Thread: bijective

If $\displaystyle f: A \to B $ is bijective show that $\displaystyle f^{-1} $ is unique.

Suppose that $\displaystyle f^{-1} $ and $\displaystyle h $ are inverses of $\displaystyle f $. Let $\displaystyle y = f(x) $. We need to show that $\displaystyle f^{-1}(y) = h(y) $ for all $\displaystyle y \in B $. So $\displaystyle f^{-1}(f(x)) = h(f(x)) = x $ for all $\displaystyle y \in B $, thus $\displaystyle f^{-1} $ is unique?

Originally Posted by particlejohn

If $\displaystyle f: A \to B $ is bijective show that $\displaystyle f^{-1} $ is unique?

Let $\displaystyle g,h: B\mapsto A$ such that $\displaystyle g(f(a)) =h(f(a))= a$ for all $\displaystyle a\in A$. To show that $\displaystyle g=h$ we need to show to things, (i) they have the same domain (ii) $\displaystyle g(x)=h(x)$ for all $\displaystyle x$ in domain. The first part is straightforward, it is given that $\displaystyle g,h$ have domain $\displaystyle B$. Now let $\displaystyle x\in B$. Then since $\displaystyle f$ is bijective it means $\displaystyle x=f(a)$ for some $\displaystyle a\in A$. Thus, $\displaystyle g(x) = g(f(a)) = a = h(f(a)) = h(x)$. And thus these are the same functors.