The set of $C$-ideals is a frame freely generated by a site $(A, C)$ The proposition in Section II.2.11 in Johnstone's Stone Spaces states

For any site $(A, C)$ [the set of $C$-ideals] is a frame,
  and is freely generated by $(A, C)$.

Here a site is a pair of a $\land$-semilattice and a covering,
a function that assigns to each element of the semillatice a family of subsets of $\mathop{\downarrow} a$ that satisfies the condition called meet-stability.
For a frame $B$ to be freely generated by a site $(A, C)$ means that there is a semilattice homomorphism $f \colon A \to B$ that "transforms covers into joins" and is universal among such homomorphisms.
The proof is by exhibiting a nucleus $j$ on the free frame $DA$ (where the functor $D$ is left adjoint to the forgetful functor $\mathbf{Frm} \to \mathbf{SLat}$ and $DA$ is the set of downsets in $A$)
such that the sublocale $(DA)_j$ is the set of $C$-ideals.
To show that the composite $j \circ \mathord{\downarrow} \colon A \to (DA)_j$ is the universal map as directed above, it suffices to see that a given map $f\colon A \to B$ uniquely factors through $j : DA \to (DA)_j$.
The following is the part I do not understand in Johnstone's proof: to see the last point, first he takes the unique extension $\bar f : DA \to B$. So far so good.
Then, he sees that the right adjoint $g\colon B \to DA$ of $f$ has a range in $(DA)_j$ and factors through $(DA)_j$, which I take to mean that $g$ factors through the inclusion map $(DA)_j \to DA$.  Anyway, he concludes that $\bar f$ factors uniquely through $j: DA \to (DA)_j$.
What does he do in the last step? Since I am new to categorical techniques,
I believe I'm missing some fact in category theory.
 A: To show that $\bar{f}$ factors through $j$, you need to show that $j(a)=j(b)$ implies $\bar{f}(a)=\bar{f}(b)$.  Since $j(j(a))=j(a)$, it suffices to show that $\bar{f}(j(a))=\bar{f}(a)$.  Now since $g$ is the adjoint of $\bar{f}$, $\bar{f}(a)\leq b$ iff $a\leq g(b)$.  Since the image of $g$ is contained in $(DA)_j$, $j(g(b))=g(b)$ so $\bar{f}(a)\leq b$ iff $a\leq j(g(b))$.  But $a\leq j(g(b))$ iff $j(a)\leq j(g(b))$ since $j(a)$ is the least element of $(DA)_j$ which is above $a$.  This says $\bar{f}(a)\leq b$ iff $\bar{f}(j(a))\leq b$, so $\bar{f}(a)=\bar{f}(j(a))$ since $b$ is arbitrary.
(You also need to check that when you factor $\bar{f}$ through $j$, the resulting map is a frame-homomorphism.  However, this is automatic from the properties of $j$.  The factoring map is just the restriction of $\bar{f}$ to $(DA)_j$, and thus preserves finite meets since meets in $(DA)_j$ are the same as meets in $DA$.  To see that it preserves joins, just use the fact that joins in $(DA)_j$ are computed by taking joins in $DA$ and then applying $j$, and we know applying $j$ won't change what $\bar{f}$ does.)
Here's what's going on at a higher level.  We know that $g$ factors as a composition $B\stackrel{h}\to (DA)_j\stackrel{i}\to DA$, where $i$ is the inclusion (which is the right adjoint of $j$).  Since $g$ and $i$ preserve meets (being right adjoints) and $i$ is injective, $h$ must preserve meets as well.  Thus $h$ has a left adjoint $h^*:(DA)_j\to B$.  Since $ih$ is the right adjoint of $\bar{f}$ and $i$ and $h$ are right adjoint to $j$ and $h^*$, we have $\bar{f}=h^*j$.  That is, $\bar{f}$ factors through $j$.
