Definition 10.85.1. Let $(A_ i, \varphi _{ji})$ be a directed inverse system of sets over $I$. Then we say $(A_ i, \varphi _{ji})$ is *Mittag-Leffler* if for each $i \in I$, the family $\varphi _{ji}(A_ j) \subset A_ i$ for $j \geq i$ stabilizes. Explicitly, this means that for each $i \in I$, there exists $j \geq i$ such that for $k \geq j$ we have $\varphi _{ki}(A_ k) = \varphi _{ji}( A_ j)$. If $(A_ i, \varphi _{ji})$ is a directed inverse system of modules over a ring $R$, we say that it is Mittag-Leffler if the underlying inverse system of sets is Mittag-Leffler.

## 10.85 Mittag-Leffler systems

The purpose of this section is to define Mittag-Leffler systems and why this is a useful notion.

In the following, $I$ will be a directed set, see Categories, Definition 4.21.1. Let $(A_ i, \varphi _{ji}: A_ j \to A_ i)$ be an inverse system of sets or of modules indexed by $I$, see Categories, Definition 4.21.4. This is a directed inverse system as we assumed $I$ directed (Categories, Definition 4.21.4). For each $i \in I$, the images $\varphi _{ji}(A_ j) \subset A_ i$ for $j \geq i$ form a decreasing directed family of subsets (or submodules) of $A_ i$. Let $A'_ i = \bigcap _{j \geq i} \varphi _{ji}(A_ j)$. Then $\varphi _{ji}(A'_ j) \subset A'_ i$ for $j \geq i$, hence by restricting we get a directed inverse system $(A'_ i, \varphi _{ji}|_{A'_ j})$. From the construction of the limit of an inverse system in the category of sets or modules, we have $\mathop{\mathrm{lim}}\nolimits A_ i = \mathop{\mathrm{lim}}\nolimits A'_ i$. The Mittag-Leffler condition on $(A_ i, \varphi _{ji})$ is that $A'_ i$ equals $\varphi _{ji}(A_ j)$ for some $j \geq i$ (and hence equals $\varphi _{ki}(A_ k)$ for all $k \geq j$):

Example 10.85.2. If $(A_ i, \varphi _{ji})$ is a directed inverse system of sets or of modules and the maps $\varphi _{ji}$ are surjective, then clearly the system is Mittag-Leffler. Conversely, suppose $(A_ i, \varphi _{ji})$ is Mittag-Leffler. Let $A'_ i \subset A_ i$ be the stable image of $\varphi _{ji}(A_ j)$ for $j \geq i$. Then $\varphi _{ji}|_{A'_ j}: A'_ j \to A'_ i$ is surjective for $j \geq i$ and $\mathop{\mathrm{lim}}\nolimits A_ i = \mathop{\mathrm{lim}}\nolimits A'_ i$. Hence the limit of the Mittag-Leffler system $(A_ i, \varphi _{ji})$ can also be written as the limit of a directed inverse system over $I$ with surjective maps.

Lemma 10.85.3. Let $(A_ i, \varphi _{ji})$ be a directed inverse system over $I$. Suppose $I$ is countable. If $(A_ i, \varphi _{ji})$ is Mittag-Leffler and the $A_ i$ are nonempty, then $\mathop{\mathrm{lim}}\nolimits A_ i$ is nonempty.

**Proof.**
Let $i_1, i_2, i_3, \ldots $ be an enumeration of the elements of $I$. Define inductively a sequence of elements $j_ n \in I$ for $n = 1, 2, 3, \ldots $ by the conditions: $j_1 = i_1$, and $j_ n \geq i_ n$ and $j_ n \geq j_ m$ for $m < n$. Then the sequence $j_ n$ is increasing and forms a cofinal subset of $I$. Hence we may assume $I =\{ 1, 2, 3, \ldots \} $. So by Example 10.85.2 we are reduced to showing that the limit of an inverse system of nonempty sets with surjective maps indexed by the positive integers is nonempty. This is obvious.
$\square$

The Mittag-Leffler condition will be important for us because of the following exactness property.

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.

**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

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

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