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OK, I am looking for no "hands waiving" proof that inverse limit over a cofinal index subset is isomorphic to the inverse limit over its superset. For instance: Given an (Abelian) category $C$ with infinite products (a sufficiently normal one), and an inverse system $f_{ij}:A_j\longrightarrow A_i$ indexed by an upward directed set $I$ that has a cofinal subset $J$, prove that $\varprojlim_I A_i\cong\varprojlim_J A_j$. I do not assume $J$ to be linearly ordered, rather it only satisfies $\forall i\in I\exists j\in J$ with $i\leq j$. If $J$ is assumed to be linearly (hence) well ordered, the proof is not too hard.

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    $\begingroup$ There is a more general definition of cofinal functor for categories, for which there is an analogous theorem. You can find a proof in e.g. [Categories for the working mathematician]. $\endgroup$ – Zhen Lin Dec 19 '15 at 0:35
  • $\begingroup$ Thanks @Zhen Lin, I will try to find it $\endgroup$ – Rado Dec 19 '15 at 1:13
  • $\begingroup$ @Zhen Lin, yes Mac Lane's CWM has a notion of a cofinal functor and final subcategory that he calls "final" since the prefix "co" is used for duality purposes. And this is a generalizing notion to cofinality of partially ordered sets. He does prove my question above for direct limits, and comments in the notes that dual notions can be had when using (co)initial. However "initial" may be replaced by "final" even for inverse limits and the isomorphism will work with appropriate adjustments. So your comment was useful, but I do not know how to give you a medal here... $\endgroup$ – Rado Dec 21 '15 at 20:06
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For this particular situation, argue as follows. Let $(A,g_j)$ be the limit over $J$; let us show that $(A,f_{ij}g_j)$ is a limit over $I$. First, it's a cone: $f_{jk}f_{ij}g_j=f_{ki}g_j$ by functoriality. Now let $(B,h_i)$ be a cone over $I$. This restricts to a cone $(B,h_j)$ over $J$, and the latter gives rise to a unique $k:B\to A$ with $g_j\circ k=h_j$ for $j\in J$. Now given any $i\in I$, let $j\in J$ and $f_{ij}:A_j\to A_i$. Then $f_{ij}g_jk=f_{ij}h_j=h_i$, which shows the $h_i$ factor through $A$. And this was the only possible factorization $k$, since $k$ is determined even by the restriction to $J$.

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  • $\begingroup$ Thank you @Kevin Carlson, your proof is certainly useful. The only small thing is that one should show that the argument works, independent of the choice of $j\in J$ (from your penultimate sentence). One can invoke upward directedness of $I$ but one can likewise relax that condition and impose an appropriate condition on $J$. $\endgroup$ – Rado Dec 21 '15 at 20:12
  • $\begingroup$ Yeah, indeed. Either use directedness or the general definition of cofinality. $\endgroup$ – Kevin Carlson Dec 22 '15 at 2:32

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