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.

Alright, I have a problem to solve:

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

### Understanding the Problem
First, I need to understand what's being asked. We have a set \( A \) with 5 elements, and we need to find all functions from \( A \) to itself that are idempotent under composition. That is, applying the function twice is the same as applying it once.

### What is an Idempotent Function?
An idempotent function \( f \) satisfies \( f \circ f = f \). This means that for any \( x \) in \( A \), \( f(f(x)) = f(x) \).

### Interpretation
This condition implies that the function \( f \) is the identity on its own image. In other words, if you take any element \( y \) in the image of \( f \) (i.e., \( y = f(x) \) for some \( x \)), then \( f(y) = y \).

### Approach to the Solution
I recall that functions satisfying \( f(f(x)) = f(x) \) are called idempotent functions. To count them, I need to consider the structure of such functions.

### Step 1: Understand the Range and Fixed Points
Since \( f \) is idempotent, the image of \( f \) must consist of fixed points. That is, for any \( y \) in the image of \( f \), \( f(y) = y \).

### Step 2: Partition the Set A
The function \( f \) can be seen as a partition of \( A \) into two parts:
1. The image of \( f \), which consists of fixed points.
2. The pre-image of the fixed points.

### Step 3: Choose the Image
First, I need to choose the subset \( S \) of \( A \) that will be the image of \( f \). Since \( f \) is idempotent, \( S \) must be a set of fixed points.

The number of ways to choose \( S \) is the number of subsets of \( A \), which is \( 2^5 = 32 \), because for each element, I can decide whether to include it in \( S \) or not.

### Step 4: Define the Function on the Image
For the elements in \( S \), the function \( f \) must satisfy \( f(s) = s \) for all \( s \in S \). So, the behavior of \( f \) on \( S \) is fixed—it must be the identity on \( S \).

### Step 5: Define the Function on the Pre-Image
The remaining elements \( A \setminus S \) must be mapped to \( S \). For any \( x \notin S \), \( f(x) \) must be in \( S \), but \( f \) doesn't have to be injective or surjective on \( A \setminus S \).

### Counting the Number of Functions
For a given subset \( S \), the number of ways to define \( f \) is:
1. For each element in \( S \), \( f \) must map it to itself.
2. For each element not in \( S \), \( f \) can map it to any element in \( S \).

- The number of choices for \( S \) is \( 2^5 = 32 \).
- For each \( S \) with \( k \) elements, the number of ways to define \( f \) on \( A \setminus S \) is \( k^{5 - k} \), because there are \( 5 - k \) elements outside \( S \), and each can be mapped to any of the \( k \) elements in \( S \).

### Summing Up the Cases
I need to consider all possible sizes of \( S \), from 0 to 5.

1. **Case \( k = 0 \)**: \( S \) is empty.
   - Number of subsets \( S \): \( \binom{5}{0} = 1 \).
   - Number of functions: \( 0^5 = 0 \) (since \( 0^0 \) is undefined, but \( 0^5 \) is 0).

2. **Case \( k = 1 \)**: \( S \) has one element.
   - Number of subsets \( S \): \( \binom{5}{1} = 5 \).
   - Number of functions: \( 1^4 = 1 \) for each \( S \).
   - Total: \( 5 \times 1 = 5 \).

3. **Case \( k = 2 \)**: \( S \) has two elements.
   - Number of subsets \( S \): \( \binom{5}{2} = 10 \).
   - Number of functions: \( 2^3 = 8 \) for each \( S \).
   - Total: \( 10 \times 8 = 80 \).

4. **Case \( k = 3 \)**: \( S \) has three elements.
   - Number of subsets \( S \): \( \binom{5}{3} = 10 \).
   - Number of functions: \( 3^2 = 9 \) for each \( S \).
   - Total: \( 10 \times 9 = 90 \).

5. **Case \( k = 4 \)**: \( S \) has four elements.
   - Number of subsets \( S \): \( \binom{5}{4} = 5 \).
   - Number of functions: \( 4^1 = 4 \) for each \( S \).
   - Total: \( 5 \times 4 = 20 \).

6. **Case \( k = 5 \)**: \( S \