Suppose that $X$ and $Y$ are Banach spaces. Is $\{f\in B(X,Y):f\ \text{has a left inverse}\}$ an open subset of $B(X,Y)$?
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4$\begingroup$ In fact, both the set of left and right inverses in B(X,Y) are open sets. The set of surjective operators as well -- while, of course the injective operators do not make an open set, unless it's empty. $\endgroup$– Pietro MajerCommented Aug 20, 2014 at 15:22
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1$\begingroup$ @Pietro Would you please explain why the set of injective operators is not necessarily open? $\endgroup$– AuroraCommented Aug 20, 2014 at 19:00
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1$\begingroup$ Just think to Id/n... $\endgroup$– Pietro MajerCommented Aug 20, 2014 at 19:41
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2$\begingroup$ If $X=Y$ is of finite dimension, injective operators are surjective. One thus needs infinite dimensional spaces to have the set of injectivions not open. Anyway, the "right" question would be not for injections but for monohomomorphisms, i.e., injections with closed range (surjections between Banach spaces are always epihomomorphisms, i.e. open). $\endgroup$– Jochen WengenrothCommented Aug 21, 2014 at 12:13
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$\begingroup$ What is the topology on $B(X,Y)$? The norm topology? Or the (weaker) "strong topology" (induced by $Y^X$, with $X$ viewed as discrete)? $\endgroup$– YCorCommented Nov 25, 2022 at 12:58
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1 Answer
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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.
answered Aug 20, 2014 at 15:14