Question about the Dual Statement for Injective Modules It's well-known that there are several equivalent definitions of a projective $A$-module $P$:


*

*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 $fh=g$; 

*Every exact sequence $0 \to M' \to M \to P \to 0$ splits; 

*There exists a module $M$ such that $P \oplus M$ is a free module; and 

*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?
Thanks so much in advance.  
 A: Let's follow up on the comments.
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.
