I'm confused by what appears to be a contradiction in linear operator theory. I'm sure I'm misunderstanding something and I'd like to understand why this doesn't work.
Suppose $A$ is a bounded injective compact linear operator with dense range on some Banach space $B$. Now my understanding is that if $B$ is infinite-dimensional then the range of $A$ must be a proper subset of $B$ and the inverse $A^{-1}$ is not bounded. However, the inverse is closed. Now consider the operator given by the composition $A\circ A^{-1}$ which also must be closed. Now, because $A$ has dense range in $B$ the domain of $A^{-1}$ must be dense. Hence for any $f\in B$ there is a sequence ${f_n}$ in the domain of $A^{-1}$ so that $f_n \to f$ and then we also have that $(A\circ A^{-1})f_n \to f$ but by closedness of $A\circ A^{-1}$ this means $f$ is in the domain of $A\circ A^{-1}$. But $f$ was chosen arbitrarily, which means the domain of $A\circ A^{-1}$ is the whole of $B$, which is a contradiction because then it has a larger domain than $A^{-1}$!
If $C$ is a closed subset of $B $ $ and $ if $C$ is a subset of the image $A(B),$ which is the domain of $A^{-1}$, then we have $(AA^{-1})(C)=C$,which ( of course) is closed in $B.$
BUT if $C$ is closed in $B$ but $C$ is not a subset of $A(B)$ then $(AA^{-1})(C)=(AA^{-1})(C\cap A(B))=C\cap A(B)$, which may fail to be closed in $B$.
SO if $f\not \in dom (A^{-1})=A(B)$ then the argument "... by the closedness of $AA^{-1}... $" applied to the closed set $C=\{f\}\cup \{f_n\}_n)$ (...which claims that $(AA^{-1})(C)$ is closed in $B$...) is not valid because $C\not\subset dom (A^{-1}).$