This is exercise(7.5) from Hilton and Stambach.
Show that it is possible to choose, for each $R$-module M, a surjection $\epsilon_{M} :P(M)\rightarrow M$, where $P(M)$ is a free $R$-module, in such a way that P is a functor from $_{R} Mod $ to the category, $_{R}Free$, of free $R$-modules and $\epsilon_{M}:P(M)\rightarrow M$ is a natural transformation. If E is the embedding functor, is there an adjugant $\eta : E\dashv P$ such that $\epsilon$ is the counit.
So my process of solving this followed this steps:
Step 1 : If $E\dashv P$ holds then there must be, for evert object $X\in$ $ _{R}Mod$, a terminal morphism $\lambda _{x}$ from $E$ to $X$.
You can see the definition of such a terminal morphism here :https://en.wikipedia.org/wiki/Adjoint_functors
Step 2: $\lambda$ must equal $\epsilon$ and P must be the indentity functor on free modules.
Step 3: Define $P(M)$ to be the free module generated by a minimum set of generators of $M$. (i.e. if $S$ is a set of generators of $M$ which is of minimum order, define $P(M)=F(S)$ where $F$ is the Free functor $Sets\rightarrow$$ _{R} Mod$ ).
And define for a map $Pf: PA\rightarrow PB$, to act to the generators of $PA$ the same way $f$ does.
So my questions are the following:
1) I believe this is a fine proof if $R$ has the invariant basis number property. And we can modify the proof if $R$ does not have the desired property such $P$ is a equivelance if it is consinder a functor with domain the category $_{R} Free$. Is this true ?
2) It seems that the axiom of choice is necessary for this proof to hold true. Is there a way to not use choices?