# A Course in Universal Algebra by Stankey Burris, H. P. Sankappanavar By Stankey Burris, H. P. Sankappanavar

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From the experiences of the 1st printing of this e-book, released as quantity fifty eight of the Encyclopaedia of Mathematical Sciences:". .. This e-book can be very worthwhile as a reference and advisor to researchers and graduate scholars in algebra and and topology. " Acta Scientiarum Mathematicarum, Ungarn, 1994 ". .

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The reader will notice that all of the different kinds of algebras listed below are distinguished from each other by their fundamental operations and the fact that they satisfy certain identities. One of the early achievements of Birkhoff was to clarify the role of identities (see §11). Examples. (1) Groups. A group G is an algebra G, ·, −1, 1 with a binary, a unary, and nullary operation in which the following identities are true: G1: x · (y · z) ≈ (x · y) · z G2: x · 1 ≈ 1 · x ≈ x G3: x · x−1 ≈ x−1 · x ≈ 1.

Let S be a subuniverse of A, let f be an n-ary member of F, and let a1 , . . , an ∈ S. Then f B (αa1 , . . , αan) = αf A(a1 , . . , an ) ∈ α(S), 48 II The Elements of Universal Algebra so α(S) is a subuniverse of B. If we now assume that S is a subuniverse of B (instead of A) and α(a1 ), . . , α(an ) ∈ S then αf A (a1 , . . , an ) ∈ S follows from the above equation, so f A (a1 , . . , an ) is in α−1 (S). Thus α−1 (S) is a subuniverse of A. 4. If α : A → B is a homomorphism and C ≤ A, D ≤ B, let α(C) be the subalgebra of B with universe α(C), and let α−1 (D) be the subalgebra of A with universe α−1 (D), provided α−1 (D) = ∅.

An ) = f C (β(αa1 ), . . , β(αan)) = f C ((β ◦ α)a1 , . . , (β ◦ α)an ). 2 The next result says that homomorphisms commute with subuniverse closure operators. 6. If α : A → B is a homomorphism and X is a subset of A then α Sg(X) = Sg(αX). Proof. From the definition of E (see §3) and the fact that α is a homomorphism we have αE(Y ) = E(αY ) for all Y ⊆ A. Thus, by induction on n, αE n (X) = E n (αX) for n ≥ 1; hence α Sg(X) = α(X ∪ E(X) ∪ E 2 (X) ∪ . . ) = αX ∪ αE(X) ∪ αE 2 (X) ∪ . . = αX ∪ E(αX) ∪ E 2 (αX) ∪ .