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I'm currently reading Hilton & Stammbach's A First Course in Homological Algebra, and the following point has stumped me:

In section 1.8, they construct co-free modules ("left moodule" over some ring) as essentially coming from the right adjoint to the forgetful functor from $\Lambda$-Modules to Abelian Groups. On the other hand, the free module is constructed as the left adjoint to the forgetful functor from $\Lambda$-modules to Sets. This turns out to be equivalent to requiring free modules to be direct sums of copies of $\Lambda$ considered as a module over itself, and to requiring co-free modules as direct products of the $\Lambda^*=Hom_\mathbb{Z}(\Lambda, \mathbb{Q}/\mathbb{Z})$.

So I guess my question is: what does the right adjoint to the forgetful functor to Set look like, and why is the right adjoint to the forgetful functor to Abelian Groups more useful?

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At heart I think it's because rings are monoids in Ab. –  Qiaochu Yuan Oct 19 '10 at 15:43
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A left moodule might be co-free, but I guess it isn't cow -free! (Sorry, couldn't resist.) –  Nate Eldredge Oct 19 '10 at 16:24
    
^ lololololololo –  CSA Feb 19 at 3:34
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1 Answer

up vote 8 down vote accepted

If a functor has a right adjoint, it preserves colimits; but the forgetful functor from $R$-Mod to Set doesn't. For instance it doesn't preserve binary coproducts. So there isn't a right adjoint.

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It doesn't even preserve the initial object. –  Martin Brandenburg Feb 20 at 20:14
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