Step-by-Step Solution

Topic: Injectivity & Surjectivity

Visual Representation: Mapping f(x) = x²

The mapping shows how elements from the Domain relate to the Codomain. Notice how negative inputs in R map to positive outputs, affecting injectivity.

(I) CHECK ONE-ONE (Real Numbers)

Function f : R → R, f(x) = x².

For a function to be one-one (injective), distinct elements must have distinct images.

Counter Example:

• • f(1) = (1)² = 1
• • f(-1) = (-1)² = 1

Since f(1) = f(-1) but 1 ≠ -1, the function is Not One-One.

(II) CHECK ONTO (Real Numbers)

Function f : R → R.

Range of x² is [0, ∞) (all non-negative real numbers).

Codomain is R (includes negative real numbers).

Since -2 ∈ Codomain but has no pre-image in Domain (no real number squares to -2):

(III)(A) f : N → N

Check One-One:

Let f(x₁) = f(x₂) ⇒ x₁² = x₂².

Since inputs are Natural numbers (positive), x₁ = x₂.

So, it is One-One.

Check Onto:

Codomain is N = {1, 2, 3, 4, 5…}.

Range is perfect squares = {1, 4, 9, 16…}.

Elements like 2, 3, 5 in the codomain have no pre-image in N.

(III)(B) f : N → S (Perfect Squares)

Codomain S = {1, 4, 9, 16, ……}.

Check One-One:

Same as above, for Natural numbers, x₁² = x₂² implies x₁ = x₂.

It is One-One.

Check Onto:

Here, the Codomain is defined specifically as the set of perfect squares.

Every element y in the codomain is a perfect square n², which has a pre-image n in the domain N.

Range = Codomain.