With a supplied well-powering, a subobject classifier represents the subobject functor

By [L3], the quotient ${\mathrm{Sub}}_M(X):=M_X\mathrm{/}\sim$ by mutual factorization is a well-defined set for each $X$, and its elements correspond exactly to subobject classes of $X$ represented inside the supplied set $M_X$.

For a morphism $u:X\to Y$ and a class $[m]\in {\mathrm{Sub}}_M(Y)$, pull back $m$ along $u$. The resulting monomorphism into $X$ defines a subobject class of $X$, and [L2] provides a representative of that class in $M_X$; step 1.1 shows that the resulting element of ${\mathrm{Sub}}_M(X)$ is independent of the chosen representative. Hence $X\mapsto {\mathrm{Sub}}_M(X)$ is a contravariant functor.

For each $X$, define ${\mathrm{\Phi }}_X:\mathcal{C}(X,\mathrm{\Omega })\to {\mathrm{Sub}}_M(X)$ by sending $f:X\to \mathrm{\Omega }$ to the class of the pullback of $\mathrm{true}$ along $f$. Conversely, define ${\mathrm{\Psi }}_X:{\mathrm{Sub}}_M(X)\to \mathcal{C}(X,\mathrm{\Omega })$ by sending the class of $m$ to its unique classifying map ${\chi }_m$ from [L1].

The composites are identities. If $f:X\to \mathrm{\Omega }$, then ${\mathrm{\Psi }}_X({\mathrm{\Phi }}_X(f))=f$ because the pullback subobject of $\mathrm{true}$ along $f$ is classified by $f$ itself, and that classifying map is unique by [L1]. If $[m]\in {\mathrm{Sub}}_M(X)$, then the pullback of $\mathrm{true}$ along ${\chi }_m$ represents the same subobject class as $m$, so ${\mathrm{\Phi }}_X({\mathrm{\Psi }}_X([m]))=[m]$. Naturality follows because pullback commutes with composition of classifying maps.

Therefore $\mathrm{\Phi }$ is a natural isomorphism $\mathcal{C}(-,\mathrm{\Omega })\cong {\mathrm{Sub}}_M$. By [L4], $\mathrm{\Omega }$ represents the supplied representative-set form of the subobject functor.