Proposition 5.3.13. Let $X, Y$ be uniform spaces and $f: X \to Y$, then the following are equivalent:
$f \in UC(X; Y)$.
For every uniformly continuous pseudometric $d_{Y}$ on $Y$, there exists a uniformly continuous pseudometric $d_{X}$ on $X$ such that for all $x, y \in X$, $d_{Y}(f(x), f(y)) \le d_{X}(x, y)$.
Proof. (1) $\Rightarrow$ (2): $d_{X}(x, y) = d_{Y}(f(x), f(y))$ is a uniformly continuous pseudometric.
(2) $\Rightarrow$ (1): Let $U$ be an entourage of $Y$. By Theorem 5.3.7, there exists a uniformly continuous pseudometric $d_{Y}$ on $Y$ and $r > 0$ such that $E(d_{Y}, r) \subset U$. By assumption, there exists a uniformly continuous pseudometric $d_{X}$ on $X$ such that $d_{Y}(f(x), f(y)) \le d_{X}(x, y)$. In which case, $(f \times f)(E(d_{X}, r)) \subset U$, and the pre-image of $U$ by $(f \times f)$ is an entourage in $X$ by Definition 5.3.3, so $f \in UC(X; Y)$.$\square$