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Consider that $f: A \to B$, where $A$ and $B$ are finite sets with $|A| = |B|.$ Let us show that $f$ is one-to-one if and only if it is onto.

If $f$ is one-to-one, then $f(x)\ne f(y)$ for any $x\ne y,$ and hence the set $Im(f)=\{f(a):a\in A\}$ has the same number of different elements as $A$. It follows that $|Im(f)|=|A|=|B|.$ Since $Im(f)\subset B$ and $B$ is finite, we conclude that $Im(f)=B,$ and hence $f$ is onto.

If $f$ is onto, then the preimage $f^{-1}(b)\ne\emptyset$ for any $b\in B.$ Let us prove using the method by contradiction. Suppose that $f$ is not one-to-one. Then $f(x)=f(y)=b$ for some $x,y\in A, b\in B, x\ne y.$ Then $A\supset f^{-1}(b)\supset\{x,y\}.$ Since $|f^{-1}(b)|\ge2$ and $|B|=|A|,$ we conclude that $|f^{-1}(b')|=0$ for some $b'\in B.$ We get that $f^{-1}(b')=\emptyset,$ and so we have the contradiction with $f^{-1}(b')\ne\emptyset.$ This contraction proves that $f$ is one-to-one.