# Inductive vs projective limit of sequence of split surjections

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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## 1 Answer

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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Thank you for your answer. May I ask a follow-up question: if we have $\varinjlim A_\bullet=0$, then can we conclude that $\varprojlim A_\bullet=0$? – Rasmus Dec 9 '12 at 19:27
I'm afraid I can't think of any reason why that should be the case, nor any counterexample. – Zhen Lin Dec 9 '12 at 19:47
I posted a follow-up question for this weaker version of the question. Do you think it is possible to construct of countable counterexample for this question (question 254649)? – Rasmus Dec 11 '12 at 9:39
I don't see why not. Instead of taking the whole of $\mathbb{Z}^{\times \mathbb{N}}$, we could take the submodule generated by $\mathbb{Z}^{\oplus \mathbb{N}}$ (which is countable) and the sequence $(1, 1, 1, \ldots)$; in other words, this is the space of eventually-constant sequences. – Zhen Lin Dec 11 '12 at 9:42
Correct link to follow up question: math.stackexchange.com/questions/254822/… – Rasmus Dec 11 '12 at 9:45