Suppose that $X$ and $Y$ are Banach spaces. Is $\{f\in B(X,Y):f\ \text{has a left inverse}\}$ an open subset of $B(X,Y)$?



Left invertible operators of $B(X,Y)$
Yes.
Let $U=\{A\in B(X,Y);\text{there is }B\in B(Y,X)\text{ so that }BA=I\_X\}$.
(You probably want to have the left inverse in $B(Y,X)$. Otherwise you can get counterexamples from compact injections.)


Take any $A\in U$.
Then there is $B\in B(Y,X)$ so that $BA=I\_X$.
Suppose $C\in B(X,Y)$ satisfies $\|C\|\leq\frac12\|B\|^{-1}$.
Then $B(A+C)=I\_X+BC\in B(X,X)$.
Now $\|BC\|\leq\|B\|\cdot\|C\|\leq\frac12$, so $I\_X+BC$ has a continuous inverse by Neumann series.
Since $(I\_X+BC)^{-1}B(A+C)=I\_X$, the operator $A+C$ has a continuous left inverse.
Therefore $A$ is an interior point of $U$.


The same argument with obvious modifications shows that the set of right invertible operators is open.


