Lemma 10.85.4. Let

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

is exact.

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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