Lemma 15.86.11. 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 15.86.8 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
as an avatar of $R^1\mathop{\mathrm{lim}}\nolimits B_ n$ (Lemma 15.86.1). 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 $R^1\mathop{\mathrm{lim}}\nolimits A_ n \to R^1\mathop{\mathrm{lim}}\nolimits B_ n$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #9804 by ZL on