math-combo-01

# Counting Idempotent Self-Maps

Let $A = \{1, 2, 3, 4, 5\}$. Count the number of functions $f : A \to A$ that satisfy
$$f(f(x)) = f(x) \qquad \text{for every } x \in A.$$

(Such a function acts as the identity on its own range.)

Give your answer as a single integer.

The problem asks for the number of functions $f: A \to A$ where $A = \{1, 2, 3, 4, 5\}$ such that $f(f(x)) = f(x)$ for all $x \in A$.
This condition is equivalent to saying that $f$ is an idempotent function.
Let $R = \text{Im}(f) = \{f(x) \mid x \in A\}$ be the range of $f$.
The condition $f(f(x)) = f(x)$ means that for any $y \in R$, $f(y) = y$.
So, $f$ acts as the identity map on its range $R$.
For any $x \in A$, $f(x)$ must be an element of $R$.
Also, if $y \in R$, then $f(y) = y$.

Let $k = |R|$. Since $R \subseteq A$, $k$ can be any integer from $1$ to $5$.
Let's fix the range $R$. Let $R = \{r_1, r_2, \dots, r_k\}$.
The condition $f(y) = y$ for all $y \in R$ is fixed.
For any $x \in A$, $f(x)$ must be in $R$.
So, for each $x \in A$, $f(x)$ can be any element of $R$.
However, we must ensure that the range of $f$ is exactly $R$.
If we just choose $f(x) \in R$ for all $x$, the range of $f$ will be a subset of $R$.
To ensure the range is exactly $R$, every element $r \in R$ must be in the image of $f$.
Since $f(r) = r$ for all $r \in R$, the condition that $r \in \text{Im}(f)$ is automatically satisfied because $r = f(r)$.
So, the condition "range is exactly $R$" is satisfied if and only if $R \subseteq \text{Im}(f)$.
Since $\text{Im}(f) \subseteq R$ is always true by definition of $R$ (if we define $R$ as the set of values taken), the only constraint is that for every $r \in R$, there exists some $x$ such that $f(x) = r$.
But we know $f(r) = r$. So $r$ is in the image.
Wait, let's re-evaluate.
Let $f: A \to A$ be a function.
Let $R = \text{Im}(f)$.
The condition $f(f(x)) = f(x)$ implies that for any $y \in R$, $f(y) = y$.
Proof: Let $y \in R$. Then $y = f(x)$ for some $x$. Then $f(y) = f(f(x)) = f(x) = y$.
So, $f$ restricted to $R$ is the identity map.
Conversely, if $f$ restricted to $R$ is the identity map, then for any $x \in A$, $f(x) \in R$, so $f(f(x)) = f(x)$ holds.
So the condition is equivalent to: $f$ maps $A$ into $R$, and $f$ is the identity on $R$.
Here $R$ is the image of $f$.
Let's choose the set $R$ first.
Let $|R| = k$. There are $\binom{5}{k}$ ways to choose the set $R$.
Once $R$ is chosen, for each $r \in R$, $f(r)$ is fixed to be $r$.
For each $x \in A \setminus R$, $f(x)$ can be any element of $R$.
The number of elements in $A \setminus R$ is $5 - k$.
For each such element, there are $k$ choices.
So there are $k^{5-k}$ ways to define $f$ on $A \setminus R$.
Is it possible that the image of $f$ is a proper subset of $R$?
Let $S = \text{Im}(f)$. We assumed $R$ is the image.
We defined $f(r) = r$ for $r \in R$. Thus $R \subseteq S$.
Also, for $x \in A \setminus R$, $f(x) \in R$, so $S \subseteq R$.
Thus $S = R$.
So, for a fixed set $R$ of size $k$, the number of such functions is $k^{5-k}$.
Wait, is this correct?
Let's check.
Let $A = \{1, 2, 3\}$.
Possible sizes for $R$: 1, 2, 3.
If $|R|=1$, say $R=\{1\}$. Then $f(1)=1$. $f(2) \in \{1\}$, $f(3) \in \{1\}$.
$f(2)=1, f(3)=1$. $f(f(2))=f(1)=1=f(2)$. Correct.
Number of functions: $\binom{3}{1} \times 1^{3-1} = 3 \times 1 = 3$.
The functions are:
1. $f(1)=1, f(2)=1, f(3)=1$. Range $\{1\}$.
2. $f(1)=1, f(2)=1, f(3)=1$ (Wait, range is $\{1\}$).
Let's list them properly.
$R=\{1\}$. $f(1)=1$. $f(2)=1, f(3)=1$.
$R=\{2\}$. $f(2)=2$. $f(1)=2, f(3)=2$.
$R=\{3\}$. $f(3)=3$. $f(1)=3, f(2)=3$.
Total 3.
If $|R|=2$, say $R=\{1, 2\}$. $f(1)=1, f(2)=2$.
$x \in \{3\}$. $f(3) \in \{1, 2\}$.
So $f(3)$ can be 1 or 2.
Case 1: $f(3)=1$. Range is $\{1, 2\}$. $f(f(3))=f(1)=1=f(3)$.
Case 2: $f(3)=2$. Range is $\{1, 2\}$. $f(f(3))=f(2)=2=f(3)$.
Number of functions for a fixed $R$ of size 2 is $2^{3-2} = 2^1 = 2$.
Number of sets $R$ of size 2 is $\binom{3}{2} = 3$.
Total for $|R|=2$ is $3 \times 2 = 6$.
If $|R|=3$, say $R=\{1, 2, 3\}$. $f(1)=1, f(2)=2, f(3)=3$.
$x \in \emptyset$. $f(x)$ choices $3^0 = 1$.
Number of functions is $1$.
Total for $|R|=3$ is $\binom{3}{3} \times 1 = 1$.
Total number of functions = $3 + 6 + 1 = 10$.

Let's double check with the formula.
Total = $\sum_{k=1}^