Key Concepts and Formulas

• Function Composition: For functions $f: A \to B$ and $g: B \to C$, the composition $f \circ g: A \to C$ is defined by $(f \circ g)(x) = f(g(x))$ for all $x \in A$.
• One-to-One (Injective) Function: A function $f$ is one-to-one if distinct elements in the domain map to distinct elements in the codomain, i.e., if $f(x_1) = f(x_2)$, then $x_1 = x_2$.
• Onto (Surjective) Function: A function $f$ is onto if every element in the codomain is the image of at least one element in the domain, i.e., for every $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$.
• Fixed Point of a Function: An element $x$ in the domain of a function $f$ is a fixed point if $f(x) = x$. In function composition, if $f \circ g = f$, then for any $x$ in the domain of $g$, $f(g(x)) = f(x)$. This means that the elements in the image of $g$ are fixed points of $f$.

Step-by-Step Solution

Step 1: Analyze the structure and properties of the function $g$. The function $g: \mathbb{N} \to \mathbb{N}$ is defined for $n \ge 0$ as: $g(3n + 1) = 3n + 2$ $g(3n + 2) = 3n + 3$ $g(3n + 3) = 3n + 1$

Let's examine the behavior of $g$ on elements of the form $3n+1, 3n+2, 3n+3$. These are consecutive integers. The function $g$ essentially cycles these three types of numbers: $3n+1 \xrightarrow{g} 3n+2 \xrightarrow{g} 3n+3 \xrightarrow{g} 3n+1$

This means that $g$ partitions the natural numbers into disjoint sets of three elements, where each set is permuted cyclically by $g$. For example: If $n=0$: $g(1) = 2$, $g(2) = 3$, $g(3) = 1$. So $\{1, 2, 3\}$ is mapped as $1 \to 2 \to 3 \to 1$. If $n=1$: $g(4) = 5$, $g(5) = 6$, $g(6) = 4$. So $\{4, 5, 6\}$ is mapped as $4 \to 5 \to 6 \to 4$. In general, for any integer $k \ge 1$, the set $\{3k-2, 3k-1, 3k\}$ is mapped cyclically by $g$.

Let's also check $g \circ g$ and $g \circ g \circ g$: $g(g(3n+1)) = g(3n+2) = 3n+3$ $g(g(3n+2)) = g(3n+3) = 3n+1$ $g(g(3n+3)) = g(3n+1) = 3n+2$

$g(g(g(3n+1))) = g(3n+3) = 3n+1$ $g(g(g(3n+2))) = g(3n+1) = 3n+2$ $g(g(g(3n+3))) = g(3n+2) = 3n+3$

Thus, $g(g(g(x))) = x$ for all $x \in \mathbb{N}$. This means $g^3$ is the identity function on $\mathbb{N}$.

Step 2: Evaluate Option (C): $g \circ g \circ g = g$. From Step 1, we found that $g(g(g(x))) = x$ for all $x \in \mathbb{N}$. The identity function on $\mathbb{N}$ is $id(x) = x$. So, $g \circ g \circ g = id$. The question is whether $id = g$. This would imply $x = g(x)$ for all $x \in \mathbb{N}$. However, we know that $g(1) = 2 \ne 1$, $g(2) = 3 \ne 2$, $g(3) = 1 \ne 3$. Therefore, $g \circ g \circ g \ne g$. Option (C) is false.

Step 3: Evaluate Option (D): There exists a function $f: \mathbb{N} \to \mathbb{N}$ such that $g \circ f = f$. The condition $g \circ f = f$ means that for every $x \in \mathbb{N}$, $g(f(x)) = f(x)$. Let $y = f(x)$. Then the condition becomes $g(y) = y$. This implies that every element in the image of $f$ must be a fixed point of $g$. However, as we observed in Step 1, $g$ does not have any fixed points. For any $x \in \mathbb{N}$, $g(x) \ne x$. If such a function $f$ existed, then for any $x$, $f(x)$ would have to be a fixed point of $g$. Since there are no fixed points for $g$, the image of $f$ must be empty. But the codomain of $f$ is $\mathbb{N}$, which is not empty. Therefore, no such function $f$ can exist. Option (D) is false.

Step 4: Evaluate Option (B): There exists a one-one function $f: \mathbb{N} \to \mathbb{N}$ such that $g \circ f = f$. From Step 3, we know that for $g \circ f = f$ to hold, every element in the image of $f$ must be a fixed point of $g$. Since $g$ has no fixed points, the image of $f$ must be empty. However, the codomain of $f$ is $\mathbb{N}$, which is non-empty. Thus, such a function $f$ cannot exist. This applies whether $f$ is one-one or not. Therefore, Option (B) is false.

Step 5: Evaluate Option (A): There exists an onto function $f: \mathbb{N} \to \mathbb{N}$ such that $g \circ f = f$. The condition $g \circ f = f$ means $g(f(x)) = f(x)$ for all $x \in \mathbb{N}$. Let $y = f(x)$. Then $g(y) = y$. This implies that every element in the image of $f$, denoted as $Im(f)$, must be a fixed point of $g$. As we've established, $g$ has no fixed points. This seems to lead to a contradiction. Let's re-examine the problem statement and the options. The question asks if there exists such a function.

Consider the structure of $g$ again. $g$ permutes elements in cycles of length 3. $g(3n+1) = 3n+2$ $g(3n+2) = 3n+3$ $g(3n+3) = 3n+1$

If $g \circ f = f$, then $g(f(x)) = f(x)$ for all $x$. Let $y = f(x)$. Then $g(y) = y$. This means that the image of $f$ must be a subset of the fixed points of $g$. Since $g$ has no fixed points, the image of $f$ must be the empty set. However, $f: \mathbb{N} \to \mathbb{N}$, and $\mathbb{N}$ is not empty, so $Im(f)$ cannot be empty.

There seems to be a misunderstanding of the question or the properties. Let's carefully consider the implications of $g \circ f = f$. This means that for any $x$, $f(x)$ is an element such that $g$ leaves it unchanged. That is, $f(x)$ must be a fixed point of $g$. Since $g$ has no fixed points, the set of fixed points of $g$ is empty. So, for any $x$, $f(x)$ must belong to the empty set. This is impossible if $f$ maps to $\mathbb{N}$.

Let's re-read the question and options very carefully. Perhaps there's a nuance in the definition of $\mathbb{N}$ or the context. The problem states $\mathbb{N} = \{1, 2, 3, \dots\}$. The definition of $g$ is for $n \ge 0$. For $n=0$: $g(1)=2, g(2)=3, g(3)=1$. For $n=1$: $g(4)=5, g(5)=6, g(6)=4$. And so on.

Let's reconsider Option (A): There exists an onto function $f : \mathbb{N} \to \mathbb{N}$ such that $f \circ g = f$. The condition is $f(g(x)) = f(x)$ for all $x \in \mathbb{N}$. This means that $f$ is constant on the cycles of $g$. For any $n \ge 0$: $f(g(3n+1)) = f(3n+1)$ $f(3n+2) = f(3n+1)$

$f(g(3n+2)) = f(3n+2)$ $f(3n+3) = f(3n+2)$

$f(g(3n+3)) = f(3n+3)$ $f(3n+1) = f(3n+3)$

Combining these, we get $f(3n+1) = f(3n+2) = f(3n+3)$ for all $n \ge 0$. This means that $f$ must map all elements of the form $3n+1, 3n+2, 3n+3$ to the same value. In other words, $f$ must be constant on sets of the form $\{3n+1, 3n+2, 3n+3\}$.

Let $f(3n+1) = f(3n+2) = f(3n+3) = c_n$, where $c_n$ is some value in $\mathbb{N}$. So, $f$ is defined by specifying its value on each set of three consecutive integers. We need to check if there exists an onto function $f: \mathbb{N} \to \mathbb{N}$ satisfying this property.

Let's define $f$ such that it is onto. For $n=0$, we have the set $\{1, 2, 3\}$. Let $f(1) = f(2) = f(3) = 1$. For $n=1$, we have the set $\{4, 5, 6\}$. Let $f(4) = f(5) = f(6) = 2$. For $n=2$, we have the set $\{7, 8, 9\}$. Let $f(7) = f(8) = f(9) = 3$. In general, we can define $f(3n+1) = f(3n+2) = f(3n+3) = n+1$ for $n \ge 0$.

Let's verify if this function $f$ is onto. The image of $f$ is $\{f(1), f(2), f(3), f(4), f(5), f(6), \dots\}$. Using our definition $f(3n+1) = f(3n+2) = f(3n+3) = n+1$: $f(1)=1, f(2)=1, f(3)=1$ (for $n=0$) $f(4)=2, f(5)=2, f(6)=2$ (for $n=1$) $f(7)=3, f(8)=3, f(9)=3$ (for $n=2$) ... The image of $f$ is $\{1, 2, 3, 4, \dots\} = \mathbb{N}$. So, this function $f$ is indeed onto.

Now let's check if this $f$ satisfies $f \circ g = f$. We need to check $f(g(x)) = f(x)$ for all $x \in \mathbb{N}$.

Case 1: $x = 3n+1$. $f(g(3n+1)) = f(3n+2)$. According to our definition, $f(3n+2) = n+1$. And $f(3n+1) = n+1$. So, $f(g(3n+1)) = n+1 = f(3n+1)$. This holds.

Case 2: $x = 3n+2$. $f(g(3n+2)) = f(3n+3)$. According to our definition, $f(3n+3) = n+1$. And $f(3n+2) = n+1$. So, $f(g(3n+2)) = n+1 = f(3n+2)$. This holds.

Case 3: $x = 3n+3$. $f(g(3n+3)) = f(3n+1)$. According to our definition, $f(3n+1) = n+1$. And $f(3n+3) = n+1$. So, $f(g(3n+3)) = n+1 = f(3n+3)$. This holds.

Since $f(g(x)) = f(x)$ for all $x \in \mathbb{N}$, the condition $f \circ g = f$ is satisfied. We have constructed an onto function $f: \mathbb{N} \to \mathbb{N}$ such that $f \circ g = f$. Therefore, Option (A) is true.

Let's quickly re-examine why other options are incorrect. Option (B): There exists a one-one function $f : \mathbb{N} \to \mathbb{N}$ such that $f \circ g = f$. We found that $f \circ g = f$ implies $f(3n+1) = f(3n+2) = f(3n+3)$ for all $n \ge 0$. If $f$ were one-one, then $3n+1 = 3n+2$ would imply $f(3n+1) = f(3n+2)$, which is true. However, if $f$ is one-one, then distinct inputs must map to distinct outputs. But $f(3n+1) = f(3n+2)$. Since $3n+1 \ne 3n+2$, for $f$ to be one-one, this equality cannot hold. The only way $f(3n+1) = f(3n+2)$ can hold for a one-one function is if there are no such distinct inputs, which is not the case for $\mathbb{N}$. So, a one-one function $f$ cannot satisfy $f(3n+1) = f(3n+2)$. Thus, no one-one function exists such that $f \circ g = f$. Option (B) is false.

Option (C): $g \circ g \circ g = g$. We showed $g \circ g \circ g = id$. Since $g$ is not the identity function, this is false.

Option (D): There exists a function $f : \mathbb{N} \to \mathbb{N}$ such that $g \circ f = f$. This means $g(f(x)) = f(x)$. Let $y = f(x)$. Then $g(y) = y$. This implies that every element in the image of $f$ must be a fixed point of $g$. As shown earlier, $g$ has no fixed points. So, the image of $f$ must be a subset of the empty set, which means $Im(f)$ is empty. However, $f: \mathbb{N} \to \mathbb{N}$, so $Im(f)$ must be a non-empty subset of $\mathbb{N}$. This is a contradiction. Therefore, no such function $f$ exists. Option (D) is false.

The reasoning for Option (A) is sound and leads to the correct answer. The key was to correctly interpret $f \circ g = f$ as $f$ being constant on the cycles of $g$, and then to construct an onto function with this property.

Common Mistakes & Tips

• Confusing $f \circ g$ with $g \circ f$: Pay close attention to the order of composition. The condition $f \circ g = f$ means $f(g(x)) = f(x)$, while $g \circ f = f$ means $g(f(x)) = f(x)$.
• Misinterpreting "exists": When a question asks "there exists", you only need to find one example of such a function. If you can construct one, the statement is true.
• Assuming Properties of $f$: Do not assume $f$ is one-one or onto unless stated. Check each property separately for each option.
• Fixed Points of $g$: Carefully analyze if $g$ has any fixed points. The existence or non-existence of fixed points is crucial for conditions like $g \circ f = f$.
• Structure of $g$: Recognizing that $g$ partitions $\mathbb{N}$ into disjoint 3-cycles is key to understanding the implications for composition.

Summary

The problem requires analyzing the behavior of the function $g$ and then testing the properties of potential functions $f$ in relation to $g$. We found that $g$ acts as a 3-cycle permutation on sets of three consecutive integers. Option (C) was ruled out because $g^3$ is the identity, not $g$. Option (D) and (B) were ruled out because the condition $g \circ f = f$ requires $f(x)$ to be a fixed point of $g$, and $g$ has no fixed points. Option (A) states the existence of an onto function $f$ such that $f \circ g = f$. This condition implies that $f$ must be constant on the 3-cycles of $g$. We successfully constructed such a function by defining $f(3n+1) = f(3n+2) = f(3n+3) = n+1$, which is onto and satisfies $f \circ g = f$.

The final answer is $\boxed{A}$.