The problem requires determining whether each given function is injective (one-to-one), surjective (onto), bijective (both), or neither for the domain and codomain $\{1,2,3,4,5\}$.

Analysis of Each Function:

1. Function 1: $f = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 4 & 3 & 2 & 1 \end{pmatrix}$

   • Injective? Yes, because no two inputs map to the same output.
   • Surjective? Yes, because every value in $\{1,2,3,4,5\}$ is covered.
   • Conclusion: Bijective.

2. Function 2: $f(x) = \begin{cases} 4 - x, & x < 4 \\ 6 - x, & x \geq 4 \end{cases}$

   • Checking outputs for $x = \{1,2,3,4,5\}$: $f(1) = 3, f(2) = 2, f(3) = 1, f(4) = 2, f(5) = 1$
   • Injective? No, because multiple inputs map to the same output (e.g., $f(2) = f(4) = 2$, $f(3) = f(5) = 1$).
   • Surjective? No, because the value 5 is missing in the range.
   • Conclusion: Neither.

3. Function 3: $f = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \end{pmatrix}$

   • Injective? Yes, each input maps to a unique output.
   • Surjective? Yes, all values in $\{1,2,3,4,5\}$ appear in the output.
   • Conclusion: Bijective.

4. Function 4: $f(x) = \begin{cases} x, & x < 3 \\ x - 2, & x \geq 3 \end{cases}$

   • Checking outputs for $x = \{1,2,3,4,5\}$: $f(1) = 1, f(2) = 2, f(3) = 1, f(4) = 2, f(5) = 3$
   • Injective? No, since $f(1) = f(3) = 1$, and $f(2) = f(4) = 2$.
   • Surjective? No, since 4 and 5 are missing in the output.
   • Conclusion: Neither.

Final Answers:

1. Neither
2. Neither
3. Bijective
4. Neither

Would you like a deeper explanation of any part? Here are some related questions:

1. How do you formally prove a function is injective or surjective?
2. What is the significance of a function being bijective?
3. How can you determine if a function is surjective using a mathematical approach?
4. What are real-world examples of injective and surjective functions?
5. How can a function be modified to make it bijective?

Tip: A function is injective if different inputs give different outputs and surjective if all possible outputs in the codomain are covered. If both hold, the function is bijective!