# How do I calculate adjoint functors to forgetful functors?

Suppose I have a forgetful functor $F:Ab\hookrightarrow Grp$ where we forget that we have commutativity. I'm trying to calculate the left and right adjoint functors, so for right adjoint, we have $Hom_{Grp}(F(X),Y)\simeq Hom_{Ab}(X,R_F(Y))$ for $X\in ob(Ab)$ and $Y\in ob(Grp).$ So if we have a homomorphism $\psi:F(X)\rightarrow Y$ and map it to $\eta(\psi):=g:X\rightarrow R_F(Y),$ we know that for $x_1,x_2\in X, \ g(x_1)g(x_2)=g(x_2)g(x_1).$ I don't know how to proceed from here, but I think I'm supposed to get the free abelian group from the right adjoint and the "cofree" abelian group from the left adjoint (so I guess the trivial group?). I used the same reasoning when finding the left and right adjoints for $F:Top\rightarrow Sets,$ and I got the indiscrete and discrete topologies respectively. I feel that I can't make my argument rigorous and for $F:Ab\hookrightarrow Grp$ I don't know how to solve it. I'd really appreciate some hints.

• If $F$ had a right adjoint, it should commute with colimits. It is not the case, hence $F$ can't have a right adjoint. As for the left adjoint, it indeed exists : try to think how you can factorize a map from a group $G$ to an abelian group $A$. – Pece Sep 14 '14 at 20:20
• It is not good notation to use $\hookrightarrow$ for a forgetful functor. – user122283 Sep 14 '14 at 20:25
• The forgetful functor from abelian groups to sets has no right adjoint; you can't expect a functor has left or right adjoint, in general: having an adjoint is a strong property. – egreg Sep 14 '14 at 20:38

$F$ in your scenario does not commute with colimits; hence it does not have a right adjoint. The left adjoint to a forgetful functor is called the free functor, and in this case it is the the abelianization functor $\mathrm{Grp}\to\mathrm{Ab}$ that maps $G\mapsto G^{\mathrm{ab}}$, with $G^{\mathrm{ab}}:=G/[G,G]$ (here $[G,G]$ is the commutator subgroup). It is the universal abelian group induced by $G$'' (nlab, abelianization).