I am trying to understand projective modules while avoiding as much as possible the use of exact sequences and category theory.
The definition I am using is that a module $P$ is projective if it is the direct summand of a free module, that is, there exists $M$ so that $M \oplus P = R^m$ for some $m$.
I would like to prove that for every surjective homomorphism $f: N \to P$ there exists injective $g: P \to N$ so that $f \circ g = id$. In fact, all I really need is that (taking $N = R^n$) the map $f: R^n \to P$ has an inverse to use the fact quoted here.
This seems to almost obviously be true without the assumption that $M$ is projective (and indeed is not true, the inverse to the surjection above won't necessarily be injective), but I doubt it. All the proofs that I have found rely on either going through the definition of split exact sequences, or uses some very involved category theory. My question: Is there a way to prove this statement and avoid all of the above?
Best Answer
Welcome to MSE!
Let $f : N \to P$ and assume $P$ projective. We want to find a section $g : P \to N$.
Now, since $P$ is projective, we have $R^m = M \oplus P$. Then we can view $f$ as a map $N \to R^m$ by composing with the inclusion $ \iota : P \hookrightarrow M \oplus P$.
Now we use the fact that $R^m$ is free: If $f : N \to R^m$, then let $\{e_i\}$ be the basis elements in $\text{Im}(f) = P$. Since they are in the image, we can find $\{n_i\} \subseteq N$ so that $f(n_i) = e_i$ for each $i$. Now we can define $g : R^m \to N$ by sending the relevant $e_i$ to the $n_i$ and sending the remaining elements of the basis to $0$ (or anywhere really). Now it is clear that $\iota \circ f \circ g = \text{id}$. Since $\iota$ is an injection, we thus have $f \circ g = \text{id}$ too.
As an aside, this is how theorems regarding projective modules usually go when we unpack the category-theoretic language. The only thing we really know is that $P$ is a summand of a free module, so we will use the inclusion $\iota : P \to R^m$ and the projection $\pi : R^m \to P$ as necessary to transfer our problem to the free-module setting (which is very easy to work with) and then transfer back.
Edit:
As for why this theorem doesn't work for every module: Consider the surjection $\pi : \mathbb{Z} \to \mathbb{Z}/2$ viewed as $\mathbb{Z}$-modules. There can be no section $g : \mathbb{Z}/2 \to \mathbb{Z}$ since there are no elements of order $2$ in $\mathbb{Z}$!
I hope this helps ^_^