I’ve found today earlier stated conjecture that lattices $\mathrm{Compl}\mathsf{FCD}(A;B)$ and $\mathrm{Compl}\mathsf{RLD}(A;B)$ are co-brouwerian.

Exercise: Prove this fact.

I’ve proved the following conjecture:

Theorem Let $f$ be a staroid such that $(\mathrm{form}\, f)_i$ is an atomic lattice for
each $i \in \mathrm{arity}\, f$. We have

$\displaystyle L \in \mathrm{GR}\, f \Leftrightarrow \mathrm{GR}\, f \cap \prod_{i \in \mathrm{dom}\, \mathfrak{A}} \mathrm{atoms}\, L_i \neq \emptyset$

for every $L \in \prod_{i \in \mathrm{arity}\, f} (\mathrm{form}\, f)_i$ (where upgrading is taken on the primary filtrator).

The proof is based on transfinite recursion. See this online article for the proof.

The above proof was with an error. Now there is a counter-example in the same article.

Let $\mathfrak{A}$ be an indexed family of sets.

Products are $\prod A$ for $A \in \prod \mathfrak{A}$.

Hyperfuncoids are filters $\mathfrak{F} \Gamma$ on the lattice $\Gamma$ of all finite unions of products.

Problem
Is $\bigsqcap^{\mathsf{FCD}}$ a bijection from hyperfuncoids $\mathfrak{F} \Gamma$ to:

1. prestaroids on $\mathfrak{A}$;
2. staroids on $\mathfrak{A}$;
3. completary staroids on $\mathfrak{A}$?

If yes, is $\mathrm{up}^{\Gamma}$ defining the inverse bijection?

If not, characterize the image of the function $\bigsqcap^{\mathsf{FCD}}$ defined on $\mathfrak{F} \Gamma$.

I have completed preliminary error checking for my online article Funcoids are Filters.

While walking home from McDonalds I conceived the following idea how we can generalize reloids and funcoids.

Let $C$ be a category with finite products, the set of objects of which is a complete lattice (for the case of funcoids as described below it is enough to be just join-semilattice).

One can argue which axioms $C$ shall obey. For example, it is yet unclear, whether we should require the product to be distributive over any suprema or just over finite suprema.

By definition, generalized relation is a supremum of some set of binary products of objects of our category.

We can define generalized reloids as filters on generalized relations.

Similarly we can define generalized funcoids replacing arbitrary suprema in the definition of generalized relation with finite suprema. (See this online article for a proof that in the case $C=\mathbf{Set}$ this is equivalent with funcoids.)

Note, that I am going to do this research in the second volume of my research monograph, having said that the first volume is yet in process of rewriting and is not yet published.

Thin groupoid is an important but a heavily overlooked concept.

When I did Google search for “thin groupid” (with quotes), I found just ${7}$ (seven) pages (and some of these pages were created by myself). It is very weird that such an important concept was overlooked by the mathematical community.

By definition of thin category, thin groupoid is a groupoid for every pair ${A}$, ${B}$ of objects of which there are at most one morphism ${A \rightarrow B}$.

I recall that a groupoid is a category all morphisms of which are isomorphisms. Moreover, in all examples below objects are sets and (iso)morphisms are isomorphisms of ${\mathbf{\mathrm{Set}}}$ that is bijections.

So, roughly, “thin groupoid” means: Between every two sets in consideration there is considered at most one bijection. In other words, all objects in consideration are equivalent up to an isomorphism.

1. Equivalent definitions of thin groupoid

Theorem 1 The following definitions of thin groupoid are equivalent:

1. a groupoid with at most one morphism ${A \rightarrow B}$ for given objects ${A}$, ${B}$;
2. a groupoid with each cycle of morphisms being identity.

Proof: The only thing we need to prove (as all the rest is obvious) is that for thin groupoid each cycle of morphisms is identity. But really, composition of a cycle of morphisms is an endomorphism, but because our category is thin, there are be just one such morphism, the identity morphism. $\Box$

“Each cycle of morphisms is identity” intuitively means: Every object is equivalent to itself in exactly one way.

2. Examples

2.1. Filters, ideals, etc.

For a lattice ${\mathfrak{Z}}$ I denote meets and joins correspondingly as $({\sqcap})$ and $({\sqcup})$.

Filters and ideals are well known concepts:

Filters are subsets ${F}$ of ${\mathfrak{A}}$ such that:

1. ${F}$ does not contain the least element of ${\mathfrak{A}}$ (if it exists).
2. ${A \sqcap B \in F \Leftrightarrow A \in F \wedge B \in F}$ (for every ${A, B \in \mathfrak{Z}}$).

Ideals are subsets ${F}$ of ${\mathfrak{A}}$ such that:

1. ${F}$ does not contain the greatest element of ${\mathfrak{A}}$ (if it exists).
2. ${A \sqcup B \in F \Leftrightarrow A \in F \wedge B \in F}$ (for every ${A, B \in \mathfrak{Z}}$).

I also introduce free stars and mixers:

Free stars are subsets ${F}$ of ${\mathfrak{A}}$ such that:

1. ${F}$ does not contain the least element of ${\mathfrak{A}}$ (if it exists).
2. ${A \sqcup B \in F \Leftrightarrow A \in F \vee B \in F}$ (for every ${A, B \in \mathfrak{Z}}$).

Mixers are subsets ${F}$ of ${\mathfrak{A}}$ such that:

1. ${F}$ does not contain the greatest element of ${\mathfrak{A}}$ (if it exists).
2. ${A \sqcap B \in F \Leftrightarrow A \in F \vee B \in F}$ (for every ${A, B \in \mathfrak{Z}}$).

I will denote ${\mathrm{dual}\, A}$ where ${A \in \mathfrak{Z}}$ the corresponding element of the dual poset ${\mathfrak{Z}^{\ast}}$. Also I denote

$\displaystyle \langle \mathrm{dual} \rangle X \overset{\mathrm{def}}{=} \left\{ \mathrm{dual}\, x \mid x \in X \right\} .$

It is easy to show that filters, ideals, free stars, and mixers are related by the bijections presented in the following diagram:

(where ${\neg}$ denotes set-theoretic complement).

This diagram is a ${4}$-elements thin groupoid (which is a subcategory of ${\mathbf{\mathrm{Set}}}$). These bijections are order isomorphisms if we define order in the right way.

In the case if ${\mathfrak{Z}}$ is a boolean lattice, there is also an alternative diagram (also a ${4}$-elements thin groupoid (which is a subcategory of ${\mathbf{\mathrm{Set}}}$)):

(here ${\langle \neg \rangle X \overset{\mathrm{def}}{=} \left\{ \bar{x} \mid x \in X \right\}}$).

2.2. Funcoids

Funcoids, funcoidal reloids, and filters on lattices ${\Gamma}$ (don’t worry if you don’t know meanings of these terms, see my Web site for a book on this topic) are isomorphic as presented by the following diagram which is also a thin groupoid. The isomorphisms preserve order and composition.

I’ve proved the theorem:

Theorem

1. $f \mapsto \bigsqcap^{\mathsf{RLD}} f$ and $\mathcal{A} \mapsto \Gamma (A ; B) \cap \mathcal{A}$ are mutually inverse bijections between $\mathfrak{F} (\Gamma (A ; B))$ and funcoidal reloids.
2. These bijections preserve composition.

(The second items is the previously unknown fact.)

and its consequence:

Theorem $(\mathsf{RLD})_{\Gamma} g \circ (\mathsf{RLD})_{\Gamma} f = (\mathsf{RLD})_{\Gamma} (g \circ f)$ for every composable funcoids $f$ and $g$.

See this online article for the proofs.