Show that an onto function $f: \{1, 2, 3\} \rightarrow \{1, 2, 3\}$ is always one-one.

(A) Let $f: A \rightarrow A$ be a function where $A = \{1, 2, 3\}$.
Suppose $f$ is not one-one. Then there exist at least two distinct elements in the domain $A$ that have the same image in the co-domain $A$.
Since the domain has $3$ elements,if $f$ is not one-one,at most $2$ distinct elements can be mapped to the co-domain.
Specifically,if $f(x_1) = f(x_2) = y$ for $x_1 \neq x_2$,then the range of $f$ can contain at most $2$ elements (the image $y$ and the image of the third element).
However,for $f$ to be onto,the range must be equal to the co-domain,which has $3$ elements.
Since the range has at most $2$ elements,it cannot be equal to the co-domain $\{1, 2, 3\}$.
This contradicts the assumption that $f$ is onto.
Therefore,$f$ must be one-one.