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For $X/{\sim}$ a quotient space, $$ Map(X/{\sim},Y)\subset Map(X,Y). $$ But is this inclusion always a homeomorphism on its image? (Assuming compact-open topology on the mapping spaces.) If not what would be the most general setting to make it true? We can also assume that $X$ and $Y$ are compactly generated.

A related question: if $q\colon X\to X/{\sim}$ is a quotient map and $X/{\sim}$ is compact, does always exist a compact $Y\subset X$ such that $q(Y)=X/{\sim}$.

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    $\begingroup$ The second related question has negative answer: just consider the subspace $X=\{(x,y)\in [0,1]\times\mathbb R:x=0$ or $xy=1\}$ of the plane and let $q:X\to [0,1]$ we the projection onto the first coordinate. This map is quotient, but not compact-covering. $\endgroup$ – Taras Banakh Jul 9 '17 at 7:58
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    $\begingroup$ My previous "counterexample" is incorrect. The converse in true. Namely, the second related question has affirmative answer if $X$ is sequential and $X/\sim$ is a convergent sequence, see Lemma 3.5 in [S.Lin, P.Yan, Sequence-covering maps of metric spaces, Topology Appl. 109 (2001) 301-314]. $\endgroup$ – Taras Banakh Jul 9 '17 at 8:21
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The second question has negative answer: just consider the unit interval $[0,1]$ and let $\mathcal S$ be the family of all closed subsets with a unique non-isolated point in $[0,1]$. The family $\mathcal S$ is endowed with the discrete topology. Let $X=\{(x,S):x\in S\in\mathcal S\}\subset [0,1]\times\mathcal S$ be the topological sum of the family $\mathcal S$ and $q:X\to[0,1]$, $q:(x,S)\mapsto x$, be the natural projection. It is easy to see that the map $q$ is quotient but $q(K)\ne [0,1]$ for any compact subset $K\subset X$. So, the space $X$ is the topological sum of all convergent sequences in $[0,1]$. It is a locally compact locally countable space of density continuum.

It seems that the (metrizable locally compact locally countable) space $X$ and the equivalence relation $\sim=\{(x,y)\in X\times X:q(x)=q(y)\}$ yield also a counterexample to the first question for $Y=\mathbb R$.

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    $\begingroup$ Typo? Do you mean $\dots\subset [0;1]\times\mathcal S\ $ ? $\endgroup$ – Wlod AA Jul 9 '17 at 15:09
  • $\begingroup$ Isn't your $X$ discrete? $\endgroup$ – Wlod AA Jul 9 '17 at 15:12
  • $\begingroup$ Would you go into more details, please? $\endgroup$ – Wlod AA Jul 9 '17 at 15:16
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    $\begingroup$ @Wlod-AA My $X$ is the topological sum of all convergent sequences in $[0,1]$, so is not discrete. The map $q$ is quotient because for any non-closed subset $F\subset [0,1]$ there exists a convergent sequence $S\subset [0,1]$ such that $S\cap F$ is not closed in $S$ and then $q^{-1}(F)$ is not closed in $X$. $\endgroup$ – Taras Banakh Jul 9 '17 at 15:51
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    $\begingroup$ Taras, your example is very nice--purely conceptual, very clean. $\endgroup$ – Wlod AA Jul 9 '17 at 16:29
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@Victor has pointed out to my error in the pre-edited version--thank you, Victor. Now everything IS under control and FIXED.

Here, after @TarasBanakh, there is another example of a quotient map $\ q: X\rightarrow X/{\sim} $ such that $X$ is compact but there is no compact subset $Y\subseteq X$ such that $f(Y)=f(X)$.

Let $Q\subset\mathbb R$ be the set of all rational numbers. Let $\ J:=[0;1]:=\{x\in\mathbb R: 0\le x\le 1\},\ $ Define

$$ X\ :=\ \{(x\ y)\in J^2\,:\, |\{x\ y\}\cap Q| = 1\} $$

And let $\ p:X\rightarrow J\ $ be the projection $\ p(x\ y)\ := x.\ $ Then $p$ is onto, and for every $A\subseteq J$ we have:

  1. $p^{-1}(A)$ is open in $X$ when $A$ is open in $J$ because $p$ is induced by the Cartesian projection;
  2. $p^{-1}(A)$ is not open in $X$ when $A$ is not open in $J$ because $p^{-1}(x)$ is dense in $\ \{x\}\times J\ $ for every $\ x\in J.\ $

Thus, $\ p\ $ is topologically equivalent to the respective quotient map.

More than this, $p$ is an open map. Indeed, sets

$$ B_{abcd}\ :=\ ((a;b)\times(c;d))\,\cap\, X$$

form a topological base of $X$, and $\ p(B_{abcd}) = (a;b)\cap J.\ $ Thus $p$ is an open map.

 

Let $\ Y\subseteq X\ $ be a compact subset such that $\ p(Y)=[0;1]\ $ (a proof by contradiction). Then $\ Y\ $ is a countable union of its compact subsets $\ C_a:=(\{a\}\times\mathbb R)\cap Y\ $ and $\ D_a:=(\mathbb R\times\{a\})\cap Y,\ $ where $\ a\ $ runs over rational numbers. Then sets $\ p(C_a)\ $ and $\ p(D_a)\ $ are compact and they cover $\ [0;1].\ $ Thus, by Baire's theorem one of these projections must have a non-empty interior in $\ [0;1]\ $ -- a contradiction. Thus such $\ Y\ $ does not exist.

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    $\begingroup$ By the way, open maps between completely metrizable spaces are compact-covering (this follows from the 0-dimensional Michael Selection Theorem). So, the space $X$ in the example of @Wlod-AA is not (and cannot be) complete (unlike to the topological sum of all convergent sequences which is completely metrizable). $\endgroup$ – Taras Banakh Jul 9 '17 at 15:58
  • $\begingroup$ My example has its modest advantage, it is metric separable (i.e. a kind of small). $\endgroup$ – Wlod AA Jul 9 '17 at 16:34
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    $\begingroup$ and the map is open, not just quotient! $\endgroup$ – Taras Banakh Jul 9 '17 at 16:38
  • $\begingroup$ What happens with the sets $\{(t,0):\, t\in J\}$ and $\{(0,t):\, t\in J\}$? Are they in $X$? If yes, their union is the required $Y$. Or, perhaps, you meant $\{x,y\}$ not $\{xy\}$ in your formula? $\endgroup$ – Victor Jul 11 '17 at 18:50
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    $\begingroup$ I have fixed my $X$ already. Give me a couple minutes extra for LaTeX. $\endgroup$ – Wlod AA Jul 14 '17 at 3:16

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