$X$ and $Y$ are Banach spaces and suppose that $T$ be a closed subspace of $X$.
$A:X \to Y$, $B:Y\to X$ and $X_{0}:Y\to X$ are linear bounded operators. It is given that the operator $BA$ is invertible on its range space denoted by $R(BA)$, also we have $R(BA)\subseteq T$ and $R(X_0) \subset T$
My question is can we write.
$(BA)^{-1}_{|R(BA)}BA X_{0} = X_{0}$.
Could anyone help me to clear my doubt. I would be very much thankful. I need to confirm this to prove one of my result.
Thanks
Your assumptions allow for a counter example even in finite dimensions: The rank of $C=(BA)^{-1}BA$ is equal to the dimension of $R(BA)$. If $R(BA)$ is a strict subspace of $T$, we have that the rank of $CX_0$ is less than the rank of $X_0$.
A suitable condition on $BA$ would be, that $R(X_0) \cap N(BA) = \{0\}$ and $R(BA)$ closed. That would guarantee that $C$ is the identity on $R(X_0)$.