2021
DOI: 10.34198/ejms.6121.105116
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Triangular Scheme Revisited in the Light of n-permutable Categories

Abstract: The first diagrammatic scheme was developed by H.P. Gumm under the name Shifting Lemma in case to characterize congruence modularity. A diagrammatic scheme is developed for the generalized semi distributive law in Mal'tsev categories. In this paper we study this diagrammatic scheme in the context of $n$-permutable, and of Mal'tsev categories in particular. Several remarks concerning the Triangular scheme case are included.

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Cited by 5 publications
(3 citation statements)
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“…The proof of this result is based on that of Proposition 5.3 in [12] which claims that a Goursat category satisfies the Shifting Lemma, 2-scheme and 3-scheme when S is a reflexive relation and R and T are equivalence relations. We prove (1). Let R and T be equivalence relations and let S i be a reflexive relation on an object X such that R ∧ S i T .…”
Section: Let Us Begin With the Following Observationmentioning
confidence: 94%
“…The proof of this result is based on that of Proposition 5.3 in [12] which claims that a Goursat category satisfies the Shifting Lemma, 2-scheme and 3-scheme when S is a reflexive relation and R and T are equivalence relations. We prove (1). Let R and T be equivalence relations and let S i be a reflexive relation on an object X such that R ∧ S i T .…”
Section: Let Us Begin With the Following Observationmentioning
confidence: 94%
“…Following [1], given any pair (R, S) of reflexive relations on an object X in a regular category E, let us denote by (R; S) n ; n ≥ 2 the alternate composition RSRS • • • of length n which is a reflexive relation as well. Clearly we have: (R, S) n ≤ (R, S) n+1 and (S, R) n ≤ (R, S) n+1 .…”
Section: Equivalence Relations In N-permutable Categoriesmentioning
confidence: 99%
“…Consider G = Z 2 × Z 6 and let U 1 = 1, U 2 = {(0, 0), (1, 0), (0, 2), (1, 2), (0, 4), (1, 4)}, U 3 = {(0, 0), (1, 0), (0, 3), (1, 3)}, U 4 = {(0, 0), (1, 0)}, U 5 = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5)}, U 6 = {(0, 0), (0, 2), (0, 4)}, U 7 = {(0, 0), (0, 3)}, U 8 = {(0, 0), (0, 2), (0, 4),(1,1),(1,3),(1,5)}, U 9 = {(0, 0), (1, 3)}, U 10 = Z 2 × Z 6 . See Figure for the subgroup lattice of G.…”
mentioning
confidence: 99%