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Let $$ A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots $$ be an inductive sequence of countable 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 composition $A_{n+1}\rightarrowtail A_n\twoheadrightarrow A_{n+1}$ is the identity for every $n$. 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 over here, 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?

[Earlier versions of this question were posted here and here on MSE.]

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up vote 2 down vote accepted

I think it's true that $\varprojlim A_n\to\varinjlim A_n$ is always injective.

We may as well assume that

$A_1\leftarrowtail A_2\leftarrowtail A_3\leftarrowtail A_4\leftarrowtail \cdots$

is a sequence of inclusions of nested subgroups, so $\varprojlim A_n$ is just the intersection. An element of the kernel of $\varprojlim A_n\to\varinjlim A_n$ is just an element $a$ of $\bigcap A_n$ that is in the kernel of the map $A_1\twoheadrightarrow A_k$ for some $k$. But this implies $a=0$ since this map is a splitting of the inclusion $A_k\rightarrowtail A_1$.

This doesn't use countability.

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Thank you very much for your enlightening reply. –  Rasmus Bentmann Dec 12 '12 at 7:23
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