Lemma 15.85.10. Let $E \to D$ be a morphism of $D(\textit{Ab}(\mathbf{N}))$. Let $(E_ n)$, resp. $(D_ n)$ be the system of objects of $D(\textit{Ab})$ associated to $E$, resp. $D$. If $(E_ n) \to (D_ n)$ is an isomorphism of pro-objects, then $R\mathop{\mathrm{lim}}\nolimits E \to R\mathop{\mathrm{lim}}\nolimits D$ is an isomorphism in $D(\textit{Ab})$.

Proof. The assumption in particular implies that the pro-objects $H^ p(E_ n)$ and $H^ p(D_ n)$ are isomorphic. By the short exact sequences of Lemma 15.85.7 it suffices to show that given a map $(A_ n) \to (B_ n)$ of inverse systems of abelian groupsc which induces an isomorphism of pro-objects, then $\mathop{\mathrm{lim}}\nolimits A_ n \cong \mathop{\mathrm{lim}}\nolimits B_ n$ and $R^1\mathop{\mathrm{lim}}\nolimits A_ n \cong R^1\mathop{\mathrm{lim}}\nolimits B_ n$.

The assumption implies there are $1 \leq m_1 < m_2 < m_3 < \ldots$ and maps $\varphi _ n : B_{m_ n} \to A_ n$ such that $(\varphi _ n) : (B_{m_ n}) \to (A_ n)$ is a map of systems which is inverse to the given map $\psi = (\psi _ n) : (A_ n) \to (B_ n)$ as a morphism of pro-objects. What this means is that (after possibly replacing $m_ n$ by larger integers) we may assume that the compositions $A_{m_ n} \to B_{m_ n} \to A_ n$ and $B_{m_ n} \to A_ n \to B_ n$ are equal to the transition maps of the inverse systems. Now, if $(b_ n) \in \mathop{\mathrm{lim}}\nolimits B_ n$ we can set $a_ n = \varphi _{m_ n}(b_{m_ n})$. This defines an inverse $\mathop{\mathrm{lim}}\nolimits B_ n \to \mathop{\mathrm{lim}}\nolimits A_ n$ (computation omitted). Let us use the cokernel of the map

$\prod B_ n \longrightarrow \prod B_ n$

as an avatar of $R^1\mathop{\mathrm{lim}}\nolimits B_ n$ (Lemma 15.85.1). Any element in this cokernel can be represented by an element $(b_ i)$ with $b_ i = 0$ if $i \not= m_ n$ for some $n$ (computation omitted). We can define a map $R^1\mathop{\mathrm{lim}}\nolimits B_ n \to R^1\mathop{\mathrm{lim}}\nolimits A_ n$ by mapping the class of such a special element $(b_ n)$ to the class of $(\varphi _ n(b_{m_ n}))$. We omit the verification this map is inverse to the map $R^1\mathop{\mathrm{lim}}\nolimits A_ n \to R^1\mathop{\mathrm{lim}}\nolimits B_ n$. $\square$

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