As the title says, I would like to prove that $f\colon X\to Y$ bijective $\Longleftrightarrow$ $f$ has an inverse function.
Proof
$\implies$
Let $f$ be bijective. That means $\forall y\in Y\exists! x\in X: f(x)=y$. Define $f^{-1}\colon Y\to X, y\mapsto x$, then $f^{-1}(y)=f^{-1}(f(x))=x$.
$\Longleftarrow$
Let $f^{-1}\colon Y\to X$ be the inverse function of $f\colon X\to Y$. Surjectivity of $f$: Let $y\in Y$. Then $f^{-1}(y)\in X$ and $y=f(f^{-1}(y))$, so there exists a $x\in X: f(x)=y$. So $f$ is surjective.
Injectivity of $f$: Consider $f(x)=f(y)$. Then $x=f^{-1}(f(x))=f^{-1}(f(y))=y$. So $f$ is injective.
So $f$ is injective and surjective so it is bijective.
That's my proof. Is it okay?