72. Suppose that is a function from A to B, where A and B are finite sets with |A| = |B| . Show that is one-to-one if and only if it is onto.

We are asked to prove that a function \( f: A \to B \), where \( |A| = |B| \) (i.e., the cardinalities of sets \( A \) and \( B \) are equal), is **one-to-one** if and only if it is **onto**.

### Definitions:
- **One-to-one (injective)**: A function \( f \) is one-to-one if \( f(a_1) = f(a_2) \) implies \( a_1 = a_2 \). In other words, different elements in \( A \) map to different elements in \( B \).
- **Onto (surjective)**: A function \( f \) is onto if for every \( b \in B \), there exists some \( a \in A \) such that \( f(a) = b \). In other words, every element of \( B \) is hit by some element from \( A \).

### Proof Outline:
We will prove this in two parts:
1. If \( f \) is **one-to-one**, then \( f \) is **onto**.
2. If \( f \) is **onto**, then \( f \) is **one-to-one**.

---

### Part 1: If \( f \) is one-to-one, then \( f \) is onto.

- Assume that \( f \) is **injective** (one-to-one).
- Since \( f \) is injective, no two elements in \( A \) map to the same element in \( B \). Therefore, \( |f(A)| = |A| \), meaning the image of \( A \) under \( f \) has the same cardinality as \( A \).
- But \( |A| = |B| \) by assumption, so \( |f(A)| = |B| \).
- Therefore, \( f(A) = B \), meaning that \( f \) is **onto**.

---

### Part 2: If \( f \) is onto, then \( f \) is one-to-one.

- Assume that \( f \) is **surjective** (onto).
- Since \( f \) is onto, every element of \( B \) has a preimage in \( A \), meaning \( f(A) = B \).
- Because \( |A| = |B| \), the number of preimages in \( A \) must be exactly equal to the number of elements in \( B \).
- If \( f \) were not injective, there would be at least two distinct elements in \( A \) mapping to the same element in \( B \), which would contradict the fact that \( |A| = |B| \).
- Therefore, \( f \) must be injective, or **one-to-one**.

---

### Conclusion:
Since we have shown that if \( f \) is injective, it is also surjective, and if \( f \) is surjective, it is also injective, we conclude that \( f \) is one-to-one if and only if it is onto.

Let me know if you'd like further details or have any questions.

Here are some related questions to deepen your understanding:
1. What happens if \( |A| > |B| \)? Can \( f \) still be one-to-one or onto?
2. How does the proof change if \( |A| \neq |B| \)?
3. Can you find a concrete example of a bijection (both one-to-one and onto) between two sets?
4. What does the property of a function being both one-to-one and onto imply about its inverse?
5. How does the proof differ for infinite sets \( A \) and \( B \)?

**Tip**: A bijective function (both one-to-one and onto) always has an inverse function that undoes the mapping.

Suppose that f is a function from A to B, where A and B are finite sets with |A| = |B|. Show that f is one-to-one if and only if it is onto.

Explore the proof that a function f from A to B is one-to-one if and only if it is onto, assuming A and B are finite sets with equal cardinality. This detailed solution covers key concepts in set theory, including injective and surjective functions, and bijections. Ideal for advanced high school or undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation