The Mittag-Leffler condition and $\varprojlim^1$ 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... $\ddot \smile$
 A: 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.
