Concrete functors between $\mathbf S\mathbf e\mathbf t$ to $\mathbf R\mathbf e\mathbf l$. There are given two concrete$(!)$ Categories $\mathbf S\mathbf e\mathbf t$ and $\mathbf R\mathbf e\mathbf l$, whose objects are pairs $(X,\rho )$, where $X$ is a Set and $\rho$ is a binary relation on this Set. They are considered as concrete categories over $\mathbf S\mathbf e\mathbf t$. And there is a statement that “there are precisely three concrete functors from $\mathbf S\mathbf e\mathbf t$ to $\mathbf R\mathbf e\mathbf l$”.
The morphisms in $\mathbf S \mathbf e\mathbf t$ are just set-functions and in $\mathbf R\mathbf e\mathbf l$ are relation-preserving maps.
By "concrete functor", I mean a functor that creates a (strictly) commutative triangle with the two forgetful functors $\mathbf S\mathbf e\mathbf t \rightarrow \mathbf S\mathbf e\mathbf t$ and $\mathbf R\mathbf e\mathbf l \rightarrow \mathbf S\mathbf e\mathbf t$.
The question is how to find them all and if we will find them, how to prove that there are no more concrete functors between such constructs.
I realize that here I need to find a simple thought but I started to learn Category theory not so long ago so it’s difficult for me to do it now.
 A: Hint: you need to find a uniform way of defining a binary relation $\rho_X$ on $X$ for any set $X$. Consider the following three possibilities: $\rho_X = \emptyset$, $\rho_X = \{(x, x) \mid x \in X\}$ and $\rho_X = X \times X$.
A: Speaking in the abstract, questions like this are often solved by the following procedure:


*

*Identify a small number of objects and morphisms that encode all of the relevant features 

*Use the definitions of functor or natural transformation to extend what you can say about those few objects to determine everything


The key idea here is going to be the following.
Let $0$ denote the empty set, $1 = \{ 0 \}$, and $2 = \{ 0, 1 \}$.
For any set $S$ with two distinct elements $x,y \in S$, you can construct:


*

*A function $f : 2 \to S$ with $f(0) = x$ and $f(1) = y$

*A function $g : S \to 2$ with $g(x) = 0$ and $g(y) = 1$


$f$ is unique; there are many such $g$. The point is that if we've selected binary relations $\rho_2$ and $\rho_S$ on these two sets and these functions must be relation preserving, then you can prove:


*

*$(0,1) \in \rho_2$ if and only if $(x,y) \in \rho_S$


A similar argument can be made to cover the case where $x=y$.
By using this, you can reduce the whole problem to selecting binary relations on the three sets $0, 1, 2$, and checking which choices extend to a functor.
