Proposition 17

Let $X$ be a uniform space, $i(X)$ its associated Hausdorff space, and let $f$ be a mapping of $X$ onto a Hausdorff uniform space $X'$, such that the uniformity of $X$ is the inverse image under $f$ of the uniformity of $X'$. Then the mapping $g : i(X) \to X'$ such that $f = g \circ i$ is an isomorphism.

By Proposition 16, $g$ is uniformly continuous; also $g$ is obviously surjective, and is also injective because the relation $f(x) = f(y)$ implies by definition that $(x, y)$ belongs to all the entourages of $X$, and therefore that $i(x) = i(y)$ (no. 7, Proposition 12). Finally, the entourages of $X'$ are the images under $f \times f$ of the entourages of $X$ ($§ 2$, no. 4, Remark), hence they are also the images under $g \times g$ of the entourages of $i(X)$ (no. 7, Proposition 12); hence the result.