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Schauder's lemma asserts that you can always extend a uniformly continuous, uniformly open map from a dense subset of a complete metric space to a uniformly open map on the completion.

I think the converse should be false: is there an example of a uniformly continuous uniformly open map from a complete metric space $f\colon X\to Y$ whose restriction $f|_D\colon D\to f(D)$ to some dense set $D\subseteq X$ fails to be uniformly open?

Edit in regard to ${\rm id}_{\mathbb R}|_{\mathbb Q}$. Suppose that $f\colon X\to Y$ is an injective uniformly open map. Then for each $\varepsilon > 0$ there is $\delta > 0$ such that for all $x$ we have $f(B(x,\varepsilon))\supseteq B(f(x), \delta)$. Let $D$ be a dense subset of $X$. Then for $x\in D$ we have $$f(B_D(x,\varepsilon)) = f(B(x,\varepsilon)\cap D) = f(B(x,\varepsilon))\cap f(D)\supseteq B(f(x), \delta) \cap f(D) = B_{f(D)}(f(x), \delta).$$ Here we used injectivity to make the image and intersection commute.

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    $\begingroup$ What about the identity map of the real line and its restriction to the set of all rational numbers? $\endgroup$ Commented Dec 29, 2022 at 19:01
  • $\begingroup$ @IosifPinelis, looks like for injective maps this is not a problem. Or am I missing something? $\endgroup$ Commented Dec 29, 2022 at 23:00
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    $\begingroup$ For me, the Schauder lemma is the following statement for a continuous map $f:X\to Y$ from a complete metric space $X$ to a metric space $Y$: If $f$ is uniformly almost open in the sense that, for every $\varepsilon>0$, there is $\delta>0$ such that $B_Y(f(x),\delta) \subseteq \overline{f(B_X(x,\varepsilon))}$ for all $x\in X$, then $f$ is uniformly open, i.e., the same condition holds without the closure. $\endgroup$ Commented Dec 30, 2022 at 7:41
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    $\begingroup$ @IosifPinelis, I think that misunderstanding emerged from the fact you wanted to keep the codomain unchanged, which was not my intention (as otherwise this is indeed trivially false). $\endgroup$ Commented Dec 30, 2022 at 9:06
  • $\begingroup$ @user243245 : According to the standard definition (en.wikipedia.org/wiki/…) of the restriction of a function to a subset of the domain, the codomain is kept -- and why not, given that in general the codomain does not have to coincide with the image of a function? Anyhow, I see that now you have specified your notion of a restriction (this should have been done right away when you posted the question). $\endgroup$ Commented Dec 30, 2022 at 13:31

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