In the category of uniform spaces, is the completion of a quotient map also a quotient map? I asked this question about 2 months ago on math.stackexchange, but so far I received neither comments nor answers.
Let $X$ and $Y$ be two Hausdorff uniform spaces. A surjective uniformly continuous map $f: X \to Y$ is a quotient map if for every map $g$ from $Y$ to a uniform space $Z$ such that $g \circ f$ is uniformly continuous, $g$ is uniformly continuous.
Let $\hat X$ and $\hat Y$ denote the completions of $X$ and $Y$ respectively, and let
$\hat f : \hat X \to \hat Y$ be the unique uniformly continuous extension of $f$.
Is $\hat f$ a quotient map?
If the answer is positive, a reference would be appreciated.
 A: I think the answer to the question is yes, but there is a little gap (called "Gap-Statement" below) that needs to be filled - and of course there is still the possibility that the "Gap-Statement" is false.
Let's split the answer in two steps.
Step 1. Let $f:X\to Y$ be uniformly continuous, let  $\hat f: \hat X \to \hat Y$ be the uniformly continuous extension, and take a uniform space $Z$ and a non-continuous map $g:\hat Y \to Z$. Then we show that $g\circ \hat f: \hat X \to Z$ is not continuous.
I'm not entirely sure the following statement is correct - but if it is, it answers the question positively:
"Gap-Statement". Let $X$ is a Hausdorff uniform space and let $i:X\to \hat X$ be the canonical map from $X$ to its extension. Let $Z$ be a uniform space and $g:\hat X \to Z$ be a map. Then $g$ is uniformly continuous if and only if $g\circ i: X\to Z$ is uniformly continuous.
(One direction is trivial.)
So let $f:X\to Y$ be uniformly continuous, let  $\hat f: \hat X \to \hat Y$ be the uniformly continuous extension, and take a uniform space $Z$ and a non-continuous map $g:\hat Y \to Z$. Let $i_X:X\to \hat X$ and $i_Y:Y\to \hat Y$ be the canonical maps from $X,Y$ to their extension. Using the "Gap-Statement" we get that $g\circ i_Y: Y\to Z$ is non-continuous. Since $f:X\to Y$ is a quotient map, the map $g\circ i_Y \circ f: X\to Z$ is non-continuous. Note that 
$$g\circ i_Y \circ f= g\circ \hat f \circ i_X.$$
So if $g\circ \hat f$ were continuous, then so would be $g\circ \hat f \circ i_X = g\circ i_Y \circ f$. But we just proved that this map is not continuous, so $g\circ \hat f: \hat X \to Z$ is not continuous, as desired.
Step 2. We have to show that if $f:X\to Y$ is surjective, then so is the continuous extension $\hat f: \hat X \to \hat Y$.
The proof for this is as follows. Note that the elements of $\hat Y$ are minimal Cauchy filters on $Y$, so let ${\cal G} \in \hat Y$ be a minimal Cauchy filter on $Y$. Then $$\{f^{-1}(V): V\in {\cal G}\}$$ is a filter base for a Cauchy filter on $X$ which contains a minimal Cauchy filter ${\cal F}$. It is not hard to show that $\hat f({\cal F}) = {\cal G}$, so $\hat f$ is surjective.
