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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!)

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    $\begingroup$ 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. $\endgroup$
    – user144221
    Oct 13, 2016 at 17:40
  • $\begingroup$ (Well, a really similar description is impossible; some nice characterization, not just an ad hoc construction.) $\endgroup$
    – user144221
    Oct 13, 2016 at 17:51

1 Answer 1

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Duals are always dual in every correct senses if the dual construction exists.

Duality is about dual objects and reversed arrows, but not about e.g. set theoretical notations as subsets and supersets. Example:

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 $M$.

A problem with co-free modules is that it lacks the dual of the universal property for free modules:

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.

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