Question

Prove or disprove: If there is an injection \(f: A \rightarrow B\) and a surjection \(g: A \rightarrow B\), then there is a bijection \(h: A \rightarrow B\).

Short answer

The statement is false. Even if there is an injection \(f: A \rightarrow B\) and a surjection \(g: A \rightarrow B\), it does not necessarily mean there is a bijection \(h: A \rightarrow B\). A counterexample is used to disprove this claim.

Step-by-step solution

Understand the Definitions

An injection is a function where no two different input values map to the same output value. A surjection is a function that maps its domain onto its codomain, meaning every element in the codomain is the image of at least one element from the domain. A bijection combines these qualities: it's a one-to-one correspondence between the elements of the two sets.

Prove the Statement

The statement asks us to prove that the injection \(f\) and surjection \(g\) imply a bijection \(h\). However, the existence of \(f\) and \(g\) separately does not guarantee a bijection. For a counterexample, consider \(A = {1, 2}\) and \(B = {1, 2, 3}\) with \(f(a) = a\) (an injection, as each input gives a distinct output) and \(g(a) = 1\) for all \(a\) in \(A\) (a surjection, as all elements of \(B\) are covered, but not injective, as all elements of \(A\) map to the same element of \(B\)). There is no function \(h\) that is both injective and surjective from \(A\) to \(B\) because \(B\) has more elements than \(A\). So the statement 'If there is an injection \(f: A \rightarrow B\) and a surjection \(g: A \rightarrow B\), then there is a bijection \(h: A \rightarrow B\)' can be disproved.

Conclude the Result

Using a counterexample, we have shown that the existence of an injection and a surjection from \(A\) to \(B\) does not imply the existence of a bijection from \(A\) to \(B\). Thus, the original statement is false.