Question

Prove that a function has an inverse function if and only if it is one-to-one

Short answer

The exercise has been proved by bi-directional implication. Starting by assuming the existence of an inverse function, it was shown that this implies the function is injective. Following this, the function was assumed to be one-to-one, and with this assumption, an inverse function was constructed. Hence, a function has an inverse if and only if it is injective.

Step-by-step solution

Assume the function has an inverse

Suppose we have a function \( f : A \rightarrow B \) with inverse \( f^{-1} : B \rightarrow A \). Using the definition of the inverse, it follows that \(f(f^{-1}(b)) = b \) for every \( b \in B \). And also \( f^{-1}(f(a)) = a \) for every \( a \in A \).

Prove that the function is one-to-one

To show \( f \) is one-to-one or injective, we assume \( f(a) = f(a') \) for some \( a, a' \) in \( A \), and prove that \( a = a' \). Since \( f(a) = f(a') \) we can apply the inverse function to both sides to get \( f^{-1}(f(a)) = f^{-1}(f(a')) \), which simplifies to \( a = a' \), hence \( f is one-to-one or injective.

Assume the function is one-to-one

Now suppose \( f : A \rightarrow B \) is one-to-one. We need to construct an inverse.

Construction of the inverse

For each \( b \) in \( B \), if there is an \( a \) in \( A \) such that \( f(a) = b \), define \( f^{-1}(b) = a \). Since \( f \) is one-to-one, there is at most one such \( a \) so \( f^{-1} \) is well-defined. For every \( a \) in \( A \), \( f(f^{-1}(f(a))) = f(a) \), and for every \( b \) in \( B \) with \( a = f^{-1}(b) \), \( f^{-1}(f(a)) = a \). Therefore, \( f^{-1} \) is indeed an inverse of \( f \).

Conclusion

We have thus proved that if a function has an inverse then it is one-to-one, and if a function is one-to-one, it can have an inverse. Therefore, a function has an inverse if and only if it is one-to-one.