inverse systems and Mittag-Leffler condition - where is it being used?

In the following example (from the stacks project https://stacks.math.columbia.edu/tag/0594 ) where (if anywhere) is the hypothesis that $$(A_i, \phi_{ij})$$ is a Mittag-Leffler system being used to make the conclusion underlined in red? I feel both surjectivity and the equality of limits hold without assuming the system is Mittag-Leffler, but maybe I'm wrong. EDIT: My proof of surjectivity had an error. That's where the ML condition is being used.

For defining the stable image, one can define $$A_i' = \cap_{j \geq i} \phi_{ji}A_j$$ so one doesn't need the system to be Mittag Leffler. However, it may be empty even if all the $$A_i$$ are non-empty, whereas that's not the case for a ML-system.

What they are showing is simply that:

1. If a system consists of surjective maps, it is Mittag-Leffler and that,

2. Conversely, any Mittag-Leffler system with a given limit can be replaced by another system that consists of surjective maps (and which, by 1., satisfies the Mittag-Leffler condition, too) with the same limit.

To show the maps are onto, you need to be able to have a stable image. You definition of stable image cannot be correct, since it does not depend at all on the morphisms $$\varphi_{ij}$$!

• Yes, but I'm asking where, if anywhere is the Mittag-Leffler condition being used ( for your item #2)? in other words, can we drop it as a hypothesis and are the claims true without it – usr0192 Oct 21 at 23:08
• @usr0192 I see. To show the maps are onto, you need to be able to have a stable image. You definition of stable image cannot be correct, since it does not depend at all of the morphisms $\varphi_{ij}$! – Pedro Tamaroff Oct 21 at 23:14
• Thanks for pointing out my typo, yes you are right I forgot to add the $\phi_{ji}$. And I think you are right that the surjectivity needs the ML-hypothesis (as I discovered an error in my proof): given $a_i'$ in the stable image $A_i'$, for all $k>i$ there is a $a_k \in A_k$ mapping to $a_i'$, ... but these might map to different elements in $A_j$ that only become equal in $A_i$! With ML hypothesis, there exists $N$ such that for $k\geq N$ $A_k$ maps to $A_j'$ ... then $a_N$ maps to an element in the stable image $A_j'$ that maps to $a_i$. Thanks! – usr0192 Oct 21 at 23:30

To get a better understanding, I recommend to follow up with the concept of the pro-category $$\text{pro-}C$$ of a category $$C$$. This has as objects all directed inverse systems over $$C$$. The morphisms are somewhat more complicated and I shall not go into details here.

The inverse limit (if it exists for all directed inverse systems over $$C$$) turns out to be a functor $$\lim : \text{pro-}C \to C$$.

The Mittag-Leffler-condition can be defined for directed inverse systems over the category of sets and more generally over "concrete categories" (whose objects are sets with an "additional structure" like groups, abelian groups, modules etc).

It is a well-known theorem that a directed inverse system over the category of groups (abelian groups, modules, ...) is Mittag-Leffler if and only if it is isomorphic in $$\text{pro-}C$$ to a directed inverse system whose bondings are surjective.