Question

Prove that a one-one function \( f : \{2, 3, 4\} \to \{2, 3, 4\} \) is onto.

Hint:

For a function to be onto, every element of the codomain must be covered by the function. A one-one function with equal domain and codomain sizes is always onto.

Answer 1

A function \( f: A \to B \) is said to be onto if for every element in the codomain \( B \), there exists an element in the domain \( A \) such that \( f(x) = y \). Let \( f : \{2, 3, 4\} \to \{2, 3, 4\} \) be a one-one function. Since the function is one-one (injective), every element of the domain must map to a unique element in the codomain. This means no two elements in the domain map to the same element in the codomain. To prove that the function is onto, we need to show that every element in the codomain \( \{2, 3, 4\} \) has a pre-image in the domain \( \{2, 3, 4\} \). Given that the function is one-one and both the domain and codomain have the same number of elements, each element of the codomain must be mapped to by exactly one element of the domain. Hence, the function is onto. Conclusion: Since every element in the codomain \( \{2, 3, 4\} \) is mapped to by some element in the domain \( \{2, 3, 4\} \), the function is onto.