Recovering the Subobject Poset from the Subobject Classifier In topos theory, one way of phrasing the role of the subobject classifier $\Omega$ is as the object representing the subobject functor.
The nlab (and all the sources I've found via googling) refer to $\text{Sub}(-)$ as a functor from $\mathcal{E}$ to $\mathsf{Set}$. However we know that $\text{Sub}(X)$ is always a poset (indeed in a topos it is always a heyting algebra). Since $\Omega$ itself is also a heyting algebra, it feels like we should be able to recover the poset structure of $\text{Sub}(X)$ by looking only at the corresponding characteristic functions $\chi_A : X \to \Omega$.
In $\mathsf{Set}$ it is intuitively clear how to do this: the poset structure on subobjects is recovered exactly by the pointwise ordering on characteristic functions $\chi_A \leq \chi_B$. This feels like it should work in an arbitrary topos, but I am struggling to find a natural categorical phrasing.
Thanks in advance ^_^
 A: This is a little indirect but it works. By the Yoneda lemma every natural operation on subobjects corresponds to an operation on $\Omega$. In particular the intersection and union (product and coproduct) of subobjects correspond to two operations $\cap, \cup : \Omega \times \Omega \to \Omega$ making $\Omega$ an internal lattice. Either of these operations can be used to define the ordering: we have $a \le b$ iff $a \cup b = b$ iff $a \cap b = a$.
A: We can define a partial order $\le$ on $\operatorname{Hom}(U, \Omega)$ by saying $f \le g$ if and only if $(f \rightarrow g) = \operatorname{true} \circ ()$ where $\operatorname{true} : 1 \to \Omega$ comes from the definition of subobject classifier and $() : U \to 1$ is the unique morphism from the universal property of $1$ as the final object; and ${\rightarrow} : \Omega \times \Omega \to \Omega$ is the "relative complement" operator from the structure of $\Omega$ as a Heyting algebra object of the topos.
Then, for subobjects $A, B \in \operatorname{Sub}(U)$, we have that $A \subseteq B$ if and only if $\chi_A \le \chi_B$.
