Show Onto Function on Finite Set is One-One
📺 Video Explanation
📝 Question
Let:
\[ A=\{1,2,3\} \]
Show that if:
\[ f:A\to A \]
is onto, then \(f\) must be one-one.
✅ Solution
Set:
\[ A=\{1,2,3\} \]
has 3 elements.
🔹 Onto Property
Since function is onto:
- Every element of codomain has pre-image.
So all:
\[ 1,2,3 \]
must appear in range.
🔹 Suppose Function is Not One-One
Then two different inputs have same output.
So total distinct outputs would become less than 3.
That means at least one element in codomain would be missed.
This contradicts onto property.
Therefore:
function must be one-one.
🎯 Final Answer
\[ \boxed{\text{Every onto function }f:A\to A\text{ is one-one}} \]
🚀 Exam Shortcut
- Finite set with same size:
- Injection ⇔ Surjection
- Very useful theorem in functions