# A function is bijective if and only if has an inverse

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## Contents

## 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.