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This question is a follow-up of this earlier question I asked.

Let $$ A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots $$ be an inductive sequence of abelian groups, the connecting homomorphisms of which are surjective and split, that is, we have embeddings $A_{n+1}\rightarrowtail A_n$ such that the diagram \begin{array}{ccccccccc} A_n & \twoheadrightarrow & A_{n+1}\\ \uparrow & & \uparrow\\ A_n & \leftarrowtail & A_{n+1} \end{array} commutes for every $n$. Here the vertical arrows denote identity homomorphisms. This means that $A_{n+1}$ is a direct summand of $A_n$.

Let $\varinjlim A_n$ denote the inductive limit of the system $$ A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots $$ and let $\varprojlim A_n$ denote the projective limit of the system $$ A_1\leftarrowtail A_2\leftarrowtail A_3\leftarrowtail A_4\leftarrowtail \cdots. $$ We get an induced map $$ \varprojlim A_n\to\varinjlim A_n. $$ As Zhen Lin has shown in his answer to my earlier question, this map need not be surjective. Here is a weaker question:

Question: If we have $\varinjlim A_n=0$, then can we conclude that $\varprojlim A_n=0$?

This would, of course, follow if the map $\varprojlim A_n\to\varinjlim A_n$ was always injective. Is there any reason to expect this?

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I re-asked this question on MO: mathoverflow.net/questions/116060/… –  Rasmus Dec 11 '12 at 9:46

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