Understanding Bijective Functions

A function \(f\) from a set A to a set B (denoted as \(f: A \to B\)) is considered a bijective function if it is both:

• Injective (One-to-One): Every element in the codomain (set B) is mapped to by at most one element in the domain (set A). This means if \(f(a_1) = f(a_2)\), then \(a_1 = a_2\). Distinct elements in A map to distinct elements in B.
• Surjective (Onto): Every element in the codomain (set B) is mapped to by at least one element in the domain (set A). This means for every \(b \in B\), there exists at least one \(a \in A\) such that \(f(a) = b\). The range of the function is equal to its codomain.

For finite sets A and B with the same number of elements (\(|A| = |B|\)), a function \(f: A \to B\) is bijective if and only if it is injective or surjective. In this question, set A = {1, 2, 3} and set B = {4, 5, 6}. Both sets have 3 elements.

Analyzing the Given Options for Bijective Function

We need to examine each given relation to determine if it represents a bijective function from A to B.

Option 1: \(f = \{(2, 4), (2, 5), (2, 6)\}\)

• In a function, each element from the domain (set A) must be mapped to exactly one element in the codomain (set B).
• In this relation, the element '2' from set A is mapped to three different elements in set B: 4, 5, and 6.
• This violates the definition of a function. Therefore, this relation is not a function at all.

Option 2: \(f = \{(1, 5), (2, 4), (3, 6)\}\)

• The domain of this relation is {1, 2, 3}, which is equal to set A. Each element in A appears exactly once as the first element of a pair. This confirms it is a function from A to B.
• Let's check if it's injective (one-to-one):
  • 1 maps to 5
  • 2 maps to 4
  • 3 maps to 6
• The distinct elements 1, 2, and 3 from A are mapped to distinct elements 5, 4, and 6 in B. So, it is injective.
• Let's check if it's surjective (onto):
  • The elements mapped to in B are {5, 4, 6}.
  • The codomain is set B = {4, 5, 6}.
• The set of mapped elements {5, 4, 6} is the same as the codomain {4, 5, 6}. Every element in B is mapped by some element in A. So, it is surjective.
• Since the function is both injective and surjective, it is a bijective function.

Option 3: \(f = \{(1, 4), (2, 5), (3, 6)\}\)

• The domain of this relation is {1, 2, 3}, which is equal to set A. Each element in A appears exactly once as the first element of a pair. This confirms it is a function from A to B.
• Let's check if it's injective (one-to-one):
  • 1 maps to 4
  • 2 maps to 5
  • 3 maps to 6
• The distinct elements 1, 2, and 3 from A are mapped to distinct elements 4, 5, and 6 in B. So, it is injective.
• Let's check if it's surjective (onto):
  • The elements mapped to in B are {4, 5, 6}.
  • The codomain is set B = {4, 5, 6}.
• The set of mapped elements {4, 5, 6} is the same as the codomain {4, 5, 6}. Every element in B is mapped by some element in A. So, it is surjective.
• Since the function is both injective and surjective, it is a bijective function.

Option 4: More than one of the above

• Our analysis shows that both Option 2 and Option 3 represent bijective functions from set A to set B.
• Therefore, this option, stating that more than one of the given options is a bijective function, is correct.

Option 5: None of the above

• Since we found that Option 2 and Option 3 are bijective functions, this option is incorrect.

Conclusion

Options 2 and 3 both represent bijective functions from set A to set B. Thus, the correct choice is the one indicating that more than one of the given options is a bijective function.