I can't find a reference for this fact, so I am wondering if this is true: a countable inverse limit of isomorphisms of abelian groups is itself an isomorphism. Or more precisely, if $f_i : A_i \to B_i$ is a countable collection of isomorphisms of abelian groups, then the induced map $f : A \to B$ of inverse limits is an isomorphism.
Here is what I believe to be a proof: We know that $0 \to \{0\}_{i\in I} \to \{A_i\}_{i\in I} \to \{B_i\}_{i\in I} \to 0$ is a short exact sequence of $I$-indexed inverse systems, and clearly the constant 0 system obeys the Mittag-Leffler condition, so we exact sequence $0 \to 0 \to A \to B \to 0$ of the inverse limits must be exact, implying that the map $f : A\to B$ is an isomorphism.
Is this correct?
Even better:
Let $I$ be your indexing category and let $\mathsf{D} = \mathsf{Ab}^{I^\text{op}}$ be the category of $I$-indexed inverse systems.
Since $I$ is small and $\mathsf{Ab}$ is complete, every $I$-indexed inverse system has a limit in $\mathsf{Ab}$. In fact, there is a functor $\lim : \mathsf{D} \to \mathsf{Ab}$ which sends each inverse system to its limit (the functoriality comes from the universal property of the limit).
You have two diagrams $\mathbf{A}, \mathbf{B} \in \mathsf{D}$, and an isomorphism $\mathbf{f} : \mathbf{A} \to \mathbf{B}$ be between them. Functors always send isomorphisms to isomorphisms, so $f = \lim \mathbf{f}$ is an isomorphism between $A = \lim \mathbf{A}$ and $B = \lim \mathbf{B}$.