Under the ultrafilter lemma, compact Hausdorff spaces and ultrafilter algebras are recovered by the two limit constructions

Starting with an algebra $\xi$, [L4] makes its induced topology compact Hausdorff, so the limit construction of clause 1 applies to it, and [L2] says that the unique-limit map of that topology is exactly $\xi$. Thus the algebra is recovered on the nose, including on an empty or singleton carrier.

Starting with a compact Hausdorff topology $\tau$, every $\tau$-open set is open for the limit algebra because a convergent ultrafilter contains each neighbourhood of its limit. Conversely, if $O$ is not a $\tau$-neighbourhood of some $x\in O$, the neighbourhood filter at $x$ together with $X\setminus O$ has the finite-intersection property; UL/BPI extends it to an ultrafilter converging to $x$ but not containing $O$, contradicting induced openness. Hence the rebuilt topology is exactly $\tau$.

The forward morphism direction is [L3]: every continuous map preserves unique ultrafilter limits and is an algebra homomorphism.

Conversely, let $f:(X,{\xi }_X)\to (Y,{\xi }_Y)$ be an algebra homomorphism. If $O$ is induced-open in $Y$ and ${\xi }_X(\mathcal{U})\in f^{-1}[O]$, then ${\xi }_Y(\beta f(\mathcal{U}))=f({\xi }_X(\mathcal{U}))\in O$, so $O\in \beta f(\mathcal{U})$ and hence $f^{-1}[O]\in \mathcal{U}$. Thus every preimage of an induced-open set is induced-open, and $f$ is continuous.

Steps 1.1 and 1.2 recover both object structures, while steps 1.3 and 1.4 identify both morphism classes. The assignments therefore define inverse functors over $\mathbf{S}\mathbf{e}\mathbf{t}$.