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That is, is it true that if $X$ and $Y$ are topological spaces (assume they are Tychonoff, if necessary), with $X \subseteq Y$, then $\beta X$ is homeomorphic to a subspace of $\beta Y$? If so, how does one prove this? If not, what would be a counter-example?

My lecturer has asked us to come up with questions for the exam in 4 days. I've been trying to prove this, but juggling all the various spaces and topologies is making my head hurt, and I'm hitting so many dead ends that I'm beginning to think it's not true.

Thanks for any help!

Stefan Hamcke
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Theo Bendit
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  • The inclusion $X\hookrightarrow Y\hookrightarrow\beta Y$ induces a map $\beta X\to\beta Y$. Are you asking if this induced map might be not injective, or if there is no injection $\beta X\to\beta Y$ at all? – Stefan Hamcke Jun 24 '13 at 16:37

2 Answers2

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If $X=\mathbb N$ and $Y=\mathbb N^*$, the one-point compactification of $\mathbb N$, then $\beta Y\approx \mathbb N^*$, which is countable. However, $\beta\mathbb N$ is not countable, hence there cannot be an injection into $\beta\mathbb N^*$

Stefan Hamcke
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  • Thank you very much Stephan, that's really helpful. I was actually just thinking about $\mathbb{N}$ in relation to $\mathbb{R}$. Would I be right in thinking that $\mathbb{R}$ is a circle, hence metrizable, so therefore incapable of being a superspace of $\beta \mathbb{N}$? – Theo Bendit Jun 24 '13 at 17:17
  • Do you mean $\mathbb R^*$? That is a circle, right. You can also compare the local bases: The circle is first-countable, but $\beta\mathbb N$ is not. – Stefan Hamcke Jun 24 '13 at 17:25
  • I mean $\beta \mathbb{R}$ is a circle. Thanks again! Actually, while I have your attention, are there any reasonable conditions that you know of that we can place on $X$ so that we get this inclusion? Perhaps if $X$ is dense in $Y$? – Theo Bendit Jun 24 '13 at 17:27
  • To be honest, I have no idea what $\beta\mathbb R$ looks like. But if you are sure that it is homeomorphic to a circle, then you can use this reasoning. – Stefan Hamcke Jun 24 '13 at 17:28
  • I'm not sure in hindsight. My logic was faulty. – Theo Bendit Jun 24 '13 at 17:31
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    @TheoBendit: Seems like you can use the universal property of the $\beta$-functor to show that $\beta\mathbb R$ can not be homeomorphic to a circle. Note that $\mathbb R$ is homeomorphic to $(0,1)$ and the embedding of $(0,1)$ into $[0,1]$ can not be extended to a map $S^1\to[0,1]$. – Stefan Hamcke Jun 24 '13 at 17:35
  • A sufficient condition for the inclusion of a subspace $X$ of a space $Y$ to extend to an inclusion $\beta X \to \beta Y$ is that $X$ be $C^\ast$-embedded in $Y$: every bounded continuous function on $X$ extends to a bounded continuous function on $Y$. In that case $\beta X$ can be identified with the closure of $X$ in $\beta Y$. // Regarding the discussion of $\beta\mathbb{R}$: it is as big as $\beta \mathbb{N}$, so has cardinality $2^{\frak{c}}$. However it is quite different, for example it is connected. – Martin Jun 24 '13 at 17:49
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    @Theo: $\beta\Bbb R$ is definitely not a circle: its cardinality is $2^\mathfrak{c}$, while that of a circle is only $\mathfrak{c}$. See this answer for details. – Brian M. Scott Jun 24 '13 at 19:17
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A natural follow-up question here is "what does the Stone-Čech compactification respect?" A general answer is that $\beta$ is left adjoint to the inclusion functor from compact Hausdorff spaces to topological spaces, and as a left adjoint it consequently preserves all colimits. In particular, it preserves disjoint unions and coequalizers (hence epimorphisms). But subspaces are related to limits rather than colimits (since they're related to monomorphisms rather than epimorphisms) so one shouldn't expect a left adjoint to preserve them.

Qiaochu Yuan
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