I am studying category theory and I came across two questions that I could not answer about the adjoint of two forgetful functors.

Let $\mathbf{TAb}$ be the category of abelian groups such that all elements have finite order and let $\mathbf{Grp}$ be the category of groups. Why does the forgetful functor $U_1 : \mathbf{TAb}\rightarrow \mathbf{Set}$ does not have left adjoint? And why the forgetful functor $U_2: \mathbf{Grp} \rightarrow \mathbf{Set}$ does not have right adjoint?

The autor also asks about the right adjoint of the forgetful functors $\mathbf{k}\text{-}\mathbf{Mod} \rightarrow \mathbf{Set}$ and $\mathbf{Ring}\rightarrow \mathbf{Set}$, but I think that they don't have right adjoint for the same reason of $U_2$.


1 Answer 1


A very interesting property of adjoints is that they preserve (co)limits: right adjoints preserve limits, left adjoints preserve colimits.

Take for instance a family of torsion abelian groups $G_i, i\in I$, with $|I|\geq 2$ then their coproduct in $\mathbf{TAb}$ is $\displaystyle\bigoplus_{i\in I}G_i$, while the coproduct of $U_1(G_i), i\in I$ is $\displaystyle\coprod_{i\in I}U_1(G_i)$ : $U_1$ doesn't preserve colimits, therefore it can't be a left adjoint, i.e. it can't have a right adjoint.

You can do a very similar proof for $U_2$ (the coproduct looks a bit different in $\mathbf{Grp}$ but it's still not preserved by $U_2$)

You can wonder whether this is a good criterion and Freyd's adjoint theorem essentially says that it is, in that for nice categories and nice functors, preserving (co)limits is enough to be a (left) right adjoint

  • $\begingroup$ So to prove that $U_1$ doesn't have a left adjoint, do I just need to prove that it does not preserve limits? $\endgroup$
    – H R
    Nov 20, 2018 at 22:28
  • $\begingroup$ Yes for instance you can do that $\endgroup$ Nov 20, 2018 at 22:40

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