Context: This question is related to this thread: Spaces of Functions
Given a topological space $X$ and a Banach algebra with unit $B$.
Consider a continuous map $F:X\to B$ that is invertible pointwise: $$F(x)F(x)^{-1}=1=F(x)^{-1}F(x)$$ Define its inverse: $$F:X\to B:x\mapsto F(x)^{-1}$$ Then, is it continuous, too? $$x_\lambda\to x\implies F^{-1}(x_\lambda)\to F^{-1}(x)$$
Put more abstract, is inversion continuous over Banach algebras with unit: $$\iota:B^*\to B^*:a\mapsto a^{-1}$$
Note: This is not the same as the formal inverse of operators: $$T:E\to E:\quad T\text{ continuous}\implies T^{-1}\text{ continuous}$$
Yes, because if $x$ is invertible and $\|y - x \| \le \|x^{-1}\|^{-1}$, $$ y = x (1 - x^{-1} (x-y)) $$ and $1 - x^{-1} (x - y)$ is invertible with $$ (1 - x^{-1}(x-y))^{-1} = \sum_{j=0}^\infty (x^{-1} (x-y))^j$$ Moreover $$ \|y^{-1} - x^{-1}\| \le \sum_{j=1}^\infty \|x^{-1}\|^{j+1} \|x - y\|^j = \dfrac{\|x^{-1}\|^2 \|x - y\|}{1- \|x-y\| \|x^{-1}\|}$$