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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. $ Question: Is the map $\varprojlim A_n\to\varinjlim A_n$ necessarily an isomorphism?

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No. Here is a counterexample: we take $A_i = \mathbb{Z}^{\times \mathbb{N}}$, with all the homomorphisms $A_i \to A_{i+1}$ being the left shift operator and the splittings $A_{i+1} \to A_i$ the right shift operator. Then $\varinjlim A_\bullet \ne 0$, because the sequence $(1, 1, 1, \ldots)$ cannot be annihilated after finitely many steps, but $\varprojlim A_\bullet = 0$ because we can rewrite the inverse chain $A_1 \leftarrowtail A_2 \leftarrowtail A_3 \leftarrowtail \cdots$ as a decreasing chain of subspaces of sequences that have the first non-zero entry at position $i$, and the only sequence that is in all of these subspaces is the sequence $(0, 0, 0, \ldots)$.

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    Indeed. Thank you very much.2012-12-11