Proposition 16

Let $X$ be a uniform space and $i$ the canonical mapping of $X$ into its Hausdorff completion $\hat{X}$. For each uniformly continuous mapping $f$ of $X$ into a Hausdorff uniform space $Y$, there is a unique uniformly continuous mapping $h : i(X) \to Y$ such that $f = h \circ i$.

We may identify $Y$ with a subspace of its completion $\hat{Y}$ (no. 7, Corollary to Proposition 12), and $f$ can then be considered as a uniformly continuous mapping of $X$ into $\hat{Y}$. By virtue of Theorem 3, $f$ is then of the form $f = g \circ i$, where $g$ is a uniformly continuous mapping of $\hat{X}$ into $\hat{Y}$. If $h$ is the restriction of $g$ to $i(X)$, then clearly $f = h \circ i$, and $h$ maps $i(X)$ into $Y$. The uniqueness of $h$ is trivial.

The pair $(i, i(X))$ is therefore the solution of a universal mapping problem (Set Theory, Chapter IV, § 3, no. 1), where this time we take the $\Sigma$-sets to be Hausdorff uniform spaces, and the $\sigma$-morphisms (resp. $\alpha$-mappings) to be uniformly continuous mappings (resp. uniformly continuous mappings of $X$ into a Hausdorff uniform space).