Let $A$ be a unital $C^*$algebra and $F:H_A\rightarrow H_A$ a Fredholm operator on the standard Hilbert $A$module $H_A:=l^2(A)$. Is it true that $\mbox{ker}(F)$ and $\mbox{coker}(F)$ are finitely generated Hilbert $A$modules?
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2$\begingroup$ Could you be a bit more precise by what "Fredholm" means here? Do you mean that F is an adjointable operator on $H_A$ with finitedimensional kernel and cokernel? Or do you have in mind a definition more "adapted" to Hilbert $C^*$modules, such as definition on nLab? $\endgroup$ – Matthew Daws Mar 1 at 8:27

3$\begingroup$ @matthew Right, I mean Fredholm in the sense of Hilbert $C^*$module theory...that is, $F$ is an adjointable operators that is invertible modulo compact operators in $K(H_A)$. $\endgroup$ – ernest Mar 1 at 10:29

$\begingroup$ +1 for your question.To be honest I was thinking to question somewhat similar to your interesting question: I submited the following note to a journal then referee inform me the shift operator is not Fredholm on $\ell^1(A)$ . Then I was thinking of consideration of the later as an $A$ module rather than a complex vector space $\endgroup$ – Ali Taghavi Mar 2 at 6:18

$\begingroup$ arxiv.org/abs/1104.4196 $\endgroup$ – Ali Taghavi Mar 2 at 6:18

$\begingroup$ After that the journal reject my paper with indication that the shift operator is not fredholm, I had email communication with the editor and wrote him "what about consideration $A$ module structure rather than complex vector space. $\endgroup$ – Ali Taghavi Mar 2 at 8:43