Crossposted to mathoverflow due to low attention.
Let $\mathcal{A}$ be a $C^*$-algebra and $A \in \mathcal{A}$. I am interested in the invertibility of $A$ modulo certain (two-sided) ideals $\mathcal{J} \subseteq \mathcal{A}$, i.e. the existence of a generalized inverse $B \in \mathcal{A}$ such that $$AB = I + K \text{ and } BA = I + L,$$ where $K,L \in \mathcal{J}$. A classical example would be the case of bounded linear operators on a Hilbert space and the ideal of compact operators, in which case $A$ would be called a Fredholm operator.
My question now concerns intersections of ideals. If I'm not mistaken, the following should be true:
Let $S$ be some (possibly uncountable) index set and $\mathcal{J}_k$ an ideal of $\mathcal{A}$ for every $k \in S$. Then $A \in \mathcal{A}$ is invertible modulo $\bigcap\limits_{k \in S} \mathcal{J}_k$ if and only if $A$ is invertible modulo every $\mathcal{J}_k$ and the set of generalized inverses is bounded in $\mathcal{A}$.
For finite sets $S$ this is easy to see. Let $$AB_1 = I + K_1 \text{ and } AB_2 = I + K_2,$$ where $B_1,B_2 \in \mathcal{A}$, $K_1 \in \mathcal{J}_1$ and $K_2 \in \mathcal{J}_2$. Then $$A(B_2-B_1K_2) = I + K_2 - K_2 - K_1K_2 = I - K_1K_2 \in I + \mathcal{J}_1 \cap \mathcal{J}_2$$ and of course similarly for a left inverse. This does not seem to generalize easily to infinite sets $S$.
However, I came up with a quick and dirty $C^*$-argument that goes as follows. Consider the map: $$\Phi: \mathcal{A} \to \prod\limits_{k \in S} \mathcal{A}/\mathcal{J}_k, \quad A \mapsto \prod\limits_{k \in S} (A + \mathcal{J}_k),$$ where we equip $\prod\limits_{k \in S} \mathcal{A}/\mathcal{J}_k$ with the $\sup$-norm. The kernel of this homomorphism is obviously $\bigcap\limits_{k \in S} \mathcal{J}_k$ and therefore the isomorphism theorem implies that $A$ is invertible modulo $\bigcap\limits_{k \in S} \mathcal{J}_k$ if and only if $\prod\limits_{k \in S} (A + \mathcal{J}_k)$ is invertible in $\prod\limits_{k \in S} \mathcal{A}/\mathcal{J}_k$.
This seems to work, but is not very satisfactory because one cannot really see why this is the case. So, I guess my question is: Is there a more elementary proof similar to the finite case, e.g. if $S$ is assumed to be countable?
I would also be interested in a counterexample for Banach algebras if the $C^*$-property is essential. Thanks!
