Mostowski collapse for extensional relations

Use the collapse map and let $Y=\{\pi (x):x\in X\}$. It is injective by the extensional-collapse lemma and surjective onto this range by definition. If $z\in \pi (x)$, its defining equation gives $z=\pi (y)$ for some $yRx$; hence $z\in Y$. Thus $Y$ is transitive.

If $yRx$ then $\pi (y)\in \pi (x)$. Conversely, if $\pi (y)\in \pi (x)$, the collapse equation gives $zRx$ with $\pi (z)=\pi (y)$; injectivity gives $z=y$. This proves preservation and reflection of the relation.

For any other isomorphism $j$ onto a transitive class $Z$, each member of $j(x)$ lies in $Z$ and so is $j(y)$ for a unique $y\in X$. Relation reflection then says exactly $yRx$. Consequently $j(x)=\{j(y):yRx\}$, the same recursion as $\pi$. Uniqueness of the collapse map gives $j=\pi$, and thus $Z=Y$. When $X$ is a set, Replacement forms both its graph and image.