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.