# Properties of free and co-free modules which are not dual

I recently read that a cofree $R$-module is defined in such a way that some properties of free modules and some properties of cofree modules are dual. For example see here.

• Every free module is projective; every cofree module is injective.

• For every module $M$, there is a surjective homomorphism from a free module to $M$; for every module $M$, there is an injective homomorphism from $M$ to a cofree module.

• A module is projective iff it can be completed by a direct sum to a free module; a module is injective iff it can be completed by a direct product to a cofree module.

But it is also pointed that free and cofree modules are not exactly duals of each other in some sense.

I was thinking that there could be a property of free modules, whose dual does not hold for cofree modules and vice-versa.

Question: What are simple properties of free modules whose duals do not hold true for co-free modules? (Similar question with free and cofree interchanged!)

• The definition of cofree modules (that I am aware of) is kind of ad hoc: Hilton and Stammbach just say that $M$ is cofree if it is a product of $R$-modules $\operatorname{Hom}_\mathbb{Z} (R, \mathbb{Q}/\mathbb{Z})$. This resembles the situation with free modules, but free modules have a nice characterization: the functor $X \rightsquigarrow R \left<X\right>$ is left adjoint to the forgetful functor $R\textbf{-Mod} \to \textbf{Set}$. Do cofree modules have a similar description? Because taking direct sums/products of copies of something is not a definition but more like a construction.
– user144221
Oct 13, 2016 at 17:40
• (Well, a really similar description is impossible; some nice characterization, not just an ad hoc construction.)
– user144221
Oct 13, 2016 at 17:51

Any module $$M$$ is a subset of an injective module, but not every module $$M$$ is a superset of a projective module. What does hold is, given module $$M$$:
1. There is a monomorphism from $$M$$ to an injective module
2. There is an epimorphism from a projective module to $$M$$.
For any function $$B\to M$$ the injection $$B\to R^{(B)}$$ can uniquely be extended to a morphism $$R^{(B)}\to M$$ do works but the dual doesn't make sense at all.