Example 13
Show that an onto function \(f : \{1, 2, 3\}
\to \{1, 2, 3\}\) is always one-one.
Solution
Suppose \(f\) is not one-one. Then
there exists two elements, say 1 and 2 in the domain whose image in the
co-domain is same.
Also, the image of 3 under \(f\) can be
only one element.
Therefore, the range set can have at the most two elements of the
co-domain \(\{1, 2, 3\}\),
showing that \(f\) is not onto, a
contradiction.
Hence, \(f\) must be one-one.