Find all equivalence classes of $R_{A,B}$

Question

Let $A$ and $B$ be sets. We define $R_{A,B}$ on $B^A$ such that:

For all $f,g:A\to B$, $f\,R_{A,B}\,g$ if and only if exists $h:A\to A$ such that $h$ is bijective on $A$ and $g=f\circ h$.

First, I wrote a proof that $R_{A,B}$ is equivalence relation on $B^A$.

Second, my question is, if we define $A=\{1,2,3\}$ and $B=\{4,5\}$ and I want to find all equivalences classes of $R_{A,B}$ and all equivalences classes of $R_{B,A}$. How can I find them?

How I think and what is my misunderstanding... I draw the functions and if $h=Id_A$ I can write for example:

$$\begin{aligned} \left[f\right]_{R}=&\{\{(1,4),(2,4),(3,5)\},\{(1,4),(2,5),(3,5)\},\{(1,4),(2,4),(3,4)\},\\ &\{(1,4),(2,5),(3,4)\},\{(1,5),(2,4),(3,4)\},\{(1,5),(2,5),(3,4)\},\\ &\{(1,5),(2,4),(3,5)\},\{(1,5),(2,5),(3,5)\}\}.\end{aligned}$$

However, I don`t know if this is correct and how to write them all? What if $h$ is not identity function on $A$?

Answer 1

There are only six bijections from $A$ into itself. Let us call them $h_1,h_2,\ldots,h_6$.

And there are eight elements in $B^A$. Let them be $f_1,f_2,\ldots,f_8$.

So, consider the functions $f_1\circ h_1,f_1\circ h_2,\ldots,f_1\circ h_6$. This set of functions is the equivalence class of $f_1$. Now, take a function $f\in B^A$ outside this equivalence class and start all over again. Keep doing this until there are no more functions left.

Answer 2

You can note that a necessary condition for $f\mathrel{R_{A,B}}g$ is that $f$ and $g$ have the same image.

Indeed, if $x\in A$, then $f(x)=g(h(x))$ is in the image of $g$ and conversely because $g=f\circ h^{-1}$.

Thus you can first group maps by their images and it should be easier to find the equivalence classes. The case $B=\{1,2\}$ is easy enough, as the images can be $\{1\}$, $\{2\}$ and $B$. In the first two cases, the maps are unique. In the second case there are six of them.