The problem asks how many functions $g : \{a, b, c\} \to \{1, 2, 3, 4, 5\}$ are such that the composition $f \circ g$ is a bijection on $\{a, b, c\}$.

Understanding the problem:

• We are given a function $f : \{1, 2, 3, 4, 5\} \to \{a, b, c\}$ where $f(1) = a$, $f(2) = a$, $f(3) = b$, $f(4) = c$, and $f(5) = c$.
• We are also defining a function $g : \{a, b, c\} \to \{1, 2, 3, 4, 5\}$, and we want the composition $f \circ g$ to be a bijection from $\{a, b, c\}$ to itself.

Key Points:

• $f \circ g$ means that for each element $x \in \{a, b, c\}$, we first apply $g$ to $x$, then apply $f$ to the result of $g(x)$. So, for each element $x \in \{a, b, c\}$, $f(g(x))$ should return $a, b,$ or $c$.
• Since $f \circ g$ must be a bijection on $\{a, b, c\}$, each element of $\{a, b, c\}$ must be mapped to a unique element of $\{a, b, c\}$ by the composition.

Step-by-Step Solution:

1. Behavior of $f$:

   • $f(1) = a$ and $f(2) = a$: Both 1 and 2 are mapped to $a$.
   • $f(3) = b$: 3 is mapped to $b$.
   • $f(4) = c$ and $f(5) = c$: Both 4 and 5 are mapped to $c$.

2. Conditions for $g$:

   • To make $f \circ g$ a bijection, each $x \in \{a, b, c\}$ must be mapped to one of the preimages of $x$ under $f$. This means:
     • $g(a)$ must be either 1 or 2 (since $f(1) = a$ and $f(2) = a$).
     • $g(b)$ must be 3 (since $f(3) = b$).
     • $g(c)$ must be either 4 or 5 (since $f(4) = c$ and $f(5) = c$).

3. Counting the possibilities:

   • For $g(a)$, we have 2 choices: either 1 or 2.
   • For $g(b)$, there is only 1 choice: 3.
   • For $g(c)$, we have 2 choices: either 4 or 5.

4. Total number of functions:

   • The total number of possible functions $g$ is the product of these choices: $2 \times 1 \times 2 = 4$

Thus, there are 4 possible functions $g$ such that $f \circ g$ is a bijection.

The correct answer is 4.

Would you like a more detailed explanation or have any further questions? Here are five related questions that might help expand this understanding:

1. How would the problem change if $f$ had different mappings, for example, if $f(1) = b$?
2. What are the general conditions for a function composition to be bijective?
3. How many bijections exist from a set of 4 elements to itself?
4. Can you always create a bijection if the sets involved have different cardinalities?
5. How does the concept of injectivity differ from bijectivity?

Tip: When dealing with bijective functions, always ensure that every element in both the domain and codomain has a unique match. This can simplify understanding the restrictions on function composition.