# Proof of bijection of a function

1. Apr 3, 2014

### Panphobia

1. The problem statement, all variables and given/known data

Consider a bijection f = (A,B,F)
Show that f^(-1) (inverse of f) is a bijection from B to A and that for any element x of A we have
f^(-1)(f(x))=x

3. The attempt at a solution

For this proof can I use contradiction and the say f^(-1) is not a bijection from B to A or there exists an element x of A that f^(-1)(f(x)) != x

If this is what I am supposed to prove. How would I go about doing it? If not what proof method should I use?

2. Apr 3, 2014

### jbunniii

To show that $f^{-1}$ is a bijection, you need to show three things:
(1) $f^{-1}$ is a function. This is not necessarily true for a general $f$. For example, if $f : \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x) = x^2$, and $y > 0$, then $f^{-1}(y)$ is a set containing two distinct elements: $\sqrt{y}$ and $-\sqrt{y}$.
(2) $f^{-1}$ is an injection, meaning that if $f^{-1}(y_1) = f^{-1}(y_2)$, then $y_1 = y_2$.
(3) $f^{-1}$ is a surjection, meaning that if $x$ is in the codomain of $f^{-1}$, then $x = f^{-1}(y)$ for some $y$ in the domain of $f^{-1}$.

3. Apr 3, 2014

### Panphobia

alright I got it.