Proof of Inverse of a Bijection

Claim 1. Let $f:X\to Y$ be a bijection. By Bijection of Sets, for every $y\in Y$ there is exactly one $x\in X$ with $f(x)=y$; let $g(y)$ denote that element. This defines a map $g:Y\to X$ with $f(g(y))=y$ for every $y\in Y$, so $f\circ g=\mathrm{id}_{Y}$. For $x\in X$, the element $x$ satisfies $f(x)=f(x)$, and $g(f(x))$ is by construction the unique element of $X$ whose image under $f$ is $f(x)$; hence $g(f(x))=x$, so $g\circ f=\mathrm{id}_{X}$.

For uniqueness, let $h:Y\to X$ also satisfy $h\circ f=\mathrm{id}_{X}$ and $f\circ h=\mathrm{id}_{Y}$. For every $y\in Y$,

using $g\circ f=\mathrm{id}_{X}$ in the first equality and $f\circ h=\mathrm{id}_{Y}$ in the second. Hence $h=g$.

Claim 3. Suppose $f:X\to Y$ and $g:Y\to X$ satisfy $g\circ f=\mathrm{id}_{X}$ and $f\circ g=\mathrm{id}_{Y}$. Let $y\in Y$. Then $f(g(y))=y$, so $g(y)$ is an element of $X$ whose image under $f$ is $y$. If $x\in X$ also satisfies $f(x)=y$, then

So there is exactly one such element, and $f$ is a bijection by Bijection of Sets. Exchanging the roles of $f$ and $g$, and of $X$ and $Y$, shows in the same way that $g$ is a bijection. Since $g$ satisfies the two identities of claim 1, the uniqueness in claim 1 gives $g=f^{-1}$.

Claim 2. Let $f:X\to Y$ be a bijection. By claim 1 the map $f^{-1}$ satisfies $f^{-1}\circ f=\mathrm{id}_{X}$ and $f\circ f^{-1}=\mathrm{id}_{Y}$. Applying claim 3 with $f^{-1}$ in the role of $f$ and $f$ in the role of $g$ shows that $f^{-1}$ is a bijection from $Y$ to $X$ and that $f=(f^{-1})^{-1}$.