A uniformly continuous function between totally bounded uniform spaces

Question

Let $X$ and $Y$ is a uniform spaces. Let $f$ is a uniformly continuous surjective function $X\rightarrow Y$.

Conjecture: If $X$ is totally bounded then $Y$ is also totally bounded.

Answer 1

It seems that the conjecture is well known and can be proved straightforwardly. Let $f:(X,{\cal E})\to (Y,\cal F)$ be a surjective uniformly continuous map between uniform spaces and the space $(X,\cal E)$ is totally bounded. Let $F\in\cal F$ be an arbitrary entourage. Since the map $f$ is uniformly continuous, there exists an entourage $E\in\cal E$ such that $E\subset (f\times f)^{-1}(F)$. Since the space $(X,\cal E)$ is totally bounded, there exists a finite subset $A$ of $X$ such that $E[A]=X$. We claim that $F[f(A)]=Y$. Indeed, let $y\in Y$ be an arbitrary point. Since the map $f$ is surjective, there exists a point $x\in X$ such that $f(x)=y$. Since $E[A]=X$, there exists a point $a\in A$ such that $(a,x)\in E$. Since $E\subset (f\times f)^{-1}(F)$, we see that $(f(a),y)=(f(a),f(x))\in F$. Therefore $y\in F[f(A)]$.