For $A$ a (not necessarily commutative) algebra over a field $k$ and $P$ a left $A$-module, it easily follows from projectivity of $P$ that there exists a left $A$-module map $s:P \to A \otimes_k P$ such that id$_P = m \circ s$, where $m: A \otimes_k P \to P$ is multiplication.
I can show that the converse is also true, i.e. whenever there exists an $A$-module map $s:P \to A \otimes_k P$ such that id$_P = m \circ s$, then $P$ is a projective $A$-module.
The problem is that my proof is half a page long and uses connections and universal derivations! I am sure there exists a one line argument but I cannot see it. Can anyone help?
Notice that as a left $A$-module, $A\otimes_k P$ is free: this tensor product depends only on the $k$-module structure of $P$, so you can pick a basis $B$ for $P$ over $k$ and then the elements $1\otimes b$ for $b\in B$ will be a basis for $A\otimes_k P$ over $A$. So the existence of $s$ implies that $P$ is a direct summand of the free module $A\otimes_k P$, and hence projective.