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(A) function $h$ has an inverse if and only if it is a bijection (both one-one and onto).$1$. Check for one-one: The function $h$ maps distinct elemen

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(A) function $h$ has an inverse if and only if it is a bijection (both one-one and onto).
$1$. Check for one-one: The function $h$ maps distinct elements of the domain $\{2, 3, 4, 5\}$ to distinct elements in the codomain $\{7, 9, 11, 13\}$. Specifically,$h(2)=7, h(3)=9, h(4)=11, h(5)=13$. Since no two elements in the domain have the same image,$h$ is one-one.
$2$. Check for onto: The range of $h$ is $\{7, 9, 11, 13\}$,which is equal to the codomain. Thus,every element in the codomain has a pre-image in the domain. Hence,$h$ is onto.
Since $h$ is both one-one and onto,it is a bijection. Therefore,the function $h$ has an inverse.

State with reason whether the following function has an inverse: $h: \{2, 3, 4, 5\} \rightarrow \{7, 9, 11, 13\}$ with $h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$.

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