Is the inverse of an adjointable Hilbert module operator always adjointable?

Question

Suppose $T:E\rightarrow F$ is a bounded adjointable bijective operator between Hilbert $A$-modules, where $A$ is some $C^*$-algebra. Then by the opening mapping theorem one knows that the inverse map $F\rightarrow E$ (let's called it tentatively $T^{-1}$) is bounded. However, is it $T^{-1}$ always adjointable? If not, what is a concrete example?

Thanks.

Answer 1

Theorem II.7.2.9 of Blackadar's book Operator Algebras: Theory of $C^*$-algebras and von Neumann Algebras states

Let $E$ and $F$ be Hilbert $B$-modules. and $T\in\mathcal L(E,F)$ with closed range. Then $T^*$ also has closed range; $\mathcal N(T)^\perp=\mathcal R(T^*)$, $\mathcal R(T^*)^\perp=\mathcal N(T)$, and hence $\mathcal N(T^*)^\perp=\mathcal R(T)$ and $\mathcal R(T)^\perp=\mathcal N(T^*)$.

(Although a proof is not given in the book, the author gives references to other sources with proofs.) From this result, we can see that if $T$ is invertible, then we have \begin{align*} \mathcal R(T^*)&=\mathcal N(T)^\perp=\{0\}^\perp=E,\\ \mathcal N(T^*)&=R(T)^\perp=F^\perp=\{0\}, \end{align*} and thus $T^*$ is invertible. Given $x\in E$, $y\in F$, we have \begin{align*} \langle T^{-1}y,x\rangle&=\langle T^{-1}y,T^*(T^*)^{-1}x\rangle\\ &=\langle TT^{-1}y,(T^*)^{-1}x\rangle\\ &=\langle y, (T^*)^{-1}x\rangle. \end{align*} Therefore $T^{-1}$ is adjointable, and $(T^{-1})^*=(T^*)^{-1}$.