finite dimensionality of $V$ or $W$ (so it is a generalized version of the Fredholm alternative).  That is, I claim that for any linear map $f:V\to W$,
$$\bigcap_{\phi\in\ker (f^*)}\ker\phi=\operatorname{im}f.$$
We have the following exact sequence of vector spaces:
$$\{0\}\to \ker f \hookrightarrow V \overset{f}{\longrightarrow} W\twoheadrightarrow \operatorname{coim}f\to \{0\},$$
where $\operatorname{coim}f$ is the coimage $W/\operatorname{im} f$.   Dualizing this exact sequence, we obtain the exact sequence
$$\{0\}\to (\operatorname{coim}f)^*\hookrightarrow W^*\overset{f^*}{\longrightarrow} V^*\twoheadrightarrow (\ker f)^*\to\{0\},$$
where $(\operatorname{coim}f)^*$ is considered a subspace of $W^*$ via the identification that sends ${\psi} \in (\operatorname{coim}f)^*$ to $\hat\psi\in W^*$ such that
$$\hat\psi(w)={\psi}(w+\operatorname{im}f)\ \forall w\in W.$$
Similarly, $(\ker f)^*$ is a quotient of $V^*$ via the restriction map sending $\sigma\in V^*$ to $\sigma|_{\ker f}\in (\ker f)^*$ (precisely, $(\ker f)^*\cong V^*/(\operatorname{cokr} f)^*$, where $\operatorname{cokr}f$ is the cokernel $V/\ker f$).  By the exactness, we get
$$\ker (f^*)=(\operatorname{coim}f)^*.$$
That is, if $w\notin \operatorname{im}f$, then $w+\operatorname{im}f$ is non-zero in $\operatorname{coim}W$.  Therefore, there exists ${\phi}_w\in (\operatorname{coim} W)^*$ that does not vanish at $w+\operatorname{im} f$.  That is, $\hat\phi_w(w)\ne 0$.  Therefore, $w\notin \bigcap_{\phi\in\ker(f^*)}\ker\phi$.  This shows that $\bigcap_{\phi\in\ker(f^*)}\ker\phi\subseteq\operatorname{im}f$.   It is also obvious that $\bigcap_{\phi\in\ker(f^*)}\ker\phi\supseteq\operatorname{im}f$.  The claim is now evident.
