Subobjects and quotient objects form oppositely oriented partially ordered collections

If $m^{\mathrm{\prime }}=m\circ a$ and $m=m^{\mathrm{\prime }}\circ a^{\mathrm{\prime }}$ represent the same subobject, and similarly $n^{\mathrm{\prime }}$ and $n$, then any factorisation $m=n\circ u$ transports by composition to a factorisation of $m^{\mathrm{\prime }}$ through $n^{\mathrm{\prime }}$, and conversely. Thus the relation is independent of representatives by [L1].

Identity factorisations give reflexivity and composites give transitivity. If $[m]\le [n]$ and $[n]\le [m]$, then the representatives mutually factor, so [L1] makes their classes equal. The subobject relation is therefore reflexive, transitive and antisymmetric — the three conditions [L2] names — and by step 1.1 each is a statement about the classes rather than about chosen representatives.

For quotient representatives the same argument is dual, but the order is reversed because $[q]\le [r]$ means that $r$ factors through $q$. Reflexivity, transitivity, and antisymmetry follow by the epic half of [L1].

Nothing in steps 1.1--3.1 quantifies over a collection whose members are subobjects: each is a statement about monomorphisms into $C$ and about the factorisation relation between them, which is what the bracket notation abbreviates. Restricting that relation to a set of monomorphisms into $C$ carrying one representative per class therefore gives a relation on a set satisfying the three conditions of [L2], hence a partial order there.