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, which can be mapped to any fixed point.

### Step 3: Choose the Image
First, I need to choose the image of \( f \). The image can be any non-empty subset of \( A \) because \( f \) must map at least one element to itself (since \( f \) is idempotent, the image must be fixed points).

But, is the image necessarily non-empty? Well, if \( f \) is the empty function, but \( A \) is non-empty, \( f \) must map at least one element to itself. So, the image must be non-empty.

### Step 4: For Each Subset, Count the Functions
For a given subset \( S \subseteq A \) (which will be the image of \( f \)), the function \( f \) must satisfy:
1. For all \( s \in S \), \( f(s) = s \) (since \( S \) is the image and \( f \) is idempotent).
2. For all \( a \in A \setminus S \), \( f(a) \) can be any element of \( S \).

### Step 5: Count the Number of Choices
1. **Choose the image \( S \):** The number of non-empty subsets of \( A \) is \( 2^5 - 1 = 31 \), because there are \( 2^5 \) total subsets and one is the empty set.
2. **Define \( f \) on \( S \):** For each \( s \in S \), \( f(s) \) must be \( s \). There's only one choice for each \( s \) in \( S \).
3. **Define \( f \) on \( A \setminus S \):** For each element \( a \) not in \( S \), \( f(a) \) can be any element of \( S \). Since \( S \) has \( k \) elements, there are \( k \) choices for each \( a \).

### Step 6: Sum Over All Possible Subsets
For a subset \( S \) of size \( k \), the number of functions is:
- \( 1 \) way to define \( f \) on \( S \) (since \( f(s) = s \) for all \( s \in S \)).
- \( k^{5 - k} \) ways to define \( f \) on the remaining \( 5 - k \) elements, because each can be mapped to any of the \( k \) elements in \( S \).

But, we must consider all possible sizes \( k \) from 1 to 5.

### Step 7: Calculate for Each \( k \)
1. **k = 1:** Choose 1 element out of 5 to be in \( S \). The number of ways to choose \( S \) is \( \binom{5}{1} = 5 \). For each such \( S \), the number of functions is \( 1^{5-1} = 1 \). Total: \( 5 \times 1 = 5 \).
2. **k = 2:** Choose 2 elements out of 5. \( \binom{5}{2} = 10 \). For each \( S \), the number of functions is \( 2^{5-2} = 2^3 = 8 \). Total: \( 10 \times 8 = 80 \).
3. **k = 3:** Choose 3 elements. \( \binom{5}{3} = 10 \). For each \( S \), the number of functions is \( 3^{5-3} = 3^2 = 9 \). Total: \( 10 \times 9 = 90 \).
4. **k = 4:** Choose 4 elements. \( \binom{5}{4} = 5 \). For each \( S \), the number of functions is \( 4^{5-4} = 4^1 = 4 \). Total: \( 5 \times 4 = 20 \).
5. *