CatDat

Unknown properties

There are 2 properties for which the database doesn't have an answer if they are satisfied or not. Please help to contribute the data!

• is coregular
• is regular

Special objects

• terminal object: singleton set with the unique uniform structure
• initial object: empty set with the unique uniform structure
• products: The product of a family of uniform spaces $(X_i,\Phi_i)_{i \in I}$ is $(\prod_{i \in I} X_i,\Phi)$, where $\Phi$ consists of all subsets that contain $\bigcap_{i \in F} (p_i \times p_i)^*(U_i)$ for some finite subset $F \subseteq I$ and some family of entourages $U_i \in \Phi_i$ for $i \in F$.
• coproducts: The coproduct of a family of uniform spaces $(X_i,\Phi_i)_{i \in I}$ is $(\coprod_{i \in I} X_i,\Phi)$, where $\Phi$ consists of all subsets that contain $\coprod_{i \in I} U_i$ for some family of entourages $U_i \in \Phi_i$ for $i \in I$.

Special morphisms

• isomorphisms: uniform isomorphisms, i.e. bijective uniform maps whose inverse map is also uniform; equivalently, bijective maps such that a relation in the domain is an entourage if and only if its image is an entourage of the codomain
• monomorphisms: injective uniform maps
• epimorphisms: surjective uniform maps
• regular monomorphisms: A morphism $f : X \to Y$ is a regular monomorphism if and only if $f$ is a uniform embedding, meaning that every entourage of $X$ is the preimage of an entourage of $Y$
• regular epimorphisms: A uniform map $f : X \to Y$ is a regular epimorphism if and only if $f$ is surjective and $Y$ carries the finest uniform structure that makes $f$ uniform. Concretely, this means that if $V \subseteq Y \times Y$ is a subset for which there exists a sequence of subsets $V_1,V_2,\dotsc$ of $Y \times Y$ such that $V_1 \subseteq V$, $V_{n+1} \circ V_{n+1} \subseteq V_n$, and each $(f \times f)^*(V_n)$ is an entourage of $X$, then $V$ is an entourage of $Y$. Thus, in contrast to the situation for $\Top$ or $\Meas$, not every subset whose preimage is an entourage is an entourage in the quotient.