Definition 10.86.1. 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.
10.86 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 10.86.2. 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 10.86.3. 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 10.86.2 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 10.86.4. 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 10.86.3 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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