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  "body": "\n## Definition\n\nFix a signature and let $\\mathcal{K}$ be a class of algebras of this signature.\nAn algebra $A \\in \\mathcal{K}$ is *projective* in $\\mathcal{K}$ if the following holds.\nFor all $B, C \\in \\mathcal{K}$, every surjective [[homomorphism|KBXSZC6BV1]] $g \\colon B \\to C$ and every homomorphism $f \\colon A \\to C$, there is a homomorphism $h \\colon A \\to B$ with $g \\circ h = f$.\n\n## Remarks\n\nIn a variety, epimorphisms of the category of algebras need not be surjective.\nSo this notion can differ from a projective object of the category; universal algebra uses the surjective version.\nIn a variety, an algebra is projective exactly when it is a [[retract|68PFSGP39B]] of a [[free algebra|GTS4DRRMZR]].\n\nExample: every free algebra $F$ on a set $X$ in a variety is projective.\nGiven $g \\colon B \\to C$ surjective and $f \\colon F \\to C$, choose $b_x \\in B$ with $g(b_x) = f(x)$ for each $x \\in X$.\nThe homomorphism $h \\colon F \\to B$ with $h(x) = b_x$ satisfies $g \\circ h = f$, since both sides agree on the generators $X$.\nIn the variety of modules over a ring, the projective algebras are the [[projective modules|BFFA03GE3V]].\n\nNon-example: $\\Z/2\\Z$ is not projective in the variety of abelian groups.\nTake $g \\colon \\Z \\to \\Z/2\\Z$ the reduction map and $f = \\id$.\nEvery homomorphism $h \\colon \\Z/2\\Z \\to \\Z$ is zero, because $\\Z$ has no element of order $2$, so $g \\circ h = 0 \\ne f$.\n",
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