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.]


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.