# Left and Right adjoint of forgetful functor

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

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

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