# Question about the Dual Statement for Injective Modules

It's well-known that there are several equivalent definitions of a projective $A$-module $P$:

1. Given an exact sequence $M \xrightarrow{f} M' \to 0$ and a morphism $P \xrightarrow{g} M'$, there exists a morphism $P \xrightarrow{h} M$ such that $fh=g$;
2. Every exact sequence $0 \to M' \to M \to P \to 0$ splits;
3. There exists a module $M$ such that $P \oplus M$ is a free module; and
4. The functor $\mathrm{Hom}_A(P,-)$ is exact.

Now the dual versions (i.e., for the definition of an injective module) of 1, 2, and 4 are easy. My question is, does there exist a dual version of 3?

I would actually like to be phrasing this in the context of an arbitrary abelian category, so would 3 be stated as something like "a subobject of a free object"? Then what would the dual version of this statement be? "A quotient object of..." something?

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"Quotient of a cofree object"... You need to think a bit what "free object" really means, so as to see if you can dualize it. –  Mariano Suárez-Alvarez Dec 8 '11 at 2:17
@Mariano: Well, except that "direct summand" is more than just "submodule"... –  Arturo Magidin Dec 8 '11 at 2:39
Yeah, indeed (I was dualizing the statement in the 3rd oparagraph, really) The problem is more with the «cofree object» part, though, no? –  Mariano Suárez-Alvarez Dec 8 '11 at 2:43
@dmdmdmdmdmd: Also: keep in mind that the duality between projectives and injectives is not perfect; for example, every module has an injective hull, but there is no dual "projective" notion that exists for all modules. In other words: it may not be possible to dualize the condition. –  Arturo Magidin Dec 8 '11 at 3:21
In the category of finitely generated modules for a finite dimensional algebra $A$ over a field $k$, you can replace 3) by "there exists $M$ such that $I \oplus M$ is a direct sum of copies of $(A_A)^*$". ($I$ is a left $A$-module and * is $\hom_k(-,k)$) –  mt_ Dec 8 '11 at 9:46

A first attempt at dualizing the condition would dualize "direct summand" (it's actually self-dual, since a direct summand is either an onto homomorphism $F\to P$ that has a section, or a one-to-one homomorphism $P\to F$ that has a retraction) and "free module".

The "free module on $X$" is $\mathbf{F}(X)$, where $\mathbf{F}\colon\mathscr{Set}\to R\mathscr{mod}$ is the left adjoint of the underlying set function $\mathbf{U}\colon R\mathscr{mod}\to\mathscr{Set}$. So a "co-free module on $X$" would be $\mathbf{G}(X)$, where $\mathbf{G}\colon\mathscr{Set}\to R\mathscr{mod}$ where $\mathbf{G}$ is the right adjoint of the underlying set functor.

However, left adjoints respect colimits; in particular, if $\mathbf{G}$ existed, then $\mathbf{U}$ would be a left adjoint, and so the underlying set of a direct sum of two modules (which is a coproduct) would necessarily be (bijectable with) the disjoint union of the underlying sets of the modules (as the disjoint union is the coproduct in $\mathscr{Set}$). This fails in every category of modules, since the direct sum of two trivial modules has underlying set with a single element, but the disjoint union of the underlying sets has two elements. So the underlying set functor cannot have a right adjoint, so there is no "co-free" functor.

Alternatively, you can consider the universal property of the free object and see if that can be constructed. A co-free module on a set $X$ would be a module $C(X)$, together with a set-theoretic function $f\colon C(X)\to X$ such that for every module $M$ and every set-theoretic function $g\colon M\to X$, there exists a unique module homomorphism $\varphi\colon M\to C(X)$ such that $g = f\circ\varphi$. However, if $X$ has more than one element, then no such object can exist: let $a\in X$ be different from $f(0)$; then define $g\colon C(X)\to G$ to be given by $g(c)=a$ for all $c$. Since every module homomorphism must map $0$ to $0$, we would have $a = g(0) = f(\varphi(0)) = f(0)$, contradicting the choice of $a$. For $X$ a singleton, $C(X)$ must be the trivial module. For $X$ empty, there is no co-free object.

So we cannot dualize "direct summand of free module" into "direct summand of co-free module", because there is no object to play the role of the co-free module. Note that you would need precisely the universal property of the "co-free module" in order to dualize the proof that a direct summand of a free module is projective and vice-versa.

An alternative approach might be to think of "free module" as just short-hand for "direct sum of copies of $R$." If we do that, then the dual of "direct summand of a direct sum of copies of $R$" would be "direct summand of a direct product of copies of $R$."

But this doesn't work either: for instance, for $\mathbb{Z}$-modules, "injective" is equivalent to "divisible". We would need $\mathbb{Q}$ to be a direct summand a direct product of copies of $\mathbb{Z}$. But $\mathbb{Q}$ cannot even be embedded in a direct product of copies of $\mathbb{Z}$, because any homomorphism $\mathbb{Q}\to\mathbb{Z}$ is trivial, so any map $\mathbb{Q}\to\prod \mathbb{Z}$ is necessarily trivial (it is completely determined by the projections, and all the projections are trivial).

In short: there is no dual to the proposition that a module is projective if and only if it is a direct summand of a free module.

This is just another facet of the fact that the duality between injectives and projectives is not perfect. As I mentioned in comments, for example, every module has an injective hull but not every module has a projective cover, so the duality is not perfect.

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+1: great answer. –  Alex Youcis Dec 8 '11 at 4:28
Thank you, Arturo, for taking the time to respond with such a thoughtful answer. I am wondering something, though, now that I remembered I had this book. In Hilton & Stammbach, p. 35, they define an A-module to be cofree if it is the direct product of modules Hom_Z(A,Q/Z). Then, they state in Th. 8.4 the dual version of 3 above: I is injective if and only if I is a direct summand in a cofree module. I am wondering: if this does not have the same interpretation as being in the image of a cofree functor, why call it cofree in the first place? –  dmdmdmdmdmd Dec 8 '11 at 6:00
@dmdmdmdmdmd: They are dualizing yet another way of thinking about free modules (direct sums of Hom sets into $A$); it does not have the dual property of the universal property, nor the dual property of the left adjoint. It's related to thinking about "cogenerators" for a module. I'll take a look at my copy of Anderson-Fuller tomorrow and see if they do something similar. –  Arturo Magidin Dec 8 '11 at 6:04
Oh, I see! Thanks again for the wonderful explanation! –  dmdmdmdmdmd Dec 8 '11 at 6:27
@dmdmdmdmdmd: Well, Anderson-Fuller don't discuss this version, nor does Lam (though Anderson-Fuller does include the notion of "cogenerators"). Basically, a function from a singleton $\{x\}$ to a module $M$ is equivalent to an abelian group homomoprhism $\mathbb{Z}\to M$; so a function from a set $X$ to a module $M$ is equivalent to an abelian group homomoprhism $\oplus_{x\in X}\mathbb{Z}\to M$. A free $A$-module over $K$ is then equivalent to a direct sum of copies of $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z},A)$; dualizing gives the direct product in question. –  Arturo Magidin Dec 8 '11 at 17:22