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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.

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  • $\begingroup$ One problem with this definition of a quotient map that I have is that not every quotient map is surjective: if $Y$ is given the discrete uniformity, then function $f:X\rightarrow Y$ is automatically a quotient map since every map $g:Y\rightarrow Z$ is automatically continuous. $\endgroup$
    – Joseph Van Name
    Commented May 11, 2015 at 19:22
  • $\begingroup$ @joseph-van-name Thank you very much for this remark. I am indeed interested in surjective maps only and I have modified the definition accordingly. $\endgroup$
    – J.-E. Pin
    Commented May 11, 2015 at 19:38
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    $\begingroup$ The mapping $\widehat{f}$ may not map $\widehat{X}$ onto $\widehat{Y}$. For example, take any complete space with a quotient that is not complete. So perhaps OP wanted to ask if the mapping $\widehat{f}$ taken as a mapping $\widehat{X}\to\widehat{f}(\widehat{X})$ is a quotient. $\endgroup$
    – user95282
    Commented Aug 19, 2022 at 18:14

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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.

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    $\begingroup$ Step 2 rang an alert bell in my head and I found the following example in Bourbaki. Take $X = \mathbb{R}$ equipped with the usual Euclidian metric and let $Y = \mathbb{R}$ equipped with the uniform structure inherited from the compact space $\overline{\mathbb{R}}$. Then the identity map $X \to Y$ is uniformly continuous and surjective, but its continuous extension from $\hat X$ to $\hat Y$ is not surjective. $\endgroup$
    – J.-E. Pin
    Commented May 14, 2015 at 11:06
  • $\begingroup$ Oh I see.. Should I better delete my answer? $\endgroup$
    – Dominic van der Zypen
    Commented May 15, 2015 at 6:34

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