Let $R$ be a commutative ring with unity. Let $R$-Mod denote the category of $R$-modules, and $Ab$ denote the category of Abelian groups. Now, it is known that a covariant additive functor

$T: R$-Mod $\to $ Ab preserves direct limits (resp. inverse limits) if and only if for some $B\in R$-Mod, we have $T(-)\cong B \otimes_R (-)$ (resp. $T(-)\cong Hom_R(B,-)$ ).

My question is the following : Let $T:R$-Mod $\to R$-Mod be a covariant linear functor. Is it true that $T$ preserves direct limits (resp. inverse limits) if and only if for some $B\in R$-Mod, we have $T(-)\cong B \otimes_R (-)$ (resp. $T(-)\cong Hom_R(B,-)$ ) ?

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    $\begingroup$ It’s not true that an additive functor $R\!\!-\!\!\text{Mod}\to\text{Ab}$ that preserves direct limits is a tensor product. For example, $\text{Hom}_R(X,-)$ preserves direct limits if $X$ is finitely presented, but is not right exact (and so can’t be a tensor product functor) unless $X$ is projective. Maybe you meant general colimits/limits instead of direct/inverse limits? $\endgroup$ – Jeremy Rickard May 19 at 12:46
  • $\begingroup$ @Jeremy Rickard: I am using the definition of direct limits, as given in Chapter 5, Homological algebra, by Rotman (indexing sets are just partially ordered sets, perhaps you're thinking about indexing set which are also directed ? ) The facts I mention are Theorem 5.51 and Theorem 5.52 of chapter 5 in Rotman's book $\endgroup$ – user102248 May 19 at 22:45
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    $\begingroup$ Ah, OK. I think it's much more common (at least these days) to reserve the terms "direct limit" and "inverse limit" for the directed case. But the fact that the usage is not universal is probably an argument for just avoiding the terms altogether, and using "colimit" and "directed colimit". $\endgroup$ – Jeremy Rickard May 20 at 8:24
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    $\begingroup$ ncatlab.org/nlab/show/Eilenberg-Watts+theorem $\endgroup$ – Qiaochu Yuan May 20 at 8:51

Yes, if $F:R\text{-mod}\to R\text{-mod}$ is an $R$-linear functor, then it preserves all colimits iff it is isomorphic to a functor of the form $B\otimes_R-$, and preserves all limits iff it is isomorphic to a functor of the form $\text{Hom}_R(B,-)$. The proofs are pretty much the same as for additive functors $R\text{-mod}\to\text{Ab}$. Briefly:

For the colimit-preserving case, let $B=FR$. There is a natural transformation $\alpha:FR\otimes_R-\to F$ given by the chain of natural maps $$\alpha_X:FR\otimes_RX\cong FR\otimes_R\text{Hom}_R(R,X)\stackrel{\text{id}\otimes F}{\longrightarrow}FR\otimes_R\text{Hom}_R(FR,FX) \stackrel{\text{eval}}{\longrightarrow}FX.$$

It is straightforward to check that $\alpha_X$ is an isomorphism for $X=R$. Since both $FR\otimes_R-$ and $F$ preserve coproducts, $\alpha_X$ is an isomorphism for $X$ any free module. Since any module $X$ fits in an exact sequence $P_1\to P_0\to X\to0$ with $P_1$ and $P_0$ free, and since both $FR\otimes_R-$ and $F$ are right exact, $\alpha_X$ is an isomorphism for all $X$, and so $\alpha$ is an isomorphism of functors.

For the limit-preserving case, this proof works:

The conditions of the Special Adjoint Functor Theorem are satisfied, and so $F$ has a left adjoint $L$. Then there are natural isomorphisms $$FX\cong\text{Hom}_R(R,FX)\cong\text{Hom}_R(LR,X),$$ and so $F\cong\text{Hom}_R(LR,-)$.

  • $\begingroup$ But the OP doesn't require $R$-linearity. Without this what you get is functors given by tensoring with an $(R, R)$-bimodule. $\endgroup$ – Qiaochu Yuan May 20 at 8:52
  • $\begingroup$ @QiaochuYuan He wrote "linear", which I interpreted to mean "$R$-linear" (especially since he used a different term "additive" for "$\mathbb{Z}$-linear"). $\endgroup$ – Jeremy Rickard May 20 at 9:12
  • $\begingroup$ by linear, I did mean $R$-linearity ... thanks $\endgroup$ – user102248 May 20 at 22:39

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