A generator detects comparison of subobjects

If $B\le C$, then any map $G\to A$ factoring through $B$ also factors through $C$ by composition.

To prove the converse, assume $B\nleqq C$. By [L2], let $p:P\rightarrowtail B$ be the pullback subobject of $b$ and $c$. If $p$ were epic, then [L3] would make it an isomorphism, forcing $b$ to factor through $c$. So $p$ is not epic.

Let $q:B\to Q$ be a cokernel of $p$, which exists by [L5]. Since $p$ is not epic, [L4] implies $q\mathrm{\ne }0$.

The morphisms $q$ and $0_{B,Q}$ are therefore distinct, so [L1] gives some $u:G\to B$ with $qu\mathrm{\ne }0$. If $bu$ factored through $c$, the pullback property in [L2] would force $u$ to factor through $p$, hence $qu=0$, impossible. Thus $bu:G\to A$ factors through $B$ but not through $C$.

Step 1.1 proves the forward implication, and steps 1.2, 2.1, and 3.1 prove the contrapositive of the reverse implication. Therefore $B\le C$ exactly when every morphism $G\to A$ factoring through $B$ also factors through $C$.