Q.Show that an onto function $f: \{1, 2, 3\} \to \{1, 2, 3\}$ is always one-one.

Concept: Onto Function Definition

A function $f: A \to B$ is onto (surjective) if every element of $B$ has at least one preimage in $A$. Here, both sets have exactly 3 elements.

Reasoning

1. Since $f$ is onto, each of the 3 elements in the codomain $\{1,2,3\}$ must be mapped to by at least one element from the domain $\{1,2,3\}$. …

Why This Works — The Intuition

Think of a function as a matching between two sets of three labelled balls. If the function is onto (surjective), every ball in the codomain $\{1,2,3\}$ gets hit at least once. Since the domain also has exactly three balls, the only way to cover all three targets is to hit each one exactly once. If any target were hit twice, some other target would be left untouched — violating onto-ness. That “exactly once” property is precisely what it means to be one-one (injective).

This is a special case of a general fact: for finite sets of equal size, onto ⇔ one‑one. The proof below makes this precise.

Step‑by‑Step Proof

1. State what we know.

We have $f: A \to A$ where $A = \{1,2,3\}$.

• $f$ is onto: for every $y \in A$, there exists some $x \in A$ with $f(x) = y$.
• $|A| = 3$ (the size of the set).

2. Count the images.

Because $f$ is onto, the image set $f(A)$ must be the whole codomain $A$. So $|f(A)| = 3$.

3. Relate domain size to image size.

For any function from a finite set $X$ to a set $Y$, the number of elements in the image cannot exceed the number of elements in the domain:

Here $|f(A)| = 3$ and $|A| = 3$, so the inequality is actually an equality: $|f(A)| = |A|$.

4. Interpret the equality.

The only way a function from a finite set to itself can have $|f(A)| = |A|$ is if no two domain elements map to the same codomain element. Why?

• Suppose two different elements $x_1 \neq x_2$ in $A$ satisfy $f(x_1) = f(x_2)$. Then the image set would have at most $|A|-1$ distinct values, because one output is “shared”.
• Since we already know the image has exactly $3$ distinct values, sharing cannot happen. Hence $f$ is one‑one. …

Method: Onto implies one-one for a finite set mapped to itself

Use this counting argument whenever the domain and codomain are the same finite set (or two finite sets of equal size).

Steps

Step 1: Note the sizes are equal and finite

Record $|A|=|B|=n$. This equality is what makes the argument work — it fails for infinite sets.

Step 2: Use onto to fix the image size

Onto means the image equals the whole codomain, so the image has $n$ distinct elements — all $n$ targets are hit.

Step 3: Deduce one-one by pigeonhole …

Common Mistakes

Mistake 1: Assuming onto $\Rightarrow$ one-one for any sets

Why it's wrong: this equivalence needs finite sets of equal size; for infinite sets an onto map can fail to be one-one (e.g. $n\mapsto\lfloor n/2\rfloor$ on $\mathbb{N}$). Correct approach: state and use that $|A|=|B|=3$ is finite.

Mistake 2: Not giving the counting/pigeonhole reason …