Suppose $X$ is a Hausdorff space (no other assumptions on it) and denote by $\eta:X\to \beta X$ the canonical map to its Stone–Čech compactification (compactification). Consider a continuous function $f:A\times X\to B$ where both $A$ and $B$ are compact Hausdorff. Is it true that there exist an extension $\beta (f): A\times \beta X\to B$ (even not unique), such that $f=\beta(f)\circ (id_A\times \eta)$ ?

I know that if $A$ is the one point space the answer is yes, as it reduces to the standard universal property of the Stone–Čech compactification. I wonder if this can be generalized for any compact Hausdorff space. Does $\beta$ commute with finite products in this setting? It would be sufficient.


Here are some quotations from “General topology” by Ryszard Engelking (Heldermann Verlag, Berlin, 1989).

I recall that Engelking calls a Tychonoff space $X$ pseudocompact provided each continuous real-valued function $f$ on $X$ is bounded. If I remember it right, by $D$ he denotes a discrete space $\{0,1\}$ and by $D(\frak c)$ a discrete space of cardinality continuum.

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  • $\begingroup$ Dear Alex, many thanks for the quotations. I understand that i can not use the fact that $\beta$ commutes with products to prove the extension above. Nevertheless it might $\endgroup$ – Tolbiac Oct 12 '16 at 7:44
  • $\begingroup$ .... still be provable (sorry I send an incomplete comment accidentally :-)) $\endgroup$ – Tolbiac Oct 12 '16 at 7:46
  • $\begingroup$ @Tolbiac I don’t quite understand what means ‘it might still be provable’. If for any Tychonoff space $X$ and any compact Hausdorff spaces $A$ and $B$ we can extend any continuous function $f:A\times X\to B$ to a continuous function $\beta(f):A\times \beta X\to B$, this means that $A\times \beta X= \beta A\times \beta X $ is equivalent to $\beta (A\times X)$, and the last means that $\beta$ commutes with a product of two spaces, one of which is compact, which not true in general (by Ex. 3.6.D), but is true in some partial cases. $\endgroup$ – Alex Ravsky Oct 12 '16 at 8:06
  • $\begingroup$ True, I was stupid. Many thanks once again. $\endgroup$ – Tolbiac Oct 12 '16 at 8:25

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