Question

Let B be a finite set and $\mathit{f}\mathbf{:}\mathit{B}\mathbf{\to }\mathit{B}$is a function. Prove that f is injective if and only if f is surjective.

Short answer

Thus, the map$f:B\to lmf$ is surjective.

Step-by-step solution

Showing that the function is surjective

Consider B is a set and$f:B\to B$ is a function. If B is a finite set,

Let$a,b\in B$ such that$f\left(a\right)=f\left(b\right)$ but $a\ne b$. Now, since this is a finite set mapping to itself then there are not enough elements in X to map to X since two elements were used to map to one element, thus the function can’t be surjective.

Therefore, if the function is surjective it is also injective.

Showing that the function not injective

Let$a\in X$ such that$f\left(b\right)\ne a$ for all $b\in X$. Since the function assumed to be injective each$a\in X$ must map to an element of X, thus two elements in X mapping to the same element in X.

Hence, the function can’t be injective,

Therefore if the function is injective it must be surjective.