Q.Given $A = \{2, 3, 4\}$, $B = \{2, 5, 6, 7\}$. Construct an example of each of the following:

Concept: One One (Injective) Function — each element of the domain maps to a distinct element of the codomain; no two domain elements share the same image.

Step 1 — Injective mapping from $A$ to $B$:

We need each of $2,3,4$ to map to different elements in $B$. Choose three distinct images, e.g.

$f(2)=2,\; f(3)=5,\; f(4)=6$.

Step 2 — Non-injective mapping from $A$ to $B$:

At least two elements of $A$ must map to the same element of $B$. For instance,

$g(2)=2,\; g(3)=2,\; g(4)=5$.

Step 3 — Mapping from $B$ to $A$: …

The core idea

A mapping (function) from a set $X$ to a set $Y$ assigns to each element of $X$ exactly one element of $Y$.

An injective (one‑one) mapping has the extra property: different inputs give different outputs.

So to build an injective map from $A$ to $B$, we just need to pick three distinct elements of $B$ (since $A$ has three elements) and pair them up. To build a non‑injective map, we deliberately repeat an image. For a map from $B$ to $A$, we assign each of the four elements of $B$ to some element of $A$ — no restriction at all.

Let’s do each part.

(a) An injective mapping from $A$ to $B$

1. $A$ has three elements: $2, 3, 4$.

   $B$ has four elements: $2, 5, 6, 7$.

   For injectivity, the three images must be distinct.

2. Choose any three distinct elements of $B$. For instance: $2, 5, 6$.

   Then define:

3. Check: $2 \neq 3 \neq 4$ and $f(2) \neq f(3) \neq f(4)$. So $f$ is injective.

(b) A mapping from $A$ to $B$ which is not injective

1. To fail injectivity, at least two different elements of $A$ must map to the same element of $B$.

2. A simple way: send two elements to the same image. For example:

3. Here $g(2) = g(3)$ even though $2 \neq 3$, so $g$ is not injective. (It is still a valid mapping because every element of $A$ gets exactly one output.) …

Method: Constructing mappings between two finite sets

Use this when asked to build a mapping (function) between finite sets with a stated injectivity requirement — injective, non-injective, or 'any mapping'.

Steps

Step 1: Confirm it is a function first.

Every element of the domain must be assigned exactly one image in the codomain. That single rule is all a plain 'mapping' needs — no restriction on how images are shared.

Step 2: For an injective (one-one) map, hand out distinct images.

List the domain elements and assign each a different element of the codomain. This is only possible when the codomain has at least as many elements as the domain ($|\text{domain}| \le |\text{codomain}|$).

Step 3: For a non-injective map, reuse one image on purpose. …

Common Mistakes

Mistake 1: Thinking 'not injective' means every image must repeat.

Why it's wrong: injectivity fails as soon as one image is shared by two inputs; the other assignments can be distinct. Correct approach: for part (b) send just two elements of $A$ to the same element of $B$ (e.g. $g(2)=g(3)=2$) and map the rest freely.

Mistake 2: Trying to build an injective map from $B$ to $A$. …