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..

• HINT: Look up "Fredholm alternative", the finite dimensional version. – Giuseppe Negro Jan 8 at 13:45
• You saved my day! Thank you SO much!! – user625722 Jan 8 at 13:56

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.