# 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.

Apr 11, 2017

This is not true for infinite sets.

#### Explanation:

Counterexample 1:
Let $f \left(x\right) = {e}^{x}$, 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 $f \left(A\right) \subseteq A$, 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 $| A | \le | {f}^{- 1} \left(A\right) |$. Since A is finite we must have $| A | = | {f}^{- 1} \left(A\right) |$. 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.