Proof of The Only Map from the Empty Set

By Axiom of Extensionality for Classes it suffices to show, for every set $F$, that $F\in B^{\emptyset}$ if and only if $F\in\{\emptyset\}$. By The Elements of the Class of Functions from a Set to a Set, and This Class Is a Set (i), $F\in B^{\emptyset}$ if and only if $F:\emptyset\to B$ is a map, and by The Empty Set, the Unordered Pair and the Singleton (iii), $F\in\{\emptyset\}$ if and only if $F=\emptyset$.

Let $F:\emptyset\to B$ be a map. By Functions, Values of a Function, and Functions from One Class to Another (iii), $F$ is a function, hence a relation, with $\operatorname{dom}F=\emptyset$. If $F$ had an element $p$, then $p=(u,v)$ for sets $u$ and $v$ by Relations, Domain, Range, Inverse and Composition (i), and $u\in\operatorname{dom}F=\emptyset$ by Relations, Domain, Range, Inverse and Composition (ii), which is impossible. So $F$ has no element, and $F=\emptyset$.

Conversely, $\emptyset:\emptyset\to B$ is a map. Having no element, $\emptyset$ is a relation, and it is a function, the condition of Functions, Values of a Function, and Functions from One Class to Another (i) on its pairs holding vacuously. By Relations, Domain, Range, Inverse and Composition (ii) and Relations, Domain, Range, Inverse and Composition (iii), a set lies in $\operatorname{dom}\emptyset$ or in $\operatorname{ran}\emptyset$ only if some pair lies in $\emptyset$; so both classes have no element. Hence $\operatorname{dom}\emptyset=\emptyset$, and $\operatorname{ran}\emptyset\subseteq B$, so $\emptyset:\emptyset\to B$ by Functions, Values of a Function, and Functions from One Class to Another (iii).

Thus $F\in B^{\emptyset}$ if and only if $F=\emptyset$, that is, if and only if $F\in\{\emptyset\}$.