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 $f_{\ast }\mathcal{M}:=\{B\subseteq Y:f^{-1}[B]\in \mathcal{M}\}$ 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 $W^{\circ 3}\subseteq V$, 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 $D^{\circ 3}\subseteq E$, then a symmetric target entourage $V$ whose pullback lies in $D$ and a symmetric $W$ with $W^{\circ 3}\subseteq V$. If $(z,z^{\mathrm{\prime }})\in W$, then $A_W(z)\times A_W(z^{\mathrm{\prime }})$ is $D$-small across the two filters; enlarging these sets by $D$ gives members of their associated minimal filters whose cross product lies in $D^{\circ 3}\subseteq E$. Hence $(S(z),S(z^{\mathrm{\prime }}))\in \widehat{E}$.

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.