It is given that h: {2, 3, 4, 5} → {7, 9, 11, 13} with

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

We can see that all distinct elements of the set (2, 3, 4, 5} have distinct images under h.

⇒ h is one–one.

Also, h is onto since for every element of the set {7, 9, 11, 13},

there exists an element x in the set {2, 3, 4, 5} such that h(x) = y.

Therefore, h is a one–one and onto function.

Therefore, h has an inverse.