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The following definition generalizes set-theoretic disjointess:

Definition 0. (Autonomy). Given a Lawvere theory $\mathsf{T}$, a $\mathsf{T}$-algebra $X$, and an indexed family $S$ of subalgebras of $X$, call $S$ autonomous iff the obvious map $$\bigsqcup_{i:I} S_i \rightarrow X$$ is injective, where $I$ denotes the index set of $S$.

(Also: I'm searching for standardized terminology for this condition.)

This can be generalized so that it makes sense inside any category equipped with a notion of "injective morphism" that has all small coproducts.

Anyway, I want to understand this order-theoretically. So define:

Definition 1. (Order-theoretic autonomy). Given a complete lattice $L$ and an indexed family $s$ of elements of $L$, call $s$ (order-theoretically) autonomous iff the join function $$\prod_{i:I} \downarrow (s_i) \rightarrow L$$ is injective.

(Notation: $\downarrow x$ is the lowerset generated by $x$, for any $x \in L$.)

Observe that (genuine) autonomy implies order-theoretic autonomy. Presumably, the converse doesn't hold.

Questions.

Q0. I'd like a counterexample to the backwards implication, or else a proof that it holds.

Q1. Has anyone ever proposed a definition of "complete lattice equipped with a notion of autonomy" such that for every algebraic structure $X$, the complete lattice of subalgebras of $X$ is canonically equipped with a notion of autonomy as per Definition 0?

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  • $\begingroup$ +1 for this nice (and natural) question! $\endgroup$ Aug 23, 2015 at 9:02
  • $\begingroup$ @DominicvanderZypen, thanks for the kind words :) $\endgroup$ Aug 23, 2015 at 13:42

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