Every uniformly continuous map into a complete Hausdorff uniform space extends uniquely across the Hausdorff completion; consequently completions are unique up to a unique uniform isomorphism

First use the canonical completion $X_c$. For a minimal Cauchy filter $\mathcal M\in X_c$, its image filter $f_*\mathcal M:=\{B\subseteq Y:f^{-1}[B]\in\mathcal M\}$ is Cauchy: for a target entourage $V$, uniform continuity supplies a source entourage $E$ whose $E$-related pairs have $V$-related images, and an $E$-small member of $\mathcal M$ has $V$-small image. Completeness gives a limit, which is unique by separatedness. Define $\widehat f_c(\mathcal M)$ to be that limit.

For $x\in X$, the image under $f$ of the minimal point filter $\eta_c(x)$ converges to $f(x)$: for a neighbourhood ball $V[f(x)]$, uniform continuity supplies a source ball at $x$ whose image lies in it. Therefore $\widehat f_c\eta_c=f$.

The map $\widehat f_c$ is uniformly continuous. Given a target entourage $V$, choose a symmetric $W$ with $W^{\circ3}\subseteq V$, and a source entourage $E$ whose $E$-related pairs have $W$-related images. If $\mathcal M\,\widehat E\,\mathcal N$, take witnesses $A\in\mathcal M$ and $B\in\mathcal N$ with $A\times B\subseteq E$. Since the image filters converge to $\widehat f_c(\mathcal M)$ and $\widehat f_c(\mathcal N)$, respectively, their members $f[A]$ and $f[B]$ meet the corresponding $W$-balls. Thus the two limits are related by $W\circ W\circ W\subseteq V$.

Any two uniformly continuous extensions across $\eta_c$ agree on the dense set $\eta_c[X]$, hence agree everywhere by [L4]. Thus the canonical completion has the asserted extension property.

Now let $\eta:X\to\widehat X$ be an arbitrary Hausdorff completion. Step 3.1 applied to $\eta$ gives a uniformly continuous $T:X_c\to\widehat X$ with $T\eta_c=\eta$. For $z\in\widehat X$, let $\mathcal F_z$ be the filter on $X$ generated by the sets $A_V(z):=\{x\in X:(\eta(x),z)\in V\},$ where $V$ ranges over symmetric entourages of $\widehat X$. Density makes these sets nonempty; intersections are refined by intersecting entourages. The pullback condition in [A1], together with a symmetric square root in $\widehat X$, shows that $\mathcal F_z$ is Cauchy. Define $S(z):=m(\mathcal F_z)\in X_c$.

The same pullback calculation gives $S\eta=\eta_c$. It also proves that $S$ is uniformly continuous: for a basic $\widehat E$ of $X_c$, choose a symmetric source entourage $D$ with $D^{\circ3}\subseteq E$, then a symmetric target entourage $V$ whose pullback lies in $D$ and a symmetric $W$ with $W^{\circ3}\subseteq V$. If $(z,z')\in W$, then $A_W(z)\times A_W(z')$ is $D$-small across the two filters; enlarging these sets by $D$ gives members of their associated minimal filters whose cross product lies in $D^{\circ3}\subseteq E$. Hence $(S(z),S(z'))\in\widehat E$.

The maps $ST:X_c\to X_c$ and $TS:\widehat X\to\widehat X$ agree with the respective identity maps on the dense images of $X$. By [L4] they are the identity maps. Thus $T$ and $S$ are inverse uniform isomorphisms, uniquely so because any competing map agrees with $T$ on the dense image.

For the original map $f:X\to Y$, the composite $\widehat f:=\widehat f_c\circ S:\widehat X\to Y$ is uniformly continuous and satisfies $\widehat f\eta=f$. Uniqueness follows from density and [L4]. Consequently every Hausdorff completion has the extension property, and step 6.1 proves uniqueness of completions up to the unique stated uniform isomorphism.