# A function is bijective if and only if has an inverse It has been suggested as of August 17, 2021 that this page or section be merged into Equivalence of Definitions of Bijection. (Discuss)

## Theorem

A function is bijective if and only if has an inverse

## Proof

### We'll first prove that if a function is bijective, then has an inverse.

Let $f$ be a bijective function.

Since $f$ is surjective, then an $a$ such that $f(a)=b$ will exist for sure (since the inverse image of a surjective function can not be the empty set).

Since $f$ is injective, then $a$ is unique, so $f^{-1}$ is well defined.

### Now let's prove the other implication:

Let $f$: $A \to B$ be a function which has an inverse.

Let's prove $f$ is injective:

Let $a$1, $a$1 $\in$ $A$ and let $f(a$1$)$ = $f(a$2$)$
We must prove that $a$1 = $a$2:

$a$1 = $Id$A($a$1)
= $(f^{-1} \circ f)(a$1$)$
= $f^{-1} ( f(a$1$))$
= $f^{-1} ( f(a$2$))$
= $(f^{-1} \circ f)(a$2$)$
= $Id$A($a$2)
= $a$2
So we proved $a$1=$a$2.

Let's prove its surjectivity:
Let $b \in B$
$b = Id$A$(b)$
= $(f \circ f^{-1})(b)$
= $f ( f^{-1}(b))$
Since $f^{-1}(b) \in A$ by definition, this proves that $a \in A$ such that $f(a)=b$ exists, thus the surjectivity has been proved.