An operator between $\mathcal{L}(X, Y)$ and $\mathcal{L}(Y, X)$ Please, I need help with this problem.

Let $X$, $Y$ be two vector normed spaces. Let $A_0\in\mathcal{L}(X, Y)$ such that $A^{-1}_0\in\mathcal{L}(Y,X)$. Show that there's an operator $\mathcal{T}_0\in\mathcal{L}(\mathcal{L}(X, Y), \mathcal{L}(Y, X))$ such that $\mathcal{T}_0A_0 = A_0^{-1}$ and $\|\mathcal{T}_0\| = \|A^{-1}_0\| / \|A_0\|$.


Thanks in advance.
 A: It's clear that $A_0 \neq 0$ ($A_0$ is inyective), then by Hahn-Banach Theorem on $\mathcal{L}(X,Y)$, there's $F\in[\mathcal{L}(X,Y)]^{\prime}$ such that
$$F(A_0)\ =\ \|A_0\|_{\mathcal{L}(X,Y)}\quad \mbox{ and }\quad \|F\|_{[\mathcal{L}(X,Y)]^{\prime}}\ =\ 1.$$
So, define the operator
$$\begin{array}{lcrcl}
\mathcal{T}_0 & : & \mathcal{L}(X,Y) & \longrightarrow & \mathcal{L}(Y,X)\\
& & B & \longmapsto & \mathcal{T}_0(B)\ :=\ \frac{F(B)}{\|A_0\|_{\mathcal{L}(X,Y)}}A_0^{-1}
\end{array}$$
Note that $\mathcal{T}_0$ is linear ($F$ is linear) and also $\mathcal{T}_0(A_0) = A_0^{-1}$. Besides,
$$\|\mathcal{T}_0(B)\| = \frac{\|A_0^{-1}\|_{\mathcal{L}(Y,X)}}{\|A_0\|_{\mathcal{L}(X,Y)}}|F(B)| \leq \frac{\|A_0^{-1}\|_{\mathcal{L}(Y,X)}}{\|A_0\|_{\mathcal{L}(X,Y)}}\|F\|\cdot\|B\| = \frac{\|A_0^{-1}\|_{\mathcal{L}(Y,X)}}{\|A_0\|_{\mathcal{L}(X,Y)}}\|B\|$$
then $\mathcal{T}_0\in\mathcal{L}(\mathcal{L}(X,Y),\mathcal{L}(Y,X))$.
Finally,
\begin{eqnarray*}
\|\mathcal{T}_0\| & = & \sup_{B\neq 0}\frac{\|\mathcal{T}_0(B)\|}{\|B\|}\ =\ \sup_{B\neq 0}\frac{\left\|\frac{F(B)}{\|A_0\|_{\mathcal{L}(X,Y)}}A_0^{-1}\right\|}{\|B\|}\\
& = & \frac{\|A_0^{-1}\|_{\mathcal{L}(Y,X)}}{\|A_0\|_{\mathcal{L}(X,Y)}}\sup_{B\neq 0}\frac{|F(B)|}{\|B\|}\\
& = & \frac{\|A_0^{-1}\|_{\mathcal{L}(Y,X)}}{\|A_0\|_{\mathcal{L}(X,Y)}}\cdot\|F\|\\
& = & \frac{\|A_0^{-1}\|_{\mathcal{L}(Y,X)}}{\|A_0\|_{\mathcal{L}(X,Y)}}.
\end{eqnarray*}
