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... $\ddot \smile$