Proof of Maps between Classes, Continued: Agreement of Values

We prove both directions of (i).

Suppose first that $v$ is a set with $(u,v)\in F$ and $(u,v)\in G$. Since $u\in A$ and $v$ is a set, Ordered Pairs in a Map between Classes: Membership as a Value, and the Components Lie in the Domain and the Codomain (i) applied to the map $F:A\to C$ gives $F(u)=v$. Likewise, since $u\in B$, the same result applied to $G:B\to D$ gives $G(u)=v$. Hence $F(u)=G(u)$.

Conversely, suppose $F(u)=G(u)$, and put $v=F(u)$. By Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (i), applied to $F$ and $u\in A$, $v$ is a set and $(u,v)\in F$. Since $v=G(u)$, $u\in B$ and $v$ is a set, Ordered Pairs in a Map between Classes: Membership as a Value, and the Components Lie in the Domain and the Codomain (i) applied to $G:B\to D$ gives $(u,v)\in G$. So $v$ is a set with $(u,v)\in F$ and $(u,v)\in G$.