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..

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

  • $\begingroup$ This is a little beyond the scope of my course but a very nice answer. Thank you! $\endgroup$ – user625722 Jan 8 at 14:58

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