how many elements does Ia have? Let $A=\{1,2,3,4\}$. Let $F$ be the set of all functions from $A$ to $A$. Let $R$ be the relation on $F$ defined by $f,g \in F$

$f R g  \Leftrightarrow |f(A)|=|g(A)|$

$f(A)=\{f(x): x\in A\}$ which is range of $f$.

Find two elements of $I_A$ ($I_A = \{(1,1),(2,2),(3,3),(4,4)\}$)(the equivalence class of $I_A$),

[(1,1),(2,2)]?

how many elements does [$I_A$] have? 

Can some one help me? or give me some hints?
 A: HINT: $I_A$ is the identity function: $I_A(a)=a$ for each $a\in A$. Thus, $I_A[A]=A$, so $|I_A[A]|=|A|=4$. This means that $[I_A]$ is the set of all functions $f:A\to A$ such that $|f[A]|=4$; and since $|A|=4$, $[I_A]$ is just the set of all functions that map $A$ onto $A$, i.e., the set of surjections from $A$ to $A$. Thus, one member of $[I_A]$ is the function 
$$\{\langle 1,2\rangle,\langle 2,1\rangle,\langle 3,3\rangle,\langle 4,4\rangle\}$$
that switches $1$ and $2$ but leaves $3$ and $4$ unchanged. You should be able to find quite a few more members of $[I_A]$ without much trouble.
To determine how many members $[I_A]$ has, think about how you might build a function $f\in[I_A]$. Imagine doing this by choosing $f(1),f(2),f(3)$, and $f(4)$ one at a time. You can choose $f(1)$ to be any of the $4$ members of $A$. If you choose $f(2)$ to be equal to $f(1)$, your function $f$ won’t have all of $A$ as its range: it will ‘hit’ at most $3$ members of $A$. Thus, you’ll have to choose $f(2)$ to be one of the other $3$ members of $A$, the ones different from $f(1)$. How many choices will you have for $f(3)$ after you’ve chosen $f(1)$ and $f(2)$? How many for $f(4)$ when you’ve already chosen the first three values of $f$? Now put the pieces together: in how many different ways can you make this string of $4$ choices? That’s the number of different functions in $[I_A]$.
