Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For a unitary ring $R$ let us denote by $\mathfrak{M}_R$ the category of right $R$-modules. Let $F:\mathfrak{M}_R\rightarrow\mathfrak{M}_S$ be a functor which commutes with all inverse limits (by commuting I mean that the natural transformation inherited from the universal property of the functor $\lim$, namely $nat:F\circ\lim\rightarrow\lim\circ F$, is an isomorphism). Then is there a direct economical proof which shows that $F$ is a right adjoint functor?

share|improve this question
A minor correction: Your definition of "commuting with limits" is not correct. There is always a canonical transformation $F \circ \mathrm{lim} \to \mathrm{lim} \circ F$. By definition $F$ commutes with limits (or $F$ is continuous) if this specific transformation is an isomorphism. –  Martin Brandenburg Jan 10 '12 at 14:10
may that my terminology "natural isomorphism" confused you. By natural isomorphism I meant a natural transformation which is an isomorphism. –  Hugo Chapdelaine Jan 10 '12 at 15:48
This is not the point. Please read my comment again and compare with "existence of a natural isomorphism". –  Martin Brandenburg Jan 10 '12 at 15:50
Ok, I got it, that was a bit subtle. –  Hugo Chapdelaine Jan 10 '12 at 16:21
add comment

2 Answers

I don't think that there is a more direct proof.

First remark that by the (easy part of) Theorem of Eilenberg-Watts a (continuous) functor $F : \mathrm{Mod}(R) \to \mathrm{Mod}(S)$ has a left adjoint iff it is isomorphic to $\mathrm{Hom}_R(B,-)$ for some some $(S,R)$-bimodule $B$. If $B$ was just a right $R$-module, then a factorization of $\mathrm{Hom}_R(B,-) : \mathrm{Mod}(R) \to \mathrm{Ab}$ through $\mathrm{Mod}(S)$ corresponds to a left $S$-module structure on $B$. Thus we may replace $\mathrm{Mod}(S)$ by $\mathrm{Ab}$, i.e. we may asume $S=\mathbb{Z}$. Next observe that every continuous functor $\mathrm{Mod}(R) \to \mathrm{Set}$ factors in a unique way through $\mathrm{Ab}$. The reason is simply that in $\mathrm{Mod}(R)$ every object is an abelian group object, the addition being $M \times M \to M, (m,n) \to m + n$ etc., and continuous functors between categories with products preserve abelian group objects. Thus we may even replace $\mathrm{Ab}$ by $\mathrm{Set}$.

This reduction shows: Our special case of SAFT (Special Adjoint Functor Theorem) is equivalent to the assertion that every continuous functor $\mathrm{Mod}(R) \to \mathrm{Set}$ is representable.

But already this was a remarkable Theorem in the first years of category theory; I speculate that it is due to Freyd and that this specific assertion motivated his general SAFT. Observe that it already has many specific applications and that - a priori - it is not clear at all how to write down the representing object. For convenience, let us assume that $R$ is commutative.

a) If $\{M_i\}$ is some diagram of $R$-modules, then $\lim_i \mathrm{Hom}(M_i,-)$ is trivially continuous. By the Theorem it is representable. But this means precisely that $\mathrm{colim}_i M_i$ exists. Thus the Theorem gives us all colimits for free.

b) Let $M,N \in \mathrm{Mod}(R)$ be fixed and consider the functor of bilinear maps $M \times N \to (-)$. It is trivial to check that it is continuous. By the Theorem it is representable. A representing object is the tensor product $M \otimes_R N$.

c) Let $M \in \mathrm{Mod}(R)$ and $n \geq 0$ be fixed and consider the functor of alternating multilinear maps $M^n \to (-)$. Again it is trivial to check continuity. By the Theorem it is representable. A representing object is the exterior power $\wedge^n M$.

d) Let $R$ be some $k$-algebra. Consider the functor $\mathrm{Der}_k : \mathrm{Mod}(R) \to \mathrm{Set}$ of $k$-derivations $R \to (-)$. Again it is trivial to check continuity, so it is representable by the Theorem. A representing object is the module of differentials $\Omega^1_{R/k}$.

You can image many more examples. Is there any canonical way to produce the representing object out of the functor? I doubt it. We should appreciate Freyd's SAFT, it is even nontrivial for module categories and has many explicit applications.

In the above examples we may actually apply Freyd's GAFT (General Adjoint Functor Theorem) whose proof is a little bit more direct than for SAFT. It is even possible to spell out the proof directly, let's do it for the tensor products in b). Consider the category of all pairs $(T,\alpha)$, where $T$ is some $R$-module and $\alpha : M \times N \to T$ is bilinear. We look for an initial object in this category. It is enough to consider those $(T,\alpha)$ where the image of $\alpha$ generates $T$, we then we have $|T| \leq |(M \times N)^{\aleph_0}|$. Consider the set of all $(T,\alpha)$, where the underlying set of $T$ is an honest subset of $(M \times N)^{\aleph_0}$. Take the product of all these pairs in our category to get a pair $P$. Now take the equalizer of all endomorphisms of $P$. This will be an initial object, i.e. the tensor product $M \otimes_R N$. Of course this explicit construction is not enlightening at all, but neither is the usual one: Only the universal property matters. Even the structure of the elements of the representing object can be deduced from the universal property (see here for some examples).

share|improve this answer
Nice answer. :) –  B. Bischof Jan 10 '12 at 16:24
Thanks Martin :)! –  Hugo Chapdelaine Jan 10 '12 at 16:49
add comment

Let $U$ a cogenarator (i.e. $h_U$ is faithful) on $Mod_R$ (example $U=AbGr(R, Q/ Z)$ with $AbGr$ the category of abelians groups, and $Q$ and $Z$ the ring of rationals and integers).

From the Freyd theorem is enought show a set of objects $A_i\in Mod_R\ i\in I$ such that for any morphism $f: B\to F(A)$ has a factorization type: $f=F(h)\circ f': B\to F(A_i)\to F(A)$ with $h: A_i\to A$. For $X\in Set$ let $U^X:=\prod_{x\in X}\ U_x$ (with $U_x=U$). For $A\in Mod_R$ let there is a natural tautological morphis $\rho_A: A\to U^{(A, U)} $ with $\pi_a\circ\rho_A=a$, and this is a monomorphis because $U$ is a cogenerator. Let fix a morphism $f: B\to F(A)$ and let $\mu_f: U^{(B, F(U))}\to U^{(A, U)}$ naturally induced by the map $(A, U)\to (B, F(U)): g\mapsto F(g)\circ f$.

Consider the pullback of $\mu_f$ by $\rho_A$, this is a monomorphism $h_f: P\to U^{(B, F(U))}$. The composition $F(\rho_A)\circ f$ is equal to $F(\mu_f)\circ \rho_B$ where $\rho_B: B\to F(U)^{(B, T(C))}$ is the tautological morphism definite in similar way as above (and we use that $F$ preserve prodots, and we "equalizing" the canonical isomorphism of "limits commuting"). Then the morphism $f$ as a factorization on the (monomorfism) $F(h_f)$. Then is enought chose for $A_i\ i\in I$ a representing set of subobject of $U^{(B, F(U))}$ .

share|improve this answer
This is just the usual, general proof. It does not have anything specific to do with the module categories. –  Martin Brandenburg Jan 11 '12 at 21:36
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.