Probably $\beta \mathbb N$ is not an absolute retract (is there an easy argument for this?), but I'd be interested to know what happens in the class of extremally disconnected (compact) spaces. Is it an absolute retract therein?
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asked Jan 17, 2022 at 16:09
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1$\begingroup$ Could you define "absolute retract"? I guess it means "is a retract of any space in which it is embedded homeomorphically" but what kind of spaces are allowed? In any case, if $A$ is a free Boolean algebra with a surjective homomorphism onto $2^\omega$, this induces an embedding $\beta\omega\to X$ which is not split (i.e. $2^\omega$ is not a retract in $X$). So $2^\omega$ is not an absolute retract within Stone spaces. $\endgroup$– YCorCommented Jan 17, 2022 at 16:43
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$\begingroup$ $\beta \mathbb{N}$ is the spectrum of the abelian $W^*$-algebra $\ell^\infty(\mathbb{N})$, so it is a hyperstonean space, in particular, itself is already stonean, i.e. extremally disconnected and compact. See, e.g. section III.1 of Takesaki's book on operator algebras. $\endgroup$– Hua WangCommented Jan 17, 2022 at 17:08
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1 Answer
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$\beta\mathbb{N}$ is not a retract of a Tychonoff cube because it is not connected; it also not a retract of a Cantor cube, not even a continuous image, see problem 3.12.12 in Engelking's book.
It is an absolute retract for ED spaces: if it is embedded in the compact ED space $X$ then $\mathbb{N}$ is relatively discrete subspace of $X$ and we have pairwise disjoint open sets $O_n$ in $X$ with $O_n\cap\beta\mathbb{N}=\{n\}$. Then the closure $K$ of the union $O=\bigcup_nO_n$ is clopen and by extremal disconnectedness $K=\beta O$. Now extend the retraction $r$ from $O$ onto $\mathbb{N}$, defined by $r(x)=n$ if $x\in O_N$, to $\beta r:K\to\beta\mathbb{N}$. As $r$ is the identity on $\mathbb{N}$ the extension $\beta r$ is the identity on $\beta\mathbb{N}$.
