Suppose $K$ is a bounded above complex of free abelian groups, and take its derived $\ell$-adic completion $K^{\wedge,\ell} = R\lim (K/\ell^n)$ in the derived category, for $\ell$ a prime.
If $K\to L$ is a map in the derived category, such that $K^{\wedge,\ell}\to L^{\wedge,\ell}$ is an isomorphism for every $\ell$, is $K\to L$ an isomorphism? (in the derived category)
As I mentioned in a comment, $K \to L$ must also be an isomorphism after rationalizing. For example, if $0 \to R \to F \to \Bbb Q \to 0$ is a free resolution of $\Bbb Q$, then let $$ K = \dots \to 0 \to 0 \to R \to F \to 0 \to 0 \to \dots $$ and $L$ be the zero complex. The map $K \to L$ is an isomorphism in the derived category after completing at any prime ($K/\ell^n$ has trivial homology for any $n$), but is not an isomorphism in the derived category itself.
However, if you add the condition that you have an isomorphism after rationalizing, the statement is true. Here is one way that you can prove that.
For any free abelian group $F$, there is an arithmetic square $$ \require{AMScd} \begin{CD} F @>>> \prod_\ell F^{\wedge,\ell}\\ @VVV @VVV\\ F \otimes \Bbb Q @>>> \left(\prod_\ell F^{\wedge,\ell}\right) \otimes \Bbb Q \end{CD} $$ which we can re-express as a natural sequence $$ 0 \to F \longrightarrow (F \otimes \Bbb Q) \oplus \prod_\ell F^{\wedge,\ell} \longrightarrow \left(\prod_\ell F^{\wedge,\ell}\right) \otimes \Bbb Q \to 0 $$ that is exact whenever $F$ is free. Applying this to any complex $K$ of free abelian groups, we get a natural exact triangle in the derived category $$ K \longrightarrow (K \otimes \Bbb Q) \oplus \prod_\ell K^{\wedge,\ell} \longrightarrow \left(\prod_\ell K^{\wedge,\ell}\right) \otimes \Bbb Q \to K[1]. $$ If $K \to L$ is a map of two such complexes in the derived category and it induces isomorphisms $K^{\wedge,\ell} \to L^{\wedge,\ell}$ and $K \otimes \Bbb Q \to L \otimes \Bbb Q$ in the derived category, then the map $K \to L$ must then also be an isomorphism in the derived category.
