Continuity of the operator "inverse of" in the space of linear, bounded and bijective operators Let $(X,\|\ \|_X),(Y,\|\ \|_Y)$ be two Banach spaces over $K$.

Definition:
$\qquad\qquad\qquad\quad I(X,Y):=\big\{A\in \mathscr B(X,Y):A\text{  is invertible and  }A^{-1}\in\mathscr B(Y,X)\big\}$
 where $\mathscr B(X,Y)$ is de space of bounded linear operators between $X$ and $Y$.

Lets define $\mathscr U:I(X,Y)\to I(Y,X)$ given by
$$
\mathscr U(A)=A^{-1}
$$
I want to prove that $\mathscr U$ is continuous.

I'm really lost trying to figure out a way for this one. I came up with the idea of a sequence $(A_n)_{n\in\Bbb N}\subset I(X,Y)$ that converged to some $A\in I(X,Y)$ but realized that $\mathscr U(A_n)$ may not always converge. Another attempt was to show that $\mathscr U$ is bounded and linear but $I(X,Y),I(Y,X)$ are not even subspaces, since the sum of invertible operators is not always invertible. Any help would be appreciated.
 A: Consider the following fact: if $A:X \rightarrow Y$ is a bounded linear operator and $\|A\| < 1$, then $I-A \in I(X,Y)$ and $$(I-A)^{-1} = \sum_{n=0}^\infty A^n.$$ Using this, one can establish that if $A \in I(X,Y)$, then for every bounded linear operator $B:X \rightarrow Y$, the relation $\|B-A\| < \|A^{-1}\|^{-1}$ implies $B \in I(X,Y)$. For in this case, we have $$\|I-A^{-1}B\|\leqq \|A^{-1}\|\|A-B\| <1,$$ so that $B= A[I-(I-A^{-1}B)] \in I(X,Y)$.
Fix $A \in I(X,Y)$ arbitrarily. If $B$ is a bounded linear operator and $\|B\| < \|A^{-1}\|^{-1}$, then $$\|-BA^{-1}\|=\|BA^{-1}\|\leqq\|B\|\|A^{-1}\|<1.$$ Thus, $A+B=[I-(-BA^{-1})]A \in I(X,Y)$. This also allows us to apply the fact I cited in the beggining (with $-BA^{-1}$ instead of $A$) to obtain $$\|(A+B)^{-1} - A^{-1}\| =\| A^{-1}(I-(-BA^{-1}))^{-1} - A^{-1}\| =\Bigg\| A^{-1} \sum_{n=0}^\infty (-BA^{-1})^n  - A^{-1}\Bigg\| =\Bigg\| A^{-1}\sum_{n=1}^\infty (-BA^{-1})^n \Bigg\| \leqq \|A^{-1}\|\sum_{n=1}^\infty(\|B\|\|A^{-1}\|)^n = \frac{\|B\|\|A^{-1}\|^2}{1-\|B\|\|A^{-1}\|}. $$ Letting $\|B\| \rightarrow 0$, it follows that $(A+B)^{-1} \rightarrow A^{-1}$ and the continuity of ${\scr U}$ at $A$ is proved. 
Note: This fact I mentioned at the beggining really uses the completeness of $(Y,\|\|_Y)$. It uses that every absolutely convergent series in ${\scr B}(X,Y)$ converges, which is equivalent to the completeness of ${\scr B}(X,Y)$, which in turn is equivalent to the completeness of $Y$. 
