Question:
If A = {1, 2, 3}, show that a one-one function f : A → A must be onto.
Solution:
A ={1, 2, 3}
Number of elements in A = 3
Number of one $-$ one functions $=$ number of ways of arranging 3 elements $=3 !=6$
So, the possible one -one functions can be the following:
(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}
Here, in each function, range = {1, 2, 3}, which is same as the co-domain.
So, all the functions are onto.