# $\mathrm{Hom}(\lim_IA_i,B)=\mathrm{colim}_I(\mathrm{Hom}(A_i,B))$?

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

$$\mathrm{Hom}(\lim_IA_i,B)=\mathrm{colim}_I(\mathrm{Hom}(A_i,B))$$

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

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.
• 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? Oct 23, 2016 at 0:57
• Probably the right dual isomorphisms are $\operatorname{Hom} (X, \operatorname{lim} Y_i) \cong \operatorname{lim} \operatorname{Hom} (X, Y_i)$ and $\operatorname{Hom} (\operatorname{colim} X_i, Y) \cong \operatorname{lim} \operatorname{Hom} (X_i, Y)$?
• @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. Oct 23, 2016 at 1:08
• @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 $\mathbb{Q}\to\mathbb{Q}$. Oct 23, 2016 at 2:27