A Universe of Sorts

$P$ is projective iff for every epimorphism $e: E \to B$ , and every morphism $f: P \to B$ , there exists a lift $\tilde{f}: P \to E$ .

     e   E ->> B   ^   ^  f~\  | f     \ |       P

Let $P$ be a free module. Suppose we have an epimorphism $e: M \to N$ and a morphism $f: P \to N$ . We must create $\tilde f: M \to N$
Let $P$ have basis $\{ p_i \}$ . A morphism from a free module is determined by the action on the basis. Thus, we simply need to define $\tilde f(p_i)$ .
For each $f(p_i): N$ , there is a pre-image $m_i \in M$ such that $e(m_i): N = f(p_i): N$ .
Thus, define $\tilde{f}(p_i) = m_i$ . This choice is not canonical since there could be many such $m_i$ .
Regardless, we have succeeded in showing that every free module is projective by lifting $f: P \to N$ to a map $\tilde f: M \to N$ .

$P$ is projective iff every exact sequence $0 \to N \to M \xrightarrow{\pi} P \to 0$ splits.
That is, we have a section $s: P \to M$ such that $\pi \circ s = id_P$ .

PROOF (1 => 2): Suppose $P$ solves the lifting problem. We wish to show that this implies that exact sequence splits.
Take the exact sequence:

            pi0 -> N -> M -> P -> 0               ^               | idP               P

            pi0 -> N -> M -> P -> 0          ^   ^       idP~\  | idP            \ |             P

This gives us the section s = idP~ such that pi . s = idP from the commutativity of the above diagram.

$P$ is projective iff it is the direct summand of a free module. So there is a another module $N$ such that $P \oplus N \equiv R^n$ .
We can always pick a surjective epi $\pi: F \to P$ , where $F$ is the free module over all elements of $P$ .
We get our ses $0 \to ker(\pi) \to F \to P \to 0$ . We know this splits because as shown above, projective splits exact sequences where $P$ is the surjective image.
Since the sequence splits, the middle term $F$ is a direct sum of the other two terms. Thus $F \simeq \ker \pi \oplus P$ .

  eE ->>B     ^    f|     P

We know that $P$ is the direct summand of a free module, so we can write a P(+)Q which is free:

  eE ->>B     ^    f|     P <<-- P(+)Q         pi

We create a new arrow f~ = f . pi which has type f~: P(+)Q -> B. Since this is a map from a free module into B, it can be lited to E. The diagram with f~ looks as follows:

  eE ->>B <--     ^    \f~    f|     \     P <<-- P(+)Q         pi

--------g~--------|                |v e              |E ->>B <--       g~     ^    \f~    |    f|     \     |     P <<-- P(+)Q         pi

From this, I create the map g: P -> E given by g(p) = g~((p, 0)). Thus, we win!

Z/pZ is not projective.
We have the exact sequence 0 -> Z -(xp)-> Z -> Z/kZ -> 0 of multiplication by p.
This sequence does not split, because Z (middle) is not a direct summand of Z (left) and Z/kZ (right), because direct summands are submodules of the larger module. But Z/pZ cannot be a submodule of Z because Z/pZis torsion while Z is torsion free.

Let $R \equiv F_2 \times F_2$ be a ring.
The module $P \equiv F_2 \times \{0\}$ is projective but not free.
It's projective because it along with the other module $Q \equiv \{0\} \times F_2$ is isomorphic to $R$ . ( $P \oplus Q = R$ ).
It's not free because any $R^n$ will have $4^n$ elements, while $P$ has only two element.
Geometrically, we have two points, one for each $F_2$ . The module $P$ is a vector bundle that only takes values over one of the points. Since the bundle different dimensions over the two points (1 versus 0), it is projective but not free.
It is projective since it's like a vector bundle. It's not free because it doesn't have constant dimension.

§ Projective Modules in Terms of Universal Property