Suppose we have two inverse systems $(A_i)$ and $(B_i)$ of abelian groups and we have a homomorphism from one system to other, say $\lambda_i:A_i\rightarrow B_i$ for all $i$ (so that respective commutativity in diagram holds).
Even if each $\lambda_i$ is surjective, the map induced between inverse limits $\lambda: \lim A_i \rightarrow \lim B_i$ need not be surjective.(*)
Are there some sufficient condition on systems so the $\lambda$ will be surjective?
[(*) Example: $A_i=\mathbb{Z}$ with identity maps $A_{i+1}\rightarrow A_i$, $B_i=\mathbb{Z}/p^i$ with natural map $B_{i+1}\rightarrow B_i$ and the morphism from $A_i$ to $B_i$ given by $\lambda_i:\mathbb{Z}\rightarrow \mathbb{Z}/p^i$, the natural map].
Your friend here is the Mittag-Leffler Condition. For surjections $\lambda_i:A_i\to B_i$ between directed inverse systems, the kernels $K_i$ again form a directed system. If the $K_i$ satisfy the Mittag-Leffler Condition, then taking the limit will be exact, and so $\lambda:A\to B$ will be surjective.