Key Concepts and Formulas

• Function Composition: For functions $f: A \to B$ and $g: B \to C$, the composite function $g \circ f: A \to C$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in A$.
• Image of a Function: The image of a function $f: A \to B$, denoted by $Im(f)$, is the set of all possible output values of $f$. That is, $Im(f) = \{f(x) \mid x \in A\}$.
• Identity Function: The identity function $id_X$ on a set $X$ is defined by $id_X(x) = x$ for all $x \in X$.

Step-by-Step Solution

Step 1: Analyze the function $f$ and determine its image.

The domain of the function $f$ is $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. The function $f$ is defined as: f(k) = \left\{ {\matrix{ {k + 1} & {if\,k\,is\,odd} \cr k & {if\,k\,is\,even} \cr } } \right.

Let's compute the values of $f(k)$ for each $k \in A$:

• For odd $k$:
  • $f(1) = 1 + 1 = 2$
  • $f(3) = 3 + 1 = 4$
  • $f(5) = 5 + 1 = 6$
  • $f(7) = 7 + 1 = 8$
  • $f(9) = 9 + 1 = 10$
• For even $k$:
  • $f(2) = 2$
  • $f(4) = 4$
  • $f(6) = 6$
  • $f(8) = 8$
  • $f(10) = 10$

The image of $f$, denoted by $Im(f)$, is the set of all distinct output values of $f$: $Im(f) = \{2, 4, 6, 8, 10\}$ This step is crucial because the condition $g \circ f = f$ imposes constraints on the behavior of $g$ specifically on the image of $f$.

Step 2: Understand the condition $g \circ f = f$.

The condition $g \circ f = f$ means that for every element $k \in A$, we must have $g(f(k)) = f(k)$. Let $y = f(k)$. Then the condition becomes $g(y) = y$. This implies that for any element $y$ that belongs to the image of $f$, the function $g$ must map $y$ to itself. In other words, $g$ must act as the identity function on the set $Im(f)$.

So, for all $y \in Im(f)$, we must have $g(y) = y$.

Step 3: Determine the constraints on the function $g$.

From Step 2, we know that $g$ must satisfy:

• $g(2) = 2$
• $g(4) = 4$
• $g(6) = 6$
• $g(8) = 8$
• $g(10) = 10$

These are 5 specific mappings that $g$ must perform.

Step 4: Determine the freedom of choice for the remaining elements in the domain of $g$.

The domain of $g$ is $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. The codomain of $g$ is also $A$. We have already fixed the values of $g$ for the elements in $Im(f) = \{2, 4, 6, 8, 10\}$. The elements in the domain of $g$ for which the value of $g$ is not yet determined are those elements in $A$ that are not in $Im(f)$. Let $S = A \setminus Im(f)$. $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \setminus \{2, 4, 6, 8, 10\}$ $S = \{1, 3, 5, 7, 9\}$ These are the odd numbers in the set $A$. There are 5 elements in $S$.

For each element $k \in S$, the value of $g(k)$ can be any element from the codomain $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. There are no further constraints on these mappings from the condition $g \circ f = f$.

Step 5: Calculate the total number of possible functions $g$.

For each of the 5 elements in $S = \{1, 3, 5, 7, 9\}$, there are 10 possible choices for its image under $g$ (since the codomain is $A$, which has 10 elements). Since the choice of the image for each element in $S$ is independent, the total number of ways to define $g$ for these elements is $10 \times 10 \times 10 \times 10 \times 10 = 10^5$.

The values of $g$ for the elements in $Im(f)$ are fixed to be the identity mapping. So there is only 1 way to define $g$ for these 5 elements.

Therefore, the total number of possible functions $g$ is the product of the number of choices for each part: Number of functions $g$ = (Number of ways to define $g$ for elements in $S$) $\times$ (Number of ways to define $g$ for elements in $Im(f)$) Number of functions $g$ = $10^5 \times 1 = 10^5$.

Common Mistakes & Tips

• Confusing Domain and Image: A common mistake is to assume that $g$ must be the identity function on its entire domain. However, the condition $g \circ f = f$ only imposes constraints on $g$ for values that are in the image of $f$.
• Misinterpreting $g(f(k)) = f(k)$: This equation means that $g$ maps the output of $f$ to itself. It does not necessarily mean $g(k) = k$ for all $k$.
• Careful Counting: Ensure that you are correctly identifying the elements for which $g$ has fixed values and the elements for which $g$ has free choices.

Summary

The problem requires us to find the number of functions $g: A \to A$ such that $g \circ f = f$. The condition $g(f(k)) = f(k)$ for all $k \in A$ implies that $g$ must act as the identity function on the image of $f$. We first determined the image of $f$ to be $Im(f) = \{2, 4, 6, 8, 10\}$. This means $g(2)=2, g(4)=4, g(6)=6, g(8)=8, g(10)=10$. For the remaining elements in the domain of $g$, which are $\{1, 3, 5, 7, 9\}$ (5 elements), the function $g$ can map each of them to any of the 10 elements in the codomain $A$. Thus, there are $10^5$ possible functions for $g$.

The final answer is $\boxed{10^5}$.