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  "body": "\n## Definition\n\nLet $A$ be a set, let $m, n \\ge 1$, and let $R \\subseteq A^m$ be an $m$-ary relation on $A$.\nAn $n$-ary [[operation|EJ8YJVM9BW]] $f \\colon A^n \\to A$ is a *polymorphism* of $R$ if for all tuples\n$$(a_{1 1}, \\dots, a_{m 1}), \\; \\dots, \\; (a_{1 n}, \\dots, a_{m n}) \\in R$$\nthe tuple\n$$\\bigl(f(a_{1 1}, \\dots, a_{1 n}), \\dots, f(a_{m 1}, \\dots, a_{m n})\\bigr)$$\nis also in $R$.\nFor a set $\\Gamma$ of finitary relations on $A$, $\\operatorname{Pol}(\\Gamma)$ denotes the set of operations that are polymorphisms of every $R \\in \\Gamma$.\nFor a set $F$ of finitary operations on $A$, $\\operatorname{Inv}(F)$ denotes the set of finitary relations on $A$ of which every $f \\in F$ is a polymorphism.\n\n## Remarks\n\nExample: on $\\set{0, 1}$, the majority operation $m$ is a polymorphism of the [[binary relation|05AVARPXA2]] $R = \\set{(0, 1), (1, 0)}$.\nCheck: given three pairs $(a_j, 1 - a_j)$ in $R$, the second coordinates are the complements of the first, so $m(1 - a_1, 1 - a_2, 1 - a_3) = 1 - m(a_1, a_2, a_3)$ and the result lies in $R$.\nNon-example: the operation $\\wedge$ on $\\set{0, 1}$ is not a polymorphism of the same $R$, since applying it to $(0, 1)$ and $(1, 0)$ gives $(0, 0) \\notin R$.\nFor a unary relation $B \\subseteq A$, the polymorphisms are the operations under which $B$ is closed, so $B$ is a subuniverse.\nFor an equivalence relation, the polymorphisms are the operations compatible with it, as in the definition of a congruence.\nIf $A$ carries a relational structure, its polymorphisms are the homomorphisms from its $n$-th power to itself.\n",
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