Question

Kindly solve this question based on Functions ?

Which of the following statement(s) is(are) correct, Explain with some example ?

(A) If `f` is a one-one mapping from set A to A , then `f` is onto.

(B) If `f` is an onto mapping from set A to A , then `f` is one-one.

Answer 1

This is not true for infinite sets.

Explanation:

Counterexample 1:
Let `f(x) = e^x`, which is defined for all x `in` R.
Then f is one-to-one (with its inverse being the natural logarithm), but f is not onto; its range is the positive numbers.

Counterexample 2:
Let f be defined on the natural numbers as follows:
f(1) = 1.
For n > 1, f(n) = n - 1.
Then f(2) = f(1), so f is not one-to-one.
However, every natural number is in the image of the function, so f is onto.

For finite sets it is true that f is one-to-one if and only if f is onto.

Let | A | be the cardinality of the finite set, A, and let |f(A)| be the cardinality of the image of A under f.
Assume f is one-to-one. Then | A | = | f(A) |, by the definition of one-to-one. Since A is finite and `f(A) sube A`, we must have f(A) = A. Thus f is onto.

Assume instead that f is onto. Then for each a `in` A, there is at least one x `in` A such that f(x) = A. Since this is true for all A, then `| A | <= |f^(-1)(A)|`. Since A is finite we must have `| A | = |f^(-1)(A)|`. That is, f is one-to-one. Alternatively, if there were at least one pair, a and b, such that f(a) = f(b), then | f(A) | would be strictly less than A, so that f would not be onto.