Is $T(T^2)^{-1}$ a left-inverse for $T$? Let $H$ be a Hilbert space and $T:\text{dom}(T)\rightarrow H$ be a densely defined, closed self-adjoint operator. Suppose that $T^2: \text{dom}(T^2)\rightarrow H$ has a bounded inverse $(T^2)^{-1}:H\rightarrow \text{dom}(T^2)$. 
Here $\text{dom}(T^2):=\{u\in\text{dom}(T):Tu\in\text{dom}(T)\}$. 
Question: Can one conclude that $T$ is invertible? That is, does there exist a bounded operator $T^{-1}:H\rightarrow\text{dom}(T)$ that is a two-sided inverse of $T$?
The natural thing to do seems to be to construct $T^{-1}$ from $(T^2)^{-1}$, and indeed a right-inverse for $T$ is $T(T^2)^{-1}$. But it is not clear to me how show it is also a left-inverse, as some commutativity seems required (and even if they "did" the domains would not make sense). Perhaps one can use the functional calculus of self-adjoint operators to show this?

For $x \in \operatorname{dom}(T)$, we have
\begin{align}
T(T^2)^{-1}Tx &= T(T^2)^{-1}T \operatorname{id}_H x \\
&= T(T^2)^{-1}TT^2(T^2)^{-1}x \\
&= T(T^2)^{-1}T^2T(T^2)^{-1}x \tag{1}\\
&= TT(T^2)^{-1}x \tag{2}\\
&= x,
\end{align}
where dropping the $(T^2)^{-1}T^2$ in the step from $(1)$ to $(2)$ is valid because $T(T^2)^{-1}x \in \operatorname{dom}(T^2)$ for $x \in \operatorname{dom}(T)$.
Thus $T(T^2)^{-1}$ is indeed a left inverse of $T$, as well as a right inverse. Furthermore, it is a globally defined closed operator, hence continuous.
