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).

  • $\begingroup$ 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? $\endgroup$ Aug 30, 2016 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, 2016 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. $\endgroup$ Aug 31, 2016 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, 2016 at 15:22
  • $\begingroup$ And what is wrong with $q(a),q(b)\in Y$? $\endgroup$ Aug 31, 2016 at 20:26

1 Answer 1


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 $\beta(A)$.

  • $\begingroup$ 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, 2016 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.