Inverse Permutation Cdc824

1. **State the problem:** We need to prove that if $f$ is a permutation on a set $A$, then its inverse function $f^{-1}$ is also a permutation on $A$. 2. **Recall definitions:** - A permutation on $A$ is a bijection from $A$ to $A$. - A bijection is a function that is both injective (one-to-one) and surjective (onto). - The inverse function $f^{-1}$ of a bijection $f$ is defined such that $f^{-1}(f(x)) = x$ for all $x \in A$. 3. **Show $f^{-1}$ is a function from $A$ to $A$:** Since $f$ is a bijection from $A$ to $A$, for every $y \in A$, there exists a unique $x \in A$ such that $f(x) = y$. Thus, $f^{-1}(y) = x$ is well-defined and maps $A$ to $A$. 4. **Show $f^{-1}$ is injective:** Suppose $f^{-1}(y_1) = f^{-1}(y_2)$ for some $y_1, y_2 \in A$. Then $x = f^{-1}(y_1) = f^{-1}(y_2)$ implies $f(x) = y_1$ and $f(x) = y_2$. Since $f$ is a function, $y_1 = y_2$. Hence, $f^{-1}$ is injective. 5. **Show $f^{-1}$ is surjective:** For any $x \in A$, since $f$ is surjective, there exists $y \in A$ such that $f(x) = y$. Then $f^{-1}(y) = x$, so every element $x$ in $A$ has a preimage under $f^{-1}$. Thus, $f^{-1}$ is surjective. 6. **Conclusion:** Since $f^{-1}$ is both injective and surjective from $A$ to $A$, it is a bijection. Therefore, $f^{-1}$ is a permutation on $A$. **Final answer:** If $f$ is a permutation on $A$, then $f^{-1}$ is also a permutation on $A$.