Subsection 8.1.1 (01R5): Foundations

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Let $A$ be a set.

A reflexive relation is equivalently:¹

An $\mathbb {E}_{0}$-monoid in $\webleft (\mathrm{N}_{\bullet }\webleft (\mathbf{Rel}\webleft (A,A\webright )\webright ),\chi _{A}\webright )$;
A pointed object in $\webleft (\mathbf{Rel}\webleft (A,A\webright ),\chi _{A}\webright )$.

¹Note that since $\mathbf{Rel}\webleft (A,A\webright )$ is posetal, reflexivity is a property of a relation, rather than extra structure.

In detail, a relation $R$ on $A$ is reflexive if we have an inclusion

\[ \eta _{R}\colon \chi _{A}\subset R \]

of relations in $\mathbf{Rel}\webleft (A,A\webright )$, i.e. if, for each $a\in A$, we have $a\sim _{R}a$.

Let $A$ be a set.

The set of reflexive relations on $A$ is the subset $\smash {\mathrm{Rel}^{\mathrm{refl}}\webleft (A,A\webright )}$ of $\mathrm{Rel}\webleft (A,A\webright )$ spanned by the reflexive relations.
The poset of relations on $A$ is is the subposet $\smash {\mathbf{Rel}^{\mathsf{refl}}\webleft (A,A\webright )}$ of $\mathbf{Rel}\webleft (A,A\webright )$ spanned by the reflexive relations.

Let $R$ and $S$ be relations on $A$.

Interaction With Inverses. If $R$ is reflexive, then so is $R^{\dagger }$.
Interaction With Composition. If $R$ and $S$ are reflexive, then so is $S\mathbin {\diamond }R$.

Proof of Proposition 8.1.1.1.4.

Item 1: Interaction With Inverses

Clear.

Item 2: Interaction With Composition

Clear.

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