A function is bijective if and only if has an inverse

Definition

Let
$$\begin{align} f &: A \rightarrow B \\ 1_A &: A \rightarrow A, \ f(a) = a\  \forall a \in A \end{align}$$
We say that $f$ is

(1) well-defined if whenever $a_1 = a_2$ for some $a_1, a_2 \in A$, then $f(a_1) = f(a_2)$.
(2) injective if whenever $f(a_1) = f(a_2)$ for some $a_1, a_2 \in A$, then $a_1 = a_2$.
(3) surjective if $\forall b \in B \  \exists a \in A  \ s.t. \  f(a)=b$.
(4) bijective if it is both injective and surjective.

Theorem 1 A function $g : B \rightarrow A$ is the inverse of $f$ if $f \circ g = 1_B$ and $g \circ f  = 1_A$

Proof

We'll proof the theorem on title by showing (a) and (b).

(a) $f : A \rightarrow B$ is bijective $\Longrightarrow$ $f$ has an inverse

Let $f$ be bijective. Since $f$ is surjective, $\forall b \in B \  \exists a \in A  \ s.t. \  f(a)=b$. So we can define a function $f^{-1} : B \rightarrow A$ and let $f^{-1}(b) = a$. Since $f$ is injective, $f^{-1}$ is well-defined automatically. Next, we'll check that $f^{-1}$ is the inverse of $f$ by showing (a-1) and (a-2) as the definition itself.

(a-1) $f^{-1} \circ f = 1_A$
Let $a\in A$ and $b = f(a)$. Then, by definition, $f^{-1}(b) = a$.
Which satisfies $f^{-1} \circ f(a) = f^{-1}(f(a)) = f^{-1}(f(a)) = a$.

(a-2) $f \circ f^{-1} = 1_B$
Let $b\in B$ and $a = f^{-1}(b)$. Then, by definition, $f(a) = b$.
Which satisfies $f \circ f^{-1}(b) = f(f^{-1}(b)) = f(a) = b$.

(b) $f$ has an inverse $\Longrightarrow$ $f : A \rightarrow B$ is bijective

Let $f : A \rightarrow B$ has an inverse $f^{-1} : B \rightarrow A$. Now, we'll check $f$ is bijective by showing (b-1) and (b-2).

(b-1) $f$ is surjective

Suppose $b \in B$ and $a = f^{-1}(b)$. Then $f(a) = f(f^{-1}(b)) = f \circ f^{-1}(b) = 1_B(b) = b$. Which implies $f$ is surjective.

(b-2) $f$ is injective

Let $a_1, a_2 \in A \ s.t. \ f(a_1) = f(a_2)$ and $a = f^{-1}(b)$. Then
$$\begin{align} a_2 &= 1_A(a_2) \\ &= f^{-1} \circ f(a_2) \\ &= f^{-1}(f(a_2)) \\ &= f^{-1}(b) \\ &= a. \end{align}$$
At the same time,
$$\begin{align} a_1 &= 1_A(a_1) \\ &= f^{-1} \circ f(a_1) \\ &= f^{-1}(f(a_1)) \\ &= f^{-1}(f(a_2)) \\ &= f^{-1}(b) \\ &= a. \end{align}$$
Therefore $a_1 = a_2$, which implies $f$ is injective.
$$Q.E.D.$$

Reference

[1] Proof of Katherine E. Stange
[2] 4th Property of Theorem 0.7 in J Gallian's "Contemporary Abstract Algebra Seventh Edition"