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Let $B$ be a topological space. Call a subset $A\subset B$ ultrafilter-like iff $A$ is dense in $B$ and each decomposition $A=A_1\cup A_2$, into the union of two open subsets extends to a decomposition of $B=B_1\cup B_2$, $A_1\subset B_1$ and $A_2\subset B_2$ into the union of two open subsets $B_1$ and $B_2$.

Is the following true? Let $f:X\rightarrow Y$ be a proper map and $A\subset B$ is ultrafilter-like. For any $g:A\rightarrow X$ and $h:B\rightarrow Y$ such that $f(g(a))=h(a)$ for each $a\in A$, there is a map $g':B\rightarrow X$ extending $g$ such that $f(g'(b))=h(b)$ for each $b\in B$.

Is this true if $Y$ is a point (and thus $X$ is an arbitrary quasi-compact space).

This is true provided $B=A\cup \{\omega\}$ where $\omega$ is closed. If $A$ is discrete, then the neighbourhoods of $\omega$ define an ultrafilter on $A$ and the property becomes the definition of a proper map via ultrafilters (see Bourbaki, General Topology, I\S10.2, Theorem I). If $A$ is not necessarily discrete, then there is an ultrafilter $F$ on $A$ such that the neighbourhoods of $\omega$ are $F$-big, and the same arguments works. (To fix the terminology: by a proper map I mean a closed map such that the preimage of any point is quasi-compact).

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  • $\begingroup$ How about the following (possible) counterexample: Let $a,b$ be two distinct free ultrafilters in $X=\beta\omega$, $Y=X/\{a,b\}$ be the quotient space and $q:X\to Y$ be the quotient map. Let $B=Y$ and $A=\beta B\setminus \{q(a),q(b)\}$ and $f:A\to X\setminus\{a,b\}$, $g:B\to Y$ be the identity maps. It seems that no map $h$ with the required properties exists. Is it Ok? $\endgroup$ – Taras Banakh Aug 30 '16 at 13:29
  • $\begingroup$ I do not understand your notation: which is the ultralifter-like map and which is the proper map ? $\beta\omega\rightarrow\beta\omega/\{a,b\}$ is the proper map and $\beta B\setminus\{q(a),q(b)\} \rightarrow \beta\omega/\{a,b\}$ is the ultrafilter-like map ? but the latter is not a subset, is it? $\endgroup$ – user97621 Aug 31 '16 at 6:50
  • $\begingroup$ $\beta\omega\to\beta/\{a,b\}$ is the proper map. $\beta B\setminus\{q(a),q(b)\}$ should be $\beta\omega\setminus\{q(a),q(b)\}$. Sorry for this misprint. $\endgroup$ – Taras Banakh Aug 31 '16 at 15:08
  • $\begingroup$ so is it $\beta\omega\rightarrow \beta\omega / \{a,b\}$ and $\beta\omega\setminus \{q(a),q(b)\} \rightarrow \beta\omega/ \{a,b\}$ ? (there is another misprint in the first map, nothing follows $\beta$) I do not see it, as $q(a),q(b)\in Y$ are elements of the quotient.. $\endgroup$ – user97621 Aug 31 '16 at 15:22
  • $\begingroup$ And what is wrong with $q(a),q(b)\in Y$? $\endgroup$ – Taras Banakh Aug 31 '16 at 20:26
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Let $C$ be a connected Tychonoff space and $a,b\in \beta C\setminus C$ be two distinct points. Let $X=\beta C$, $Y=X/\{a,b\}$ be the quotient space and $f:X\to Y$ be the quotient map. It is clear that $f$ is perfect and hence proper.

Let $B=Y$, $A=B\setminus \{q(a)\}$, $h:B\to Y$ be the identity map, $g=q^{-1}|A:A\to X\setminus\{a,b\}\subset X$ be the homeomorphism of $A=Y\setminus\{q(a)\}$ onto $X\setminus\{a,b\}$. It is easy to see that $f\circ g=h|A$. On the other hand, it can be shown that $g$ admits no continuous extension to a map $g':B\to X$ such that $f\circ g'=h$.

The space $g(A)$ is connected sinse it contains a dense connected subspace $C$. Then the space $A$ is connected as well (being homeomorphic to $g(A)$). The connectedness of $A$ implies that $A$ is ultrafilter-like in $B$ (being connected the space $A$ admits no non-trivial partitions into two open sets).

So, to get a sensible answer, we should assume that the space $A$ is disconnected, or better (strongly) zero-dimensional. In this case there is a hope for the positove answer since ultrafilter-likenes of $A$ in $B$ should imply that the identity embedding $A\to\beta A$ extends to an embedding of $B$ into $\beta(A)$.

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  • $\begingroup$ Thank you! Yes, this appears an interesting counterexample, though not exactly to what I was asking (there was no requirement that $A=A_1\cup A_2$ where $A_1$ and $A_2$ are disjoint). $\endgroup$ – user97621 Sep 27 '16 at 11:54

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