Projective algebra

Remarks

In a variety, epimorphisms of the category of algebras need not be surjective. So this notion can differ from a projective object of the category; universal algebra uses the surjective version. In a variety, an algebra is projective exactly when it is a retract of a free algebra.

Example: every free algebra $F$ on a set $X$ in a variety is projective. Given $g \colon B \to C$ surjective and $f \colon F \to C$, choose $b_x \in B$ with $g(b_x) = f(x)$ for each $x \in X$. The homomorphism $h \colon F \to B$ with $h(x) = b_x$ satisfies $g \circ h = f$, since both sides agree on the generators $X$. In the variety of modules over a ring, the projective algebras are the projective modules.

Non-example: $\Z/2\Z$ is not projective in the variety of abelian groups. Take $g \colon \Z \to \Z/2\Z$ the reduction map and $f = \id$. Every homomorphism $h \colon \Z/2\Z \to \Z$ is zero, because $\Z$ has no element of order $2$, so $g \circ h = 0 \ne f$.