Question

Suppose that f is a function from A to B, where A and B are finite sets with |A| = |B|. Show that f is one-to-one if and only if it is onto.

Short answer

f is one-to-one if and only if it is onto.

Step-by-step solution

Step: 1

Assuming that f is one-to-one.

We assume that f is not onto.

Then there exists an element$b\in B$ such that $\forall \mathrm{a}\in :\mathrm{f}\left(\mathrm{a}\right)\ne \mathrm{b}$

Since there then exists two different elements in A such that they have the same image.

So f(x) = f (y) which implies that x = y

Step: 2

Assume that f is onto.

We assume that f is not one-to-one there exists

$\begin{array}{r}x\in A,y\in A\\ f\left(x\right)=f\left(y\right)\\ x\ne y\end{array}$

which is a contradiction.

Hence f is onto.