Proof of Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages

Throughout, every element of a class is a set (Elements Are Sets, and the Empty Set and the Pair of Two Sets Exist and Are Unique (i)), and classes with the same elements are equal (Axiom of Extensionality for Classes).

For (i): $u\in A=\operatorname{dom}F$ by Functions, Values of a Function, and Functions from One Class to Another (iii), so $F(u)$ is the unique set $v$ with $(u,v)\in F$ by Functions, Values of a Function, and Functions from One Class to Another (ii); it lies in the range of $F$, which is contained in $B$ by Functions, Values of a Function, and Functions from One Class to Another (iii).

For (ii): for $u\in A$ we have $u\in\operatorname{dom}F$ by (i), and the uniqueness in Functions, Values of a Function, and Functions from One Class to Another (ii) gives the first statement. For the second, $(u,v)\in F$ forces $u\in A$ by Domains and Restrictions of Maps (i), and the first statement applies.

For (iii): $F$ and $G$ are functions with the same domain $A$ by Functions, Values of a Function, and Functions from One Class to Another (iii), and with the same values on it, so $G=F$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (i).

For (iv): by The Restriction of a Function or Relation to a Class (i), the elements of $(H|_{S})|_{T}$ are the pairs $(u,v)\in H$ with $u\in S$ and $u\in T$, and those of $H|_{T}$ are the pairs $(u,v)\in H$ with $u\in T$. Since $T\subseteq S$, the condition $u\in S$ follows from $u\in T$, so the two classes have the same elements.

For (v): $F|_{A}$ consists of elements of $F$ by The Restriction of a Function or Relation to a Class (i). Conversely, every element of $F$ is a pair $(u,v)$, as $F$ is a relation, and $u\in A$ by Domains and Restrictions of Maps (i); so it lies in $F|_{A}$.

For (vi): the first statement is Domains and Restrictions of Maps (ii). For the second, $u\in S$ and $u\in\operatorname{dom}F$ by (i), so $u\in S\cap\operatorname{dom}F$ and $(F|_{S})(u)=F(u)$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ix).

For (vii): by The Cartesian Product of Two Classes (i), every element of $A\times\{c\}$ is a pair $(u,v)$ with $u\in A$ and $v\in\{c\}$, that is, $v=c$ by The Empty Set, the Unordered Pair and the Singleton (iii); conversely $(u,c)\in A\times\{c\}$ for every $u\in A$ by (viii). So the elements of $A\times\{c\}$ are the pairs $(u,c)$ with $u\in A$. So $A\times\{c\}$ is a relation in which every first coordinate has the single second coordinate $c$, that is, a function; its domain is $A$ and its range is contained in $\{c\}$. Hence $A\times\{c\}:A\to\{c\}$ is a map, its value at every $u\in A$ is $c$ by Functions, Values of a Function, and Functions from One Class to Another (ii), and if $c\in B$ then $\{c\}\subseteq B$, so $A\times\{c\}:A\to B$ is a map.

For (viii): this is Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set (i).

For (ix): in either direction $u$ is an element of a class, hence a set, and Membership in Inverse Relations, Preimages and Images of Classes, and Enlarging the Codomain of a Map (iii) gives the equivalence.

For (x): by (ix) and The Boolean Operations on Classes, Disjointness, and the Universal Class (i), a set lies in either side exactly when it lies in $A$ and its value under $F$ lies in both $D$ and $E$.