The following theorem is well-known:

Let $A$ be a ring. Let $C$ be a cocomplete $Ab$-category and $X$ an $A$-module-object in $C$, i.e. an object endowed with a ring homomorphism $\sigma : A \to \text{End}(X)$. Then this yields a cocontinuous $Ab$-functor $F : \text{Mod}(A) \to C$ with $F(A)=X$ and the action of $F$ on endomorphisms of $A$ is given by $\sigma$. Actually this yields an equivalence of categories between the category of cocontinuous $Ab$-functors $\text{Mod}(A) \to C$ and the category of $A$-module objects in $C$.

The crucial step is the construction of $F$. It works as follows: Construct first the right adjoint of $F$ (which has to exist by abstract nonsense). For an object $Y$, obviously the abelian group $\text{Hom}(X,Y)$ carries the structure of a right $A$-module; just precompose with the action of $X$. The functor $\text{Hom}(X,-) : C \to \text{Mod}(A)$ has a left adjoint $F = X \otimes -$: Namely it is enough to define $F$ on objects and verify the corresponding universal property; the action on morphisms, functoriality and preservation of colimits is automatic. Well, $X \otimes A = X$, it is then clear what to do for free $A$-modules, and choosing a presentation of an arbitrary $A$-module $M$, it is clear how to define $X \otimes M$.

Question Is there a direct construction of the functor $X \otimes -$ which does not involve universal properties?

The background is that I want to prove some variants of this theorem where we cannot just enrich the hom-functor over $\text{Mod}(A)$ and use adjoints. So this question is not just out of curiosity. I hope that an affirmative answer to this question makes the other theorems accessible.

It's clear how to start: For every free $A$-module $M$, fix a basis $B_M$, and put $X \otimes M := X^{\oplus B_M}$ (I think if we do not want to use the global Axiom of Choice, we should use anafunctors instead). Thinking of matrices, it is easy to make $X \otimes -$ a functor on the free modules. Now if $M$ is an arbitrary $A$-module, fix once and for all a presentation

$M_1 \to M_2 \to M \to 0$

with free $A$-modules $M_1,M_2$, and define $F(M)$ to be the cokernel of $F(M_1) \to F(M_2)$, i.e. the sequence

$F(M_1) \to F(M_2) \to F(M) \to 0$

is exact (at least if $C$ is abelian, which you may assume if necessary). Now in order to show that $F$ is functorial, let $N$ be another $A$-module and $N_1 \to N_2 \to N \to 0$ its presentation. Suppose $f : M \to N$ is a homomorphism. Since $M_2$ is projective, we may lift it to a homomorphism $g : M_2 \to N_2$. However, I don't know how to go on.

Even if this works, I'm curious if we can do it without using that free objects in the presentation are projective. The motivation for this comes from this unanswered question.

  • $\begingroup$ I would think you could do the following. First, convince yourself that $C$ is a right $\rm Ab$ module, in that there's a "universal" map $\boxtimes:C\times {\rm Ab}\to C$. Now check that you have a map $X\boxtimes A \to X$ in $C$, where by $A$ I mean the underlying abelian group, and that you have $A\boxtimes M\to M$ for any $M \in \operatorname{Mod}(A)$. Make sure everything is cocontinuous, and define $X\otimes M$ to be the coequalizer of the two maps $X\boxtimes A\boxtimes M\to X\boxtimes M$. I haven't worked out any details; if I do, I'll leave an answer, but for now it's a comment. $\endgroup$ – Theo Johnson-Freyd Jan 5 '11 at 18:21
  • $\begingroup$ Sounds good! Your idea may be summarized as $X \otimes_A M = (X \otimes_\mathbb{Z} M)/(x \otimes am = ax \otimes m)$. Thus we may reduce to the case $A=\mathbb{Z}$. But here the same problems as above arise. $\endgroup$ – Martin Brandenburg Jan 5 '11 at 19:14

This is too long for a comment only, therefore I write it into an answer (all categories are assumed to be additive). This will not answer your question but maybe it will provide some insight.

Let me first show you how to construct $F$ in the specific situation of $\text{Mod}(A)$ -- this is essentially the proof of the comparison theorem for projective resolutions: You've already explained how to get $g_{2}: M_{2} \to N_{2}$ from a given $f: M \to N$.

Let $ZM$ be the image of $M_{1} \to M_{2}$ and let $ZN$ be the image of $N_{1} \to N_{2}$. Since the morphism $ZM \to N$ is zero, there is a unique morphism $h: ZM \to ZN$ such that the diagram

M_1 --->> ZM >--> M_2 ---->> M
           |       |         |
           h      g_2        f
           |       |         |
           v       v         v
N_1 --->> ZN >--> N_2 ---->> N

commutes. Now use projectivity of $M_{1}$ to lift $M_1 \to ZN$ up to $g_{1}: M_{1} \to N_{1}$. Note that $g_{2}$ and $g_{1}$ are far from unique. If $g_{2}$ and $g_{2}'$ are two lifts of $f$ then their difference will factor through a map $M_{2} \to ZN$ and you may lift this map to a morphism $M_{2} \to N_{1}$ by using projectivity of $M_{2}$ again. The non-uniqueness of $g_{1}$ need not concern us.

We are thus led to the following category $\text{mod}(\mathscr{F}(A))$: Objects are morphisms $(M_{1} \to M_{2})$ with $M_{1}$ and $M_{2}$ free modules in $\text{Mod}(A)$. Two morphisms of arrows (commutative squares) are identified if $g_{2} - g_{2}'$ factors through a morphism $M_{2} \to N_{1}$. Think of an object $(M_{1} \to M_{2})$ as representing its cokernel in $\text{Mod}{(A)}$, two morphisms are identified if and only if they induce the same morphism on the cokernel.

By construction, a functor from the category of free modules $\mathscr{F}(A)$ to a category with cokernels $C$ will extend uniquely (up to unique isomorphism) to a functor $\text{mod}(\mathscr{F}(A)) \to C$. Moreover, it is not difficult to check that the inclusion $\mathscr{F}(A) \to \text{Mod}(A)$ extends to an equivalence $\text{mod}(\mathscr{F}(A)) \to \text{Mod}(A)$. So this should be more or less a complete answer for the question of how to get $F$ and its functoriality.

The category $\text{mod}(\mathscr{A})$ described above can be defined for every additive category. The notation is motivated by thinking of it as the category of finitely presented functors on $\mathscr{A} \to \text{Mod}{(\mathbb{Z})}$ - that is to say of functors of the form $\text{Coker}(\text{Hom}(-,M_{1}) \to \text{Hom}(-,M_{2}))$. It is abelian if and only if $\mathscr{A}$ has weak kernels (weak kernels are defined as kernels but without requiring uniqueness). A discussion of all this can be found in Freyd, Representations in abelian categories, in the La Jolla proceedings 1966. A more recent exposition is in Beligiannis, On the Freyd categories of an additive category, which you can download from his homepage.

I do not know how to do this without using that free modules are projective and whether it is helpful for your original motivation.

  • $\begingroup$ Thanks! Apprently I forgot basics about projective resolutions :). $\endgroup$ – Martin Brandenburg Jan 6 '11 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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