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Let $A : X\to Y$ and $B : Y \to X$ be bounded operators between Banach spaces $X,Y$. Assume that $AB : Y\to Y$ is Fredholm. I would like to prove that $A(X)$ and $B(Y)$ are closed. Furthermore, possibly conclude that $A$ is also Fredholm.

I made some progress on it. For instant, since $\dim \ker AB = \dim B(Y)\cap \ker A$, then this intersection has finite dimension. It would be very nice to conclude that $\ker A$ is finite dimensional. However, I don't see how it is possible from here. But note that if $B$ is Fredholm, then necessarily $\dim B(Y) = \infty$, and hence, $\dim \ker A < \infty$.

On the other hand, if I show that $A(X)$ is closed, then since $\infty > \dim Y/AB(Y) \geq \dim Y/A(X)$ it would follow that $A$ is Fredholm.

However, I was not able to show that $A(X)$ and $B(Y)$ are closed, neither that is the case that $\dim B(Y) = \infty$ (in general).

Any hints?

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  • $\begingroup$ If $A: X \to Y$ and $B: Y \to X$, then $AB: Y \to Y$, not $X \to Y$. $\endgroup$ – Robert Israel May 22 at 22:51
  • $\begingroup$ @RobertIsrael this was only a typo, thank you. $\endgroup$ – L.F. Cavenaghi May 22 at 22:54

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