Correction to: Identical inclusions of semilattices (Algebra universalis, (2020), 81, 2, (26), 10.1007/s00012-020-0649-6)

In this note we refer to the paper “Identical inclusions of semilattices” [1]. The lattice L(Sl) of inclusive varieties of semilattices in [1, Figure 1] is incorrect; specifically, the inclusion I(Fn-11,0)⊂I(Fn);n≥2 does not hold. In this note, we prove that the inclusive varieties I(Fn) and I(Fn-11,0) are incomparable, so the lattice L(Sl) is as follows: (Figure presented.) Let us recall notations from [1]. Fn is the free semilattice with n free generators a1,…,an; Fn0 is Fn with externally adjoined 0; Fn1 is Fn with externally adjoined 1; Fn1,0 is Fn with externally adjoined 0 and 1; Cn is a chain with n elements; I(S) is an inclusive variety generated by a semigroup S; I(Φ) is an inclusive variety defined by a set of identical inclusions Φ; Vn-1 is the set of all words without repeated letters in the alphabet {x1,…,xn} of the length n-1; Vn-1xn is the set of all words without repeated letters in the alphabet {x1,…,xn} of the length n-1 such that each word includes xn; Vnx1 is the set of all words without repeated letters in the alphabet {x1,…,xn,xn+1} of the length n such that each word includes x1.In notations 4 and 5 we assume that n>1. Fn is the free semilattice with n free generators a1,…,an; Fn0 is Fn with externally adjoined 0; Fn1 is Fn with externally adjoined 1; Fn1,0 is Fn with externally adjoined 0 and 1; Cn is a chain with n elements; I(S) is an inclusive variety generated by a semigroup S; I(Φ) is an inclusive variety defined by a set of identical inclusions Φ; Vn-1 is the set of all words without repeated letters in the alphabet {x1,…,xn} of the length n-1; Vn-1xn is the set of all words without repeated letters in the alphabet {x1,…,xn} of the length n-1 such that each word includes xn; Vnx1 is the set of all words without repeated letters in the alphabet {x1,…,xn,xn+1} of the length n such that each word includes x1. Inclusive varieties I(Fn) and I(Fn-11,0) are incomparable. Let z∉{x1,…,xn}. We consider two identical inclusions μ=x1⋯xn∈^Vn-1 and τ=x1x2⋯xn∈^Vn-1xn∪{x1x2⋯xnz}. Fn-11,0 satisfies μ because the image of x1⋯xn is equal to 0 or 1 or it is the product of at most n-1 free generators and in any of these cases it is equal to the image of a word without repeated letters of length n-1. The substitution x1→a1,…,xn-1→an-1,xn→an proves that Fn does not satisfy μ. Thus, I(Fn)⊄I(Fn-11,0). The substitution z→0,x1→a1,…,xn-1→an-1,xn→1 proves that Fn-11,0 does not satisfy τ because a1⋯an-1∉{a1⋯an-2,…,a2⋯an-1,0}. Let f:X+→Fn be a homomorphism. If f(x1x2⋯xn) is the product of at most n-2 free generators a1,…,an then f(x1x2⋯xn)∈f(Vn-1xn) because all possible sequences of at most n-2 variables are included in the image of Vn-1xn. If f(x1x2⋯xn)=a1⋯an then f(x1x2⋯xn)=f(x1x2⋯xnz). In the remaining case f(x1x2⋯xn) is the product of n-1 free generators from the set {a1,…,an} and without a loss of generality we assume that this product is a1⋯an-1. We denote the set of free generators that are factors of f(w) by c(f(w)). {c(f(x1)),…,c(f(xn))}⊆{a1,…,an-1}. c(f(xn)) contains at least one generator of {a1,…,an-1} and, therefore, we need at most n-2 variables to get all set {a1,…,an-1} and all possible sets of n-2 variables are included in the image of Vn-1xn. Thus, Fn satisfies τ and I(Fn-11,0)⊄I(Fn). □ Fn1 as well as Fn0 contain both Fn and Fn-11,0. Bases of identities of Fn1 and of Fn0 one can find in Theorem 5 of the paper [1]. Namely, (Formula presented.) Thus, we obtain the following proposition. I(Fn)∪I(Fn-11,0)=I(Fn1)∩I(Fn0)=I(Φ), where (Formula presented.)