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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28
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Zassenhaus butterfly lemma in an abelian category

Statement

Let A0≤A1 and B0≤B1 be subobjects of an object X in an abelian category. Then there is a canonical isomorphism

A0∨(A1∧B1)A0∨(A1∧B0)  ≅  (A1∧B1)∨B0(A0∧B1)∨B0.

Facts & Assumptions

Given: Subobjects A0≤A1 and B0≤B1 of an object X.

[L1]

The subobject lattice of X is modular (The subobject lattice of an abelian category is modular).

[L2]

For subobjects U,V of a common object, one has (U∨V)/V≅U/(U∧V) (Second isomorphism theorem in an abelian category).

[L3]

Nested quotients satisfy the third isomorphism theorem (Third isomorphism theorem in an abelian category).

Proof

technique · direct
1.1L1L2construct

Put M:=A1∧B1, U:=A1∧B0, and V:=A0∧B1. Since U≤M, the second isomorphism theorem [L2] applied to A0 and M inside A0∨M gives A0∨MA0∨U≅MM∧(A0∨U). By modularity [L1] inside the interval below M, M∧(A0∨U)=(M∧A0)∨U=V∨U. So the left quotient is canonically isomorphic to M/(V∨U).

1.2L1L2construct

Similarly, since V≤M, the second isomorphism theorem [L2] applied to M and B0 inside M∨B0 gives M∨B0V∨B0≅MM∧(V∨B0). Again modularity yields M∧(V∨B0)=V∨(M∧B0)=V∨U. So the right quotient is also canonically isomorphic to M/(V∨U).

2.1L3step 1.1step 1.2∎

The two quotients in steps 1.1 and 1.2 are canonically isomorphic to the same quotient of M, hence to each other. The nested-quotient compatibility of [L3] identifies these isomorphisms with the displayed butterfly quotient comparison.

Depends on

Used by

Dependency tree · two levels

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Sources