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