Given three Banach spaces $X,Y,Z$ we say that $Z$ is an interpolation space between $X$ and $Y$ if the following holds (Where $B(X)$ stands for the set of bounded operators from $X$ to itself): $$ A \in B(X) \wedge A \in B(Y) \Rightarrow A \in B(Z)$$ I've seen that, for example, any separable Orlicz space is an interpolation space between $L¹$ and $L^{\infty}$ (Maligranda L., Orlicz spaces and interpolation).
An operator $A \in B(X) $ is said to be Fredholm if the dimensions of its kernel and cokernel are finite and the set $Im(A) = \{Ax : x \in X \}$ is closed. The set of all Fredholm operators from $X$ to itself is denoted $\Phi(X)$
My question is if there is a result as the one mentioned above, but for Fredholm Operators. So if there are any conditions on a Banach Space $Z$ such that $$A \in \Phi(X) \wedge A \in B(Y) \Rightarrow A \in \Phi(Z)$$
Thanks