Proof of Images and Preimages under a Bijection and Its Inverse

Throughout, a pair $(u,v)$ lies in the inverse $R^{-1}$ of a relation $R$ exactly when $(v,u)\in R$, so for a function $h:A\to B$ the preimage $h^{-1}[C]$ of a class $C$ consists of the $a\in A$ with $h(a)\in C$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vi), $g(f(x))=x$ for $x\in X$ and $f(g(y))=y$ for $y\in Y$.

(i): Both sides are subclasses of $Y$ (for $f[E]$ since $f:X\to Y$), and $y\in Y$ lies in $g^{-1}[E]$ exactly when $g(y)\in E$. If $y\in f[E]$, then $y=f(x)$ for some $x\in E$, and $g(y)=g(f(x))=x\in E$. Conversely, if $g(y)\in E$, then $y=f(g(y))\in f[E]$.

(ii): An element $x\in X$ lies in $f^{-1}[f[E]]$ exactly when $f(x)\in f[E]$. Since $f$ is injective and $E\subseteq X$, this holds exactly when $x\in E$, by Images of Unions, Intersections, Differences and Subclasses under a Function, and under an Injective Function (iv). Hence $f^{-1}[f[E]]=E$.