forgetful functor from RMod to Ab as left adjoint

1. Proposition

Let \(R\) be a ring and \(\mathrm{RMod}\) the category RMod and \(\mathrm{Ab}\) the Category Ab. Then the forgetful functor from RMod to Ab is a left adjoint to the covariant hom functor from Ab to RMod

2. Proof

Let \(A \in \mathrm{Ob}(\mathrm{Ab})\) and \(M \in \mathrm{Ob}(\mathrm{RMod})\) then it remains to show, that there exists a natural isomorphism

\begin{align*} \mathrm{Hom}_{\mathrm{Ab}}(\mathrm{Forget}(M),G) \cong \mathrm{Hom}_{\mathrm{RMod}}(M, \mathrm{Hom}_{\mathrm{Ab}}(R,G)) \end{align*}

We set

\begin{align*} \varphi: \mathrm{Hom}_{\mathrm{Ab}}(\mathrm{Forget}(M),G) \rightarrow& \mathrm{Hom}_{\mathrm{RMod}}(M, \mathrm{Hom}_{\mathrm{Ab}}(R,G)) \\ (f: M \rightarrow G) \mapsto& (m \mapsto (r \mapsto f(r \cdot m)) \\ \psi: \mathrm{Hom}_{\mathrm{RMod}}(M, \mathrm{Hom}_{\mathrm{Ab}}(R,G)) \rightarrow& \mathrm{Hom}_{\mathrm{Ab}}(\mathrm{Forget}(M),G) \\ (f: M \rightarrow \mathrm{Hom}_{\mathrm{Ab}}(R,G)) \mapsto& (m \mapsto f(m)(1)) \end{align*}

2.1. bijective

2.2. natural

TODO

3. alternative

Date: nil

Author: Anton Zakrewski

Created: 2024-10-14 Mo 08:54