7
$\begingroup$

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
    At heart I think it's because rings are monoids in Ab.2010-10-19
  • 6
    A left *moodule* might be co-free, but I guess it isn't *cow* -free! (Sorry, couldn't resist.)2010-10-19
  • 0
    ^ lololololololo2014-02-19

1 Answers 1

10

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.

  • 2
    It doesn't even preserve the initial object.2014-02-20