A function is bijective if and only if has an inverse

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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.