Show One-One Function on Finite Set is Onto

📺 Video Explanation

📝 Question

Let:

\[ A=\{1,2,3\} \]

Show that if:

\[ f:A\to A \]

is one-one, then \(f\) must be onto.

---

✅ Solution

Set \(A\) has:

\[ 3 \] elements.

---

🔹 One-One Property

If function is one-one:

• Different inputs have different outputs.

So images of:

\[ 1,2,3 \]

must all be distinct.

---

🔹 Since Codomain Also Has 3 Elements

Codomain:

\[ A=\{1,2,3\} \]

also has exactly 3 elements.

If three distinct outputs are chosen from a set of three elements, all elements must be used.

So:

range = codomain.

Hence:

✔ Function is onto.

---

🎯 Final Answer

\[ \boxed{\text{Every one-one function }f:A\to A\text{ is onto}} \]

---

🚀 Exam Shortcut

• Finite set + same number of elements
• Injection automatically gives surjection
• Very important finite set theorem