Proof of Extending a Map by One Point, and Extending a Bijection by One Point

By The Boolean Operations on Classes, Disjointness, and the Universal Class (i) and The Empty Set, the Unordered Pair and the Singleton (iii), a set $u$ lies in $A\cup\{p\}$ if and only if $u\in A$ or $u=p$; likewise for $C\cup\{c\}$.

Existence in (i). The map $g$ is a set by Sets and Maps: Ordinary Notation (v), the ordered pair $(p,c)$ is a set by The Ordered Pair of Two Sets and Nested Tuples (i), and its singleton $\{(p,c)\}$ is a set by The Empty Set, the Unordered Pair and the Singleton (iii); so $h=g\cup\{(p,c)\}$ is a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper (iv), and its elements are the elements of $g$ together with $(p,c)$, by The Boolean Operations on Classes, Disjointness, and the Universal Class (i) and The Empty Set, the Unordered Pair and the Singleton (iii). Since $g$ is a function, every element of $g$ is an ordered pair by Relations, Domain, Range, Inverse and Composition (i); hence $h$ is a relation.

Let $u$, $v$ and $v'$ be sets with $(u,v)\in h$ and $(u,v')\in h$. If $(u,v)\in g$, then $u\in A$ by Domains and Restrictions of Maps (i), so $u\neq p$; hence $(u,v')\neq(p,c)$ by The Characteristic Property of Ordered Pairs and Nested Tuples of Sets (i), so $(u,v')\in g$, and $v=v'$ because $g$ is a function (Functions, Values of a Function, and Functions from One Class to Another (i)). If $(u,v)=(p,c)$, then $u=p$ and $v=c$ by The Characteristic Property of Ordered Pairs and Nested Tuples of Sets (i); as $p\notin A$, $(u,v')\notin g$ by Domains and Restrictions of Maps (i), so $(u,v')=(p,c)$ and $v'=c=v$. Thus $h$ is a function.

For a set $u$: by Relations, Domain, Range, Inverse and Composition (ii), $u\in\operatorname{dom}h$ if and only if $(u,v)\in h$ for some set $v$, that is, $(u,v)\in g$ for some $v$ or $(u,v)=(p,c)$ for some $v$. The first holds if and only if $u\in\operatorname{dom}g$, and $\operatorname{dom}g=A$ by Functions, Values of a Function, and Functions from One Class to Another (iii); the second holds if and only if $u=p$, by The Characteristic Property of Ordered Pairs and Nested Tuples of Sets (i). Hence $\operatorname{dom}h=A\cup\{p\}$ by Axiom of Extensionality for Classes. If $v\in\operatorname{ran}h$, then by Relations, Domain, Range, Inverse and Composition (iii) either $(u,v)\in g$ for some $u$, so $v\in\operatorname{ran}g\subseteq B$ by Functions, Values of a Function, and Functions from One Class to Another (iii), or $(u,v)=(p,c)$ for some $u$, so $v=c\in B$. Hence $\operatorname{ran}h\subseteq B$, and $h:A\cup\{p\}\to B$ is a map by Functions, Values of a Function, and Functions from One Class to Another (iii).

For $u\in A$, $(u,g(u))\in g\subseteq h$ by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (ii), so $h(u)=g(u)$ by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (i), applied to $h$; and $(p,c)\in h$ gives $h(p)=c$ in the same way.

Uniqueness in (i). If $h':A\cup\{p\}\to B$ is a map with $h'(u)=g(u)$ for every $u\in A$ and $h'(p)=c$, then $h'(u)=h(u)$ for every $u\in A\cup\{p\}$, as $u\in A$ or $u=p$; hence $h'=h$ by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (iv).

Proof of (ii). For $u\in A$, $h(u)=g(u)\in C$ by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (ii), applied to $g:A\to C$; and $h(p)=c$. So $h(u)\in C\cup\{c\}$ for every $u\in A\cup\{p\}$. If $v\in\operatorname{ran}h$, then $(u,v)\in h$ for some set $u$ by Relations, Domain, Range, Inverse and Composition (iii), $u\in A\cup\{p\}$ by Domains and Restrictions of Maps (i), and $v=h(u)$ by Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (i); so $\operatorname{ran}h\subseteq C\cup\{c\}$, and $h:A\cup\{p\}\to C\cup\{c\}$ is a map by Functions, Values of a Function, and Functions from One Class to Another (iii).

Injective. Let $u,u'\in A\cup\{p\}$ with $h(u)=h(u')$. If $u,u'\in A$, then $g(u)=g(u')$, so $u=u'$ as $g$ is injective. If $u\in A$ and $u'=p$, then $h(u)=g(u)\in C$ while $h(u')=c\notin C$, which is impossible; likewise if $u=p$ and $u'\in A$. If $u=u'=p$, there is nothing to prove.

Surjective. Let $v\in C\cup\{c\}$. If $v\in C$, there is $u\in A$ with $g(u)=v$, as $g:A\to C$ is surjective, and then $u\in A\cup\{p\}$ and $h(u)=v$. If $v=c$, then $h(p)=v$.

Hence $h:A\cup\{p\}\to C\cup\{c\}$ is a bijection by Injective, Surjective and Bijective Functions between Classes (iii).