# Mapping space from a quotient space

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 $q(Y)=X/{\sim}$.

• 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. – Taras Banakh Jul 9 '17 at 7:58
• 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]. – Taras Banakh Jul 9 '17 at 8:21

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 $$Y=\mathbb R$$.

• Typo? Do you mean $\dots\subset [0;1]\times\mathcal S\$ ? – Wlod AA Jul 9 '17 at 15:09
• Isn't your $X$ discrete? – Wlod AA Jul 9 '17 at 15:12
• Would you go into more details, please? – Wlod AA Jul 9 '17 at 15:16
• @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 $X$. – Taras Banakh Jul 9 '17 at 15:51
• Taras, your example is very nice--purely conceptual, very clean. – Wlod AA Jul 9 '17 at 16:29

@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.

• 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). – Taras Banakh Jul 9 '17 at 15:58
• My example has its modest advantage, it is metric separable (i.e. a kind of small). – Wlod AA Jul 9 '17 at 16:34
• and the map is open, not just quotient! – Taras Banakh Jul 9 '17 at 16:38
• 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? – Victor Jul 11 '17 at 18:50
• I have fixed my $X$ already. Give me a couple minutes extra for LaTeX. – Wlod AA Jul 14 '17 at 3:16