Proof of Maps on a Set Enlarged by One Point

Throughout, $C\cap\{x\}=\emptyset$ because $x\notin C$; the elements of $B^{C}$ are the maps $C\to B$ and those of $\operatorname{Inj}(C,D)$ the injective maps $C\to D$ (The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (i), The Sets of Injections, Surjections and Bijections between Two Sets (i)). For a set $Y$ and $b\in Y$, let $c_{b}:\{x\}\to Y$ be the constant map with value $b$ (Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (i)). For a map $\varphi$ on $C'$, the restriction $\varphi|_{C}$ is a map on $C$ with the same values as $\varphi$ there; so two maps on $C'$ with the same restriction to $C$ and the same value at $x$ agree at every point of $C'$, and are equal.

Step 1: maps ((i)). For $\varphi\in B^{C'}$ we have $\varphi|_{C}\in B^{C}$ and $\varphi(x)\in B$, since $x\in C'$; so the map is defined. It is injective by the last remark above. It is surjective: let $(\chi,b)\in B^{C}\times B$. By Gluing Maps on Disjoint Domains, and Shrinking the Codomain (i), applied to $\chi$ and $c_{b}:\{x\}\to B$, the union $\varphi=\chi\cup c_{b}$ is a map $\varphi:C'\to B$ with $\varphi(u)=\chi(u)$ for $u\in C$ and $\varphi(x)=b$. Hence $\varphi|_{C}=\chi$, and $\varphi$ is mapped to $(\chi,b)$.

Step 2: injections ((ii)). Fix $b\in D$ and put $E=\{\varphi\in\operatorname{Inj}(C',D):\varphi(x)=b\}$.

The map is defined: let $\varphi\in E$ and $u\in C$. Then $u\neq x$, so $\varphi(u)\neq\varphi(x)=b$ by injectivity. Thus $\varphi|_{C}$ takes its values in $D\setminus\{b\}$, so it is a map $C\to D\setminus\{b\}$ (Gluing Maps on Disjoint Domains, and Shrinking the Codomain (iii)), and it is injective because $\varphi$ is.

It is injective: two maps in $E$ with the same restriction to $C$ also have the same value $b$ at $x$, so they are equal.

It is surjective: let $\chi:C\to D\setminus\{b\}$ be injective. Its values lie in $D$, so $\chi$ is also a map $C\to D$ (Gluing Maps on Disjoint Domains, and Shrinking the Codomain (iii)). By Gluing Maps on Disjoint Domains, and Shrinking the Codomain (i), applied to $\chi$ and $c_{b}:\{x\}\to D$, the union $\varphi=\chi\cup c_{b}$ is a map $\varphi:C'\to D$ with $\varphi|_{C}=\chi$ and $\varphi(x)=b$. It is injective: two points of $C$ with the same value are equal because $\chi$ is injective, and a point $u\in C$ has $\varphi(u)=\chi(u)\neq b=\varphi(x)$. So $\varphi\in E$, and it is mapped to $\chi$.