TITLE: A module is projective iff it has a projective basis
QUESTION [7 upvotes]: I have a question related to this: Projective modules
I'm trying to understand the "philosophy" of the statement, because it seems too similar to the statement "a module is free iff every element can be written uniquely as a finite linear combination of elements of a basis".
Is this "projective basis" property saying this:
a module P is projective iff every element in P can be written as a finite linear combination of some elements of P?
We lose uniqueness in the expression as a sum: in the elements of P, in the elements of R, and in the number of terms (so the concept of "rank" wouldn't make sense). Is this all, or am I misunderstanding the statement?
Any other intuition related to that property is also appreciated.

REPLY [12 votes]: The statement you're linking to is: A module $P$ is projective if and only if there is a family $\{x_{i}\}_{i \in I} \subset P$ and morphisms $f_{i}: P \to R$ such that for each $x \in P$ we have $x = \sum_{i \in I} f_{i}(x) x_{i}$. The last statement says three things:

In order for the sum to make sense we must have that for all $x$ the set  $\{i\,:\,f_{i}(x) \neq 0 \}$ is finite. Or, as stated there: for all $x$ we have $f_{i}(x) = 0$ for almost all $i$.
The set $(x_{i})_{i \in I}$ generates $P$. In other words, the map $g: \bigoplus_{i \in I} R \to P$ sending $(r_{i})$ to $\sum_{i \in I} r_{i} x_{i}$ is an epimorphism (it suffices to take $r_{i} = f_{i}(x)$ to see that this map is onto).
The epimorphism $g$ splits: there is a right inverse $f: P \to \bigoplus_{i \in I} R$ of $g$, i.e. $gf = \operatorname{id}_{P}$ (this morphism $f$ is of the form $(f_{i})_{i \in I}$ with morphisms $f_{i}: P \to R$ and by the definition of a direct sum we have $f_{i}(x) \neq 0$ only for finitely many $i$).

In my opinion its just an extremely explicit way of phrasing the much more catchy "a module is projective if and only if it is a direct summand of a free module".