Lemma 15.87.4. Let $(A_ n)$ and $(B_ n)$ be inverse systems of abelian groups. A morphism of pro-systems $\varphi : (A_ n) \to (B_ n)$ determines maps $\mathop{\mathrm{lim}}\nolimits A_ n \to \mathop{\mathrm{lim}}\nolimits B_ n$ and $R^1\mathop{\mathrm{lim}}\nolimits A_ n \to R^1\mathop{\mathrm{lim}}\nolimits B_ n$. These maps are isomorphisms if $\varphi $ is a pro-isomorphism.
Proof. Please see Categories, Example 4.22.6 for a discussion of morphisms of pro-systems. The map $\varphi $ is given by $1 \leq m_1 < m_2 < m_3 < \ldots $ and maps $\varphi _ n : A_{m_ n} \to B_ n$ compatible with transition maps. Set $\varphi ', \varphi '' : \prod A_ n \to \prod B_ n$ equal to $\varphi '((a_ n)) = (\varphi _ n(a_{m_ n}))$ and
\[ \varphi ''((a_ n)) = (\varphi _ n(a_{m_ n} + f(a_{m_ n + 1}) + \ldots + f(a_{m_{n + 1} - 1}))) \]
where each occurence of $f$ denotes a suitable transition map of the inverse system $(A_ n)$. Then the diagram
\[ \xymatrix{ \prod A_ n \ar[r]_\delta \ar[d]^{\varphi '} & \prod A_ n \ar[d]^{\varphi ''} \\ \prod B_ n \ar[r]^\delta & \prod B_ n } \]
where the horizontal arrows are as in Lemma 15.87.1 is commutative. In this way we obtain the desired maps. The construction is functorial in the sense that if we're given an inverse system $(C_ n)$ and $1 \leq m'_1 < m'_2 < m'_3 < \ldots $ and maps $\psi _ n : B_{m'_ n} \to C_ n$ compatible with transitition maps, then $(\psi \circ \varphi )' = \psi ' \circ \varphi '$ and $(\psi \circ \varphi )'' = \psi '' \circ \varphi ''$ where the composition $\psi \circ \varphi $ refers to the integers $1 \leq m_{m'_1} < m_{m'_2} < \ldots $ and the maps $\psi _ n \circ \varphi _{m'_ n} : A_{m_{m'_ n}} \to C_ n$. We will show that if $B_ n = A_ n$ and $\varphi _ n$ is the transition map, then the resulting maps are the identity maps. This will both prove that the construction is independent of the choice of the representative $(m_ n, \varphi _ n)$ of $\varphi $ and the final statement of the lemma.
Thus we let $B_ n = A_ n$ and $\varphi _ n$ be the transition map. Let $(a_ n) \in \mathop{\mathrm{lim}}\nolimits A_ n$ be an element. Then $\varphi '((a_ n))$ is the element which has in degree $n$ the image of $a_{m_ n}$ which is equal to $a_ n$. This proves the statement for $\mathop{\mathrm{lim}}\nolimits A_ n$. Let $\xi \in R^1\mathop{\mathrm{lim}}\nolimits A_ n$ be the class of the element $(a_ n)$ in $\prod A_ n$. Consider the element $(b_ n)$ with
\[ b_ i = a_ i + f(a_{i + 1}) + \ldots + f(a_{m_ n - 1}) \]
if $m_{n - 1} < i < m_ n$ and $0$ if $i = m_ n$ for some $n$. Then $(a'_ n) = (a_ n) - \delta ((b_ n))$ defines the same class in $R^1\mathop{\mathrm{lim}}\nolimits A_ n$ and a computation shows that $a'_ i = 0$ unless $i = m_ n$ for some $n$. Thus we may and do assume $a_ i = 0$ unless $i = m_ n$ for some $n$. Note that
\[ (0, \ldots , -f(a_{m_ j}), 0, \ldots , 0, a_{m_ j}, 0, \ldots ) \]
with nonzero entries in spots $j$ and $m_ j$, is the image under $\delta $ of
\[ c_ j = (0, \ldots , 0, f(a_{m_ j}), f(a_{m_ j}), \ldots , a_{m_ j}, 0, \ldots ) \]
with nonzero entries in spots $j + 1, \ldots , m_ j$. The sum $c = \sum c_ j$ makes sense in $\prod A_ n$. Recalling that $\varphi ''((a_ n)) = (a_{m_1}, a_{m_2}, \ldots )$ we see that
\[ (a_ n) - \varphi ''((a_ n)) = \delta (c) \]
and the proof is complete. $\square$
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