The functor $D(X)=X\bigsqcup X$ on $\mathbf{S}\mathbf{e}\mathbf{t}$ is not covariantly representable

Given: The category $\mathbf{S}\mathbf{e}\mathbf{t}$ and the tagged doubling assignment $D$ in the statement.

Sets and functions form the locally small category $\mathbf{S}\mathbf{e}\mathbf{t}$, and a functor must preserve identities and composition (Sets and functions form the large locally small category $\mathbf{S}\mathbf{e}\mathbf{t}$, Covariant functor, identity functor, composite functor, and contravariant functor).

The Cartesian product consists of its ordered pairs, and binary union contains exactly the elements in either of its two members (The Cartesian product $A\times B:=\{z\in \mathcal{P}(\mathcal{P}(A\cup B)):\mathrm{\exists }a\in A\mathrm{\exists }b\in Bz=(a,b)\}$, The union $\bigcup x$ of a set, and the binary union $a\cup b:=\bigcup \{a,b\}$, $\bigcup \mathrm{\varnothing }=\mathrm{\varnothing }$, $\bigcup \{a\}=a$, $\bigcup \{a,b\}=a\cup b$, $\bigcap \{a\}=a$, and $\bigcap \{a,b\}=a\cap b$).

Ordered pairs satisfy $(x,s)=(y,t)$ if and only if $x=y$ and $s=t$; the naturals $0$ and $1$ are distinct (The Kuratowski ordered pair $(a,b):=\{\{a\},\{a,b\}\}$, $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$, The natural numbers $\mathbb{N}$ (von Neumann)).

The functions $R\to 1$ form the set $1^R$, and a function assigns exactly one value to each element of its domain. Hence for every set $R$, including $R=\mathrm{\varnothing }$, there is exactly one function $R\to 1$ (The set $B^A$ of all functions $A\to B$, A function is a relation $f$ with $(a,b)\in f$ and $(a,c)\in f$ implying $b=c$; $f:A\to B$, the value $f(a)$, domain and codomain).

Representability by $R$ would give bijections $\mathbf{S}\mathbf{e}\mathbf{t}(R,X)\cong D(X)$ for every set $X$, and a bijection must be both injective and surjective (Presheaves, covariantly and contravariantly representable functors, and representations, Injection, surjection, bijection).