Let $R$ be a ring, and $R\text{-Mod}$ its category of all left modules. There is a "forgetful" functor $\operatorname{Forget}: R\text{-Mod} \to \text{AbGp}$, which is additive, continuous, and cocontinuous (in particular, exact). Since $R\text{-Mod}$ is both complete and cocomplete, $\operatorname{Forget}$ has both a left adjoint $\operatorname{Free}: \text{AbGp} \to R\text{-Mod}$ and a right adjoint $\operatorname{Cofree}: \text{AbGp} \to R\text{-Mod}$.

You can see what these functors are explicitly. Let me write $_R R_{\mathbb Z}$ for "$R$ as a left module" and $_{\mathbb Z} R _R$ for "$R$ as a right module". The $\operatorname{Forget}$ functor is (isomorphic to) the functor $\operatorname{Hom}_R({_R R_{\mathbb Z}},-)$ — this description makes it clearly continuous, and its left adjoint is $\operatorname{Free} \cong {_R R_{\mathbb Z}} \otimes_\mathbb Z (-)$. But we also have $\operatorname{Forget} \cong {_{\mathbb Z} R _R}\otimes_R (-)$, whence its right adjoint is $\operatorname{Cofree} \cong \operatorname{Hom}_{\mathbb Z}({_{\mathbb Z} R _R},-)$.

I feel like I have some positive amount of experience with free modules. (I would say, given the above, that the correct definition of "free module" is "object in the essential image of $\operatorname{Free}$", although what's actually used is "object of the form $\operatorname{Free}(\mathbb Z^{\oplus \kappa})$ for some cardinal $\kappa$.) But I hardly ever come across the essential image of $\operatorname{Cofree}$, or indeed the cofree functor at all. (Again, maybe the "standard" definition of "cofree module" is "module isomorphic to $\operatorname{Cofree}((\mathbb Q/\mathbb Z)^{\times \kappa})$," or something.) The functors are not the same: when $ R = \mathbb Z/2$, then $\operatorname{Free}(\mathbb Z) = \mathbb Z/2$, whereas $\operatorname{Cofree}(\mathbb Z) = 0$. If you would rather replace $\mathbb Z$ by a field throughout, then they are still not the same when $R$ is infinite-dimensional (for example).

So: Do people use cofree modules? If so, how? If not, why not? Are free modules just a lot nicer than cofree ones, and if so, how?

Sets, notAbGp(although of course, at least in the free case, it comes down to the same thing). $\endgroup$ – Peter LeFanu Lumsdaine Sep 8 '10 at 19:15