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?


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 $(0, 0, 0, \ldots)$.

  • 1
    $\begingroup$ Thank you for your answer. May I ask a follow-up question: if we have $\varinjlim A_\bullet=0$, then can we conclude that $\varprojlim A_\bullet=0$? $\endgroup$ – Rasmus Dec 9 '12 at 19:27
  • $\begingroup$ I'm afraid I can't think of any reason why that should be the case, nor any counterexample. $\endgroup$ – Zhen Lin Dec 9 '12 at 19:47
  • $\begingroup$ 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)? $\endgroup$ – Rasmus Dec 11 '12 at 9:39
  • $\begingroup$ 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. $\endgroup$ – Zhen Lin Dec 11 '12 at 9:42
  • $\begingroup$ Correct link to follow up question: math.stackexchange.com/questions/254822/… $\endgroup$ – Rasmus Dec 11 '12 at 9:45

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.