Using the definition, Prove that the function `f:A to B` is invertible if and only if `f` is both one-one and onto.

Explanation

To prove that the function \( f: A \to B \) is invertible if and only if \( f \) is both one-one (injective) and onto (surjective), we will break the proof into two parts:

1. **If \( f \) is invertible, then \( f \) is one-one and onto.**
2. **If \( f \) is one-one and onto, then \( f \) is invertible.**

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

**Step 1:** Assume that \( f \) is invertible. This means there exists a function \( f^{-1}: B \to A \) such that for every \( b \in B \), \( f^{-1}(b) \) gives a unique \( a \in A \) such that \( f(a) = b \).

**Step 2:** To show that \( f \) is one-one, assume \( f(a_1) = f(a_2) \) for some \( a_1, a_2 \in A \).

- Applying \( f^{-1} \) to both sides gives:
\[ f^{-1}(f(a_1)) = f^{-1}(f(a_2)) \]
- This simplifies to:
\[ a_1 = a_2 \]
- Hence, \( f \) is one-one.

**Step 3:** To show that \( f \) is onto, take any \( b \in B \).

- Since \( f \) is invertible, there exists an \( a \in A \) such that \( f(a) = b \) (specifically \( a = f^{-1}(b) \)).
- This shows that every element in \( B \) has a pre-image in \( A \), thus \( f \) is onto.

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

**Step 4:** Assume \( f \) is one-one and onto.

**Step 5:** To define the inverse function \( f^{-1}: B \to A \), we need to show that for each \( b \in B \), there exists a unique \( a \in A \) such that \( f(a) = b \).

- Since \( f \) is onto, for every \( b \in B \), there exists at least one \( a \in A \) such that \( f(a) = b \).
- Since \( f \) is one-one, this \( a \) is unique.

**Step 6:** Define \( f^{-1}(b) = a \) where \( f(a) = b \).

- This function \( f^{-1} \) is well-defined because \( a \) is unique for each \( b \).
- We need to show that \( f(f^{-1}(b)) = b \) and \( f^{-1}(f(a)) = a \) for all \( a \in A \) and \( b \in B \), which follows from the definitions.

Thus, \( f \) is invertible.

### Conclusion

We have shown that \( f \) is invertible if and only if \( f \) is both one-one and onto.

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