Let $I$ be a cofiltered index category, $A_i$ be a directed system of $R$-modules. Is the following true?


(Note this is not true for arbitrary limits, for example if $I$ is the discrete category.)


1 Answer 1


No. For instance, let $R=\mathbb{Q}$, $I=\mathbb{N}$, and $A_n=\mathbb{Q}^n$ connected by the obvious projection maps. The inverse limit $\lim A_n$ is then $\mathbb{Q}^\mathbb{N}$, which has uncountable dimension. In particular, $\operatorname{Hom}(\mathbb{Q}^\mathbb{N},\mathbb{Q})$ is uncountable. But $\operatorname{Hom}(\mathbb{Q}^n,\mathbb{Q})$ is countable for each $n$, and so it follows that the direct limit is also countable.

In fact, you can find a counterexample for any nonzero module $B$ at all: see my answer here.

  • $\begingroup$ Ok. I think it a bit strange. There're natural isomorphisms $\mathrm{Hom}(P,\lim Q)\cong ...$ and $\mathrm{Hom}(\mathrm{colim}P,Q)\cong...$) by universal properties. Moreover, for directed index, we have $\mathrm{Hom}(M,\mathrm{colim}_IM_i)=\mathrm{colim}_I(\mathrm{Hom}(M,N_i))$, is there a "lim" counterpart for this? $\endgroup$
    – Qixiao
    Oct 23, 2016 at 0:57
  • $\begingroup$ Probably the right dual isomorphisms are $\operatorname{Hom} (X, \operatorname{lim} Y_i) \cong \operatorname{lim} \operatorname{Hom} (X, Y_i)$ and $\operatorname{Hom} (\operatorname{colim} X_i, Y) \cong \operatorname{lim} \operatorname{Hom} (X_i, Y)$? $\endgroup$
    – user144221
    Oct 23, 2016 at 1:05
  • $\begingroup$ @Qixiao: That last statement you wrote is only true if $M$ is finitely presented, so you would expect a dual statement to hold only if $M$ is "co-finitely presented". However, I think that the only "co-finitely presented" modules are $0$, though I haven't checked carefully. In any case, it shouldn't be too surprising that some statements like this can't be dualized: all that's going on is that $R\text{-Mod}^{op}$ is not a category of modules over some ring. $\endgroup$ Oct 23, 2016 at 1:08
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    $\begingroup$ @Qixiao: Consider $\mathbb{Q}$ as the filtered colimit of its cyclic subgroups and $M=\mathbb{Q}$. There is no nonzero map $\mathbb{Q}\to\mathbb{Z}$ so the right hand side of your equation is $0$, but the left hand side is nonzero since there are nonzero maps $\mathbb{Q}\to\mathbb{Q}$. $\endgroup$ Oct 23, 2016 at 2:27
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    $\begingroup$ I have posted a question about whether there can be any nontrivial "co-finitely presented" modules at math.stackexchange.com/questions/1980817/…. $\endgroup$ Oct 23, 2016 at 2:46

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