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?

  • 1
    $\begingroup$ At heart I think it's because rings are monoids in Ab. $\endgroup$ – Qiaochu Yuan Oct 19 '10 at 15:43
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    $\begingroup$ A left moodule might be co-free, but I guess it isn't cow -free! (Sorry, couldn't resist.) $\endgroup$ – Nate Eldredge Oct 19 '10 at 16:24
  • $\begingroup$ ^ lololololololo $\endgroup$ – AIM_BLB Feb 19 '14 at 3:34

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

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