In the middle of the first paragraph of page 295 of the Elephant, Johnstone writes that normalized transition morphisms satisfying the cocycle condition are necessarily isomorphisms.
The definitions are given in the previous two pages. I don't see how to prove his assertion. The normalization and cocycle conditions tell us the pullbacks to "triple intersections" of the transition morphisms are isomorphisms. How does this imply the transition morphisms are themselves isomorphisms?
Added. Perhaps I should clarify that I understand the concrete proof in the case of descent of trivializing a single bundle. That is, if we're descending a bundle $A\overset{\alpha}{\to} B$ along $B\to \bf 1$ then the monad encoding descent is "taking product with $B$", and the algebra structure is given by a morphism $A\times B\overset{g}{\longrightarrow}A$ satisfying $\alpha\circ g=\pi_2$, satisfying moreover the unit axiom $g(a,\alpha(a))=a$ and the associativity (cocycle) condition $g(g(a,b),b^\prime)=g(a,b^\prime)$. The unit axiom ensures $g(-,b)$ is onto $\alpha^{-1}(b)$ and combined with the cocycle condition this ensures that $g(-,b),g(-,b^\prime)$ are mutually inverse bijections of the fibers $\alpha^{-1}(b),\alpha^{-1}(b^\prime)$. My problem is that I don't see how the assertion follows formally.