Let $\beta X$ be the Stone-Čech compactification and $\upsilon X$ be the Hewitt real compactification of a completely regular space $X$.

It is well known that any continuous surjective map $f:X\rightarrow Y$ induces surjective map $\beta \left( f\right) :\beta X\rightarrow \beta Y$. Similarly, Is the induced map $\upsilon \left( f\right) :\upsilon X\rightarrow \upsilon Y$ surjective?


1 Answer 1


In general, no. It is known that the continuous image of a realcompact space is not necessarily realcompact (I don't know a specific example offhand, but this appears as Problem 432 in Tkachuk's C_p-theory problem book).

So let $f : X \to Y$ be a continuous surjection, where $X$ is realcompact and $Y$ is not. Since $X$ is realcompact, $\upsilon(X) = X$, so $\operatorname{im}(\upsilon(f)) = \operatorname{im}(f) = Y$. But since $Y$ is not realcompact, $\upsilon(Y) \supsetneq Y$, and so $\upsilon(f)$ is not surjective.

  • $\begingroup$ Actually, what I want to know whether the orbit map $\pi :X\rightarrow X/G$ induces surjective map between their real compactification spaces. The orbit space of a realcompact space is also realcompact. (R. Engelking, 3.11G.). Also, the orbit map is perfect open map. $\endgroup$ Aug 9, 2019 at 12:26
  • $\begingroup$ I see - sorry, I don't have any particular insight into that case. Hopefully someone else will be able to answer. $\endgroup$
    – user103663
    Aug 9, 2019 at 17:11

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