most recent 30 from http://mathoverflow.net 2013-05-22T20:02:29Z http://mathoverflow.net/feeds/question/116060 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116060/inductive-vs-projective-limit-of-sequence-of-split-surjections Rasmus 2012-12-11T09:43:30Z 2012-12-11T11:21:09Z

<p>Let $$ A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots $$ be an inductive sequence of <strong>countable</strong> abelian groups, the connecting homomorphisms of which are surjective and <strong>split</strong>, 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 $A_n$.</p> <p>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 <a href="http://math.stackexchange.com/users/5191/zhen-lin" rel="nofollow">Zhen Lin</a> has shown in <a href="http://math.stackexchange.com/a/254702/367" rel="nofollow">over here</a>, this map need not be surjective. Here is a weaker question:</p> <p><strong>Question:</strong> If we have $\varinjlim A_n=0$, then can we conclude that $\varprojlim A_n=0$?</p> <p>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?</p> <p>[Earlier versions of this question were posted <a href="http://math.stackexchange.com/questions/254649" rel="nofollow">here</a> and <a href="http://math.stackexchange.com/questions/254822" rel="nofollow">here</a> on MSE.]</p>

http://mathoverflow.net/questions/116060/inductive-vs-projective-limit-of-sequence-of-split-surjections/116068#116068 Jeremy Rickard 2012-12-11T10:52:05Z 2012-12-11T11:21:09Z

<p>I think it's true that $\varprojlim A_n\to\varinjlim A_n$ is always injective.</p> <p>We may as well assume that </p> <p>$A_1\leftarrowtail A_2\leftarrowtail A_3\leftarrowtail A_4\leftarrowtail \cdots$</p> <p>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 $A_k\rightarrowtail A_1$.</p> <p>This doesn't use countability.</p>