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In a poset, whenever the meets and joins below exist, their universal properties induce a containment $$(A\vee B)\wedge (A\vee C)\geq A\vee(B\wedge C).$$ This is an instance of codistributivity. In a lattice distributivity is equivalent to codistributivity, but this requires the above identity to hold for several ordered triples $A,B,C$.

In the categories of groups and of modules, when dealing with sufficiently disjoint subobjects, a codistributive identity holds. Below is the argument for groups (and modules).

Proposition. Let $A,B,C\vartriangleleft G$ be normal subgroups such that $B,C$ are disjoint and $A$ is disjoint from $B\vee C$ (in particular $A,B,C$ are disjoint). Then$$(A\vee B)\wedge (A\vee C)= A.$$

Proof. Let $g$ be an element of the LHS. Since $A$ is normal, $A\vee B=AB,A\vee C=AC$ where juxtaposition denotes the subset product. Thus $g\in AB\cap AC$. Hence we may write $$g=a_1b=a_2c.$$ From the equality on the right we obtain $$a_1^{-1}a_2=bc^{-1}.$$ In particular, this element belongs to $A$ and $BC$. Since $B$ is normal, $BC=B\vee C$. Thus $$a_1^{-1}a_2=bc^{-1}\in A\wedge (B\vee C).$$ By assumption the RHS is trivial, so $b=c\in B\wedge C$. By assumption $B,C$ are disjoint so $b=c=1_G$. Coming back to the first equation for $g$ we see $g\in A$, proving the $\leq $ direction of the proposition. The $\geq $ is always true.


Question. How can I prove the proposition above for a sufficiently nice protomodular category?


Update. Some possibly relevant information.

In a finitely complete pointed protomodular category, the poset of normal subobjects is isomorphic to the poset of equivalence relations. Hence we may equivalently deal with the desired identity in the poset (in fact lattice) of equivalence relations.

For this, the following two papers by Bourn seem relevant.

Congruence distributivity in Goursat and Mal'cev categories and A categorical genealogy of the congruence distributive property.

In particular, Theorem 2.1 of the former paper characterizes congruence distributivity in Goursat categories via the preservation of binary infima by the direct image functor associated to every regular epi $f:A\to B$. $$\mathrm{Rel}_{\mathsf{C}}(A)\to \mathrm{Rel}_{\mathsf{C}}(B)$$

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  • $\begingroup$ I don't know about protomodular categories, and I have forgotten much about varieties which are congruence modular. Your situation seems like you may be dealing with something like a congruence-modular variety. There are Mal'cev conditions for congruence distributivity and more complicated ones for congruence modularity. Maybe one of those will help. You show the existence of a term or set of terms obeying equations that give the property. Gerhard "Can Recall More If Needed" Paseman, 2017.06.11. $\endgroup$ – Gerhard Paseman Jun 12 '17 at 4:06
  • $\begingroup$ @GerhardPaseman I updated the question with seemingly relevant information related to your advice, but I'm still stuck. $\endgroup$ – Arrow Jun 19 '17 at 10:47
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Any regular protomodular category is also a Mal'tsev category, and thus its lattice of equivalence relation is modular, which means that the following identity holds for all equivalence relations $R,S,T$ : $$(S\vee R)\wedge T\leq S\vee (R\wedge (S\vee T)).\label{1}\tag{1}$$ Now given three relations $A,B,C$ such that $B\wedge C=0$ and $A\wedge (B\vee C)=0$, taking $S=C$, $R=A$ and $T=B$ in \eqref{1} gives you $$(C\vee A)\wedge B\leq C\vee (A\wedge (C\vee B))=C.$$ Since we also have $(C\vee A)\wedge B\leq B$, we conclude that $(C\vee A)\wedge B=0$.

Now taking $S=A$, $R=B$ and $T=A\vee C$ in the identity \eqref{1} gives you $$(A\vee B)\wedge (A\vee C)\leq A\vee (B\wedge (A\vee C))=A.$$

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