Let $R$ be a commutative ring with unit and let $M$ be an $R$-module (with $1\cdot m = m$ for all $m\in M$). Let $f:M\twoheadrightarrow M$ be a surjective morphism of $R$-modules of $M$ onto itself. Then, as a consequence of Zorn's lemma, one can construct a map $g$ such that $f\circ g = \mbox{id}_M$.
Does it follow that $g$ is also a morphism of $R$-modules? Or can one at least construt one such $g$ which is a morphism?
Thank you!
EDIT:
I should have made more clear what I'm looking for. So there are 2 questions:
- 1) Does $f\circ g=\mbox{id}_M$ (as set maps) imply that $g$ is a morphism?
- 2) Is there a morphism $g:M\rightarrow M$ such that $f\circ g = \mbox{id}_M$?
1) seems unlikely to me now, as we must have $g(0)=0$ if $g$ is a morphism. But in the construction of $g$ using Zorn's lemma, one could chose $g(0)$ to be any element of $\mbox{ker}(f)$ and still have $f\circ g=\mbox{id}_M$...
I'm really interested in an answer to 2) however...