The Mittag-Leffler condition and $\varprojlim^1$

Question

Recall that an inverse system of abelian groups $$\cdots \rightarrow G_2 \stackrel{\alpha_2}{\rightarrow} G_1 \stackrel{\alpha_1}{\rightarrow} G_0$$ is said to satisfy the Mittag-Leffler condition if, for each $i$, there exists a number $N$ such that the image of $G_{i+n} \rightarrow G_i$ is invariant for all $n \geq N$ (the map is obviously the composition $\alpha_{i+n} \circ \cdots \circ \alpha_{i+1}$).

I'm trying to show that, given a system that satisfies this condition, $\varprojlim^1G_i = 0$. Recall that $\varprojlim^1 G_i$ is defined to be the cokernel of the map $$\delta \colon \prod_i G_i \rightarrow \prod_i G_i$$ given by $(\ldots, g_i, \ldots) \mapsto (\ldots, g_i - \alpha_{i+1}(g_{i+1}), \ldots)$.

I've spent a half-hour trying to show that $\delta$ is surjective, but I haven't gotten very far with it. I don't actually need this result right now - all the morphisms in the inverse system i'm consider are surjective, and it's trivial to show that $\varprojlim^1$ dies in this case - so I'm just going to leave it and move on for now, but does anybody have a proof they could share? I feel as if it only needs a little thought brought to bear on it, but thought is a rare commodity on Friday afternoons... $\ddot \smile$

Answer 1

I don't know if there is a direct proof, but one can prove things indirectly as follows:

If one has a short exact sequence $0\to \mathcal{A}\to \mathcal{B}\to \mathcal{C}\to 0$ of inverse sequences, then the snake lemma yields an exact sequence

$$ 0 \to {\displaystyle\lim_{\longleftarrow}}\mathcal{A} \to {\displaystyle\lim_{\longleftarrow}}\mathcal{B} \to{\displaystyle\lim_{\longleftarrow}}\mathcal{C} \to{\displaystyle\lim_{\longleftarrow}}^1\mathcal{A} \to{\displaystyle\lim_{\longleftarrow}}^1\mathcal{B} \to{\displaystyle\lim_{\longleftarrow}}^1\mathcal{C} \to 0$$

You know that ${\displaystyle\lim_{\longleftarrow}}^1\mathcal{A}$ vanishes when the maps are surjective. The next step is to show that it vanishes when $\mathcal{A}$ satisfies the trivial ML-condition, that is, when the images are eventually all $0$. Finally, if $B_i$ is the eventual image inside of $A_i$, we can look at the short exact sequence

$$0\to \mathcal{B}\to \mathcal{A}\to \mathcal{A}/\mathcal{B}\to 0$$

The first term has surjective maps and the last satisfies trivial ML.

Answer 2

(Has it really been more than a decade since 2011?) I came across this question while working on a couple of exercises in section 8, part 3 of Adams' blue book. Here's a direct way to prove this (it more or less unravels the exact sequence in the answer above). I'll use a slightly different notation.

Let $\{G_n\}$ be abelian groups with morphisms $g_{n, n+1}\colon G_{n+1}\to G_n$. We have

$$\begin{align*} \delta\colon&\Pi_n G_n\to\Pi_n G_n\\ & (x_n)\mapsto (x_n-g_{n, n+1}x_{n+1}) \end{align*}$$

and $\varprojlim^1G_n$ is the cokernel and is zero iff $\delta$ is surjective. In a way, the inverse of $\delta$ would be given by tail sums (by bringing everything to the corresponding group), but the issue is with convergence.

Exercise (from Adams): If each $g_{n, n+1}$ is injective, then $G = G_1$ has a descending filtration by subgroups, making it a topological group. Then $\varprojlim^1G_n = 0$ iff $G$ is complete (i.e., Cauchy sequences have limit(s)). (When $G$ is complete, the tail sums converge and we can construct an inverse.)

Now we add the ML condition. The ML condition gives to each $G_n$ a subgroup $H_n$ which is the eventual limit of the images of $g_{n, m}, m\geq n$. The key observation is that $g_{n, n+1}(H_{n+1}) = H_n$.

The ML condition also allows us to determine the $n$th tail sum up to an element in $H_n$, i.e., given a sequence $(y_n)\in\Pi_nG_n$, the $n$th "tail sum" gives us a coset of $G_n/H_n$. What we'll do is to find elements within these cosets inductively. Now for the actual proof:

Fix an element $(y_n)\in\Pi_nG_n$. For each $n$, by the ML condition, there is an $N_n\geq n$ beyond which the images stabilize. We choose these numbers so that $N_1>N_2>\dots$. Now, let

$$x_n = y_n+g_{n, n+1}(y_{n+1})+\dots+g_{n, N_n-1}(y_{N_n-1}).$$

By construction, $x_n-g_{n, n+1}(x_{n+1})-y_n\in H_n = g_{n, n+1}H_{n+1}$.

We don't do anything to $x_1$. Since $x_1-g_{1, 2}(x_2)-y_1\in H_1 = g_{1, 2}H_2$, modify $x_2$ appropriately so that an equality holds. Say, $x_2$ becomes $x_2+s$ for some $s\in H_2$. We still have $x_2+s-g_{3, 2}(x_3)-y_2\in H_2$ because $H_2$ is a subgroup. This lets us inductively modify each $x_n$ to prove surjectivity of $\delta$.