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\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.85.4. Let

\[ 0 \to A_ i \xrightarrow {f_ i} B_ i \xrightarrow {g_ i} C_ i \to 0 \]

be an exact sequence of directed inverse systems of abelian groups over $I$. Suppose $I$ is countable. If $(A_ i)$ is Mittag-Leffler, then

\[ 0 \to \mathop{\mathrm{lim}}\nolimits A_ i \to \mathop{\mathrm{lim}}\nolimits B_ i \to \mathop{\mathrm{lim}}\nolimits C_ i\to 0 \]

is exact.

Proof. Taking limits of directed inverse systems is left exact, hence we only need to prove surjectivity of $\mathop{\mathrm{lim}}\nolimits B_ i \to \mathop{\mathrm{lim}}\nolimits C_ i$. So let $(c_ i) \in \mathop{\mathrm{lim}}\nolimits C_ i$. For each $i \in I$, let $E_ i = g_ i^{-1}(c_ i)$, which is nonempty since $g_ i: B_ i \to C_ i$ is surjective. The system of maps $\varphi _{ji}: B_ j \to B_ i$ for $(B_ i)$ restrict to maps $E_ j \to E_ i$ which make $(E_ i)$ into an inverse system of nonempty sets. It is enough to show that $(E_ i)$ is Mittag-Leffler. For then Lemma 10.85.3 would show $\mathop{\mathrm{lim}}\nolimits E_ i$ is nonempty, and taking any element of $\mathop{\mathrm{lim}}\nolimits E_ i$ would give an element of $\mathop{\mathrm{lim}}\nolimits B_ i$ mapping to $(c_ i)$.

By the injection $f_ i: A_ i \to B_ i$ we will regard $A_ i$ as a subset of $B_ i$. Since $(A_ i)$ is Mittag-Leffler, if $i \in I$ then there exists $j \geq i$ such that $\varphi _{ki}(A_ k) = \varphi _{ji}(A_ j)$ for $k \geq j$. We claim that also $\varphi _{ki}(E_ k) = \varphi _{ji}(E_ j)$ for $k \geq j$. Always $\varphi _{ki}(E_ k) \subset \varphi _{ji}(E_ j)$ for $k \geq j$. For the reverse inclusion let $e_ j \in E_ j$, and we need to find $x_ k \in E_ k$ such that $\varphi _{ki}(x_ k) = \varphi _{ji}(e_ j)$. Let $e'_ k \in E_ k$ be any element, and set $e'_ j = \varphi _{kj}(e'_ k)$. Then $g_ j(e_ j - e'_ j) = c_ j - c_ j = 0$, hence $e_ j - e'_ j = a_ j \in A_ j$. Since $\varphi _{ki}(A_ k) = \varphi _{ji}(A_ j)$, there exists $a_ k \in A_ k$ such that $\varphi _{ki}(a_ k) = \varphi _{ji}(a_ j)$. Hence

\[ \varphi _{ki}(e'_ k + a_ k) = \varphi _{ji}(e'_ j) + \varphi _{ji}(a_ j) = \varphi _{ji}(e_ j), \]

so we can take $x_ k = e'_ k + a_ k$. $\square$


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