Given, `A = {2,3,4}, B={2,5,6,7}`. Construct an example of each of the following
(i) an injective mapping from A to B.
(ii) a mapping from A to B which is not injective.
(iii) a mapping from B to A.

Explanation

To solve the problem, we need to construct examples of mappings based on the given sets \( A \) and \( B \).

Given:
- \( A = \{2, 3, 4\} \)
- \( B = \{2, 5, 6, 7\} \)

### Step-by-Step Solutions:

#### (i) An injective mapping from \( A \) to \( B \)

An injective function (or one-to-one function) means that each element in set \( A \) maps to a unique element in set \( B \) without any repetitions.

**Example of an injective mapping:**
- Let \( f: A \to B \) be defined as:
- \( f(2) = 5 \)
- \( f(3) = 6 \)
- \( f(4) = 7 \)

In this mapping:
- Each element from \( A \) (2, 3, 4) maps to a different element in \( B \) (5, 6, 7).
- Thus, this mapping is injective.

#### (ii) A mapping from \( A \) to \( B \) which is not injective

A mapping that is not injective means that at least two elements from \( A \) map to the same element in \( B \).

**Example of a non-injective mapping:**
- Let \( g: A \to B \) be defined as:
- \( g(2) = 2 \)
- \( g(3) = 5 \)
- \( g(4) = 2 \)

In this mapping:
- Both \( 2 \) and \( 4 \) from set \( A \) map to \( 2 \) in set \( B \).
- Therefore, this mapping is not injective.

#### (iii) A mapping from \( B \) to \( A \)

Now we need to create a mapping from set \( B \) to set \( A \).

**Example of a mapping from \( B \) to \( A \):**
- Let \( h: B \to A \) be defined as:
- \( h(2) = 2 \)
- \( h(5) = 3 \)
- \( h(6) = 4 \)
- \( h(7) = 4 \)

In this mapping:
- \( 2 \) from \( B \) maps to \( 2 \) in \( A \),
- \( 5 \) maps to \( 3 \),
- \( 6 \) maps to \( 4 \),
- \( 7 \) also maps to \( 4 \).

### Summary of Mappings:
1. **Injective Mapping \( f \)**:
- \( f(2) = 5 \)
- \( f(3) = 6 \)
- \( f(4) = 7 \)

2. **Non-injective Mapping \( g \)**:
- \( g(2) = 2 \)
- \( g(3) = 5 \)
- \( g(4) = 2 \)

3. **Mapping \( h \) from \( B \) to \( A \)**:
- \( h(2) = 2 \)
- \( h(5) = 3 \)
- \( h(6) = 4 \)
- \( h(7) = 4 \)