Okay, your argument seems sound. But consider the following. If Z is a free group, say Z = F(X), then you can show that for any Z-module M (if M is a pi-module a homomorphism Z --> pi makes M into a Z-module of course), then for either a right or a 2-sided module, it is trivial to see that Der(Z,M) = M^X. That is, any function X --> M extends in a unique way to a derivation Z --> M. Now resolve an arbitrary Z with G^2Z ===> GZ ---> Z and then we get an equalizer Der(Z,M) ---> Der(GZ,M) ===> Der(G^2Z,M) or Der(Z,M) ---> Der(GC,M) ===> Der(G^2Z,M) which is Der(GZ,M) ---> M^Z ===> M^{GZ} for whichever kind of module you have. But, the ismorphisms are not natural (except with respect to maps of the basis, which G\epsilon is, but \epsilon G isn't) and so the two versions of Der(Z,M) may well be different and hence the two versions of Y. I guess Jon didn't correctly describe the category of pi-modules. Michael ----- Original Message ----- From: "George Janelidze" <George.Janelidze@uct.ac.za> To: "Michael Barr" <barr@math.mcgill.ca>, "Categories list\" <Categories list>" <categories@mta.ca> Sent: Saturday, August 26, 2017 6:47:42 AM Subject: Re: categories: modules over a group Dear Michael, Although the answer to your question is "Of course not!", it is a very interesting question because it inspires very interesting questions about internal actions in semi-abelian categories. In this message I shall, however, only explain why is it "Of course not!" (with "!"). 1. The category of pi-modules can be identified with the category of Z[pi]-modules, where Z[pi] is the group ring over pi. Now, suppose, for simplicity, that pi is abelian. Then you question can be stated as: Are the rings Z[pi] and Z[pi x pi] Morita equivalent? Of course not, because two commutative rings are Morita equivalent if and only if they are isomorphic. Well, for some funny groups they are, but I know you are not asking asking about funny groups. 2. Your functor from 2-sided modules to split epis is fine, but it is not an equivalence, e.g. because your structure on Y will not allow you to recover the pi-module structure on M from the Beck module structure on Y-->pi. In fact one can prove the following: Two 2-sided pi-module structures * and # on M give the same Beck modules structures on a chosen pi-->Y-->pi if and only if x*m*(inverse of x) = x#m#(inverse of x) for every x in pi and every m in M. For example if we assume that both left actions are trivial, then the right actions will have to coincide, as in Beck's theory. Remark: but you never mention what happened after Beck - I mean Bourn protomodularity and internal actions... Best regards, George [For admin and other information see: http://www.mta.ca/~cat-dist/ ]