psie

I don't understand the second sentence.

> The algebra of all continuous functions on the compact Hausdorff space $\beta X$, denoted $C(\beta X)$, separates points. This means that if two equivalence relations on $\beta X$ are respected by the exact same subset of $C(\beta X)$, then the equivalence relations themselves must be identical.

I will quote the whole passage I'm reading:

First I don't understand why this is true, and second I don't see the relevance of $C(\beta X)$ separating points.

Consider the Stone-Cech compactification of $X$, i.e. $\beta X$, and two other compactifications $Z,Z'$ of $X$. $\beta X$ is compact Hausdorff, so that $C(\beta X)$ separates points by Urysohn's lemma. Moreover, $Z,Z'$ are realized as quotient spaces of $\beta X$. Is it true that if the two equivalence relations defining $Z,Z'$ are respected by the exact same subset of $C(\beta X)$, then the equivalence relations must be identical?

Ok, thank you. $Z,Z'$ are realized as quotient spaces of $\beta X$, where $p,q\in\beta X$ equal iff $\beta e(p)=\beta e(q)$ (or $\beta e'(p)=\beta e'(q)$). So then naturally $\beta f(p)=\beta f(q)$ under this equivalence relation.

Could someone explain why that claim is true?

> Claim. A function $f \in C(X)$ belongs to $\mathcal{A}_Z$ if and only if its Stone-Cech extension $\beta f: \beta X \to \mathbb{R}$ is constant on the fibers of the map $\beta e$. That is: $$f \in \mathcal{A}_Z \iff \forall p,q \in \beta X, \left( \beta e(p) = \beta e(q) \implies \beta f(p) = \beta f(q) \right).$$

Consider a completely regular space $X$ and two compactifications of $X$, $(Z,e)$ and $(Z',e')$. Denote all continuous functions on $C(X)$ that can be extended from $X$ to $Z$ by $\mathcal{A}_Z=\{f\circ e:f\in C(Z)\}$. Extend $e,e'$ to $\beta X$ and denote these extensions by $\beta e$ and $\beta e'$.

@leslietownes Ah ok, makes sense. So if $f$ is complex conjugation, $g\in C(Y)$ and $h=e$, then indeed $$f\circ g\circ h(x)=\overline{g\circ e(x)}=\overline{g}(e(x)).$$

@leslietownes yes! :D looks away in some other direction

Another stupid question; is $\mathcal{A}_Z$ closed under complex conjugation because $f\in C(Y)$ implies $\overline{f}\in C(Y)$ (conjugation is continuous) and $\overline{f\circ e}=\overline{f}\circ e$? Usually you take the complex conjugate of both functions in a composition if they are complex-valued, but I'm not sure about $e$ here.

(I'm not sure if this approach uses the fact that $\mathcal{A}_Z=\mathcal{A}_{Z'}$ somehow.)

We have $e:X\to Z$ and $e':X\to Z'$. We are just given that $Z,Z'$ are compactifications. If there was some theorem that allowed me to uniquely and continuously extend $e$ to $Z'$ and $e'$ to $Z$ (we denote these extensions by $E$ and $E'$), then I think I'd be able to solve this problem, since $E\circ E'$ has the property that it is the identity $i_X$ on $X$. Since the identity map on $Z$ is also the extension of $i_X$, we'd be done.

@Jakobian I'm sort of working on the converse of this proposition, which I'm sure you're aware. After looking into some related texts, my idea is this now.

Ok. 👍

(If you consider complex algebras, closed under complex conjugation too. Forgot to mention this.)

Ok.

Hmm, ok. Developed theory sounds tricky, but perhaps I'll understand some of it.

Hmm, yes, I need to use the assumption $\mathcal{A}_Z=\mathcal{A}_{Z'}$ somehow.

Hmm, upon closer thought, $e'\circ e^{-1}$ is not onto.

So $\phi=e'\circ e^{-1}$.

@Jakobian If $X=Y$, then I'd say $\Phi$ is just the identity map. The diagram collapses to this I'd say.

Yes. 👍

Ok. Thanks. 👍

@Jakobian The diagram treats the case when $Z=\beta X$, I think.

Since we want $\phi\circ e=e'$.

I need to manipulate this equality I think: $f\circ e=g\circ e'$.

A hint would be great! :)

@Jakobian Ah, ok. 👍 So image of domain is closed in codomain, makes sense. Thanks!

@Jakobian first I thought $C(X,I)$, since we have $f\circ e=g\circ e'$, where $f\in C(Y), g\in C(Y')$, and $f\circ e=g\circ e'\in C(X,I)$.

yeah probably

Oh, ok.

In order to construct the homeomorphism $\phi:Z\to Z'$, I feel like there has to be a middle man, i.e. some set (in regards to the diagram above, this would be $\beta Y$) that connects $Z$ and $Z'$, though I'm not sure what this set would be.

@Thorgott when you say closed embedding, what does this mean? I have googled this term, but nothing elementary that caught my eye shows up. Like "This is what a closed embedding is: ..." My gut says that the hooked arrows indicate simply inclusion (by closed, I assume you mean $\beta X$ is closed). Also, if you have a hint to the above problem I posted today about equivalent compactifications, I'd be very grateful.

In regards to the diagram above, if $Y$ is a compactification of $X$, does it still make sense to speak of $\beta Y$? I.e. the Stone-Cech compactification of a compactification?

I quite struggle with these problems. :( Any help is appreciated.

I'm asked to show that if $(Z',e')$ is another Hausdorff compactification of $X$ such that $\mathcal{A}_{Z}=\mathcal{A}_{Z'}$, there is a homeomorphism $\phi:Z\to Z'$ such that $\phi\circ e=e'$. I'm not sure how to set things up, and what the relevance of $\mathcal{A}_{Z}=\mathcal{A}_{Z'}$ is. Somehow I'm tempted to say we are just swapping $\beta X$ for $Z$ and $Y$ for $Z'$. How does the definition of $\Phi$ change in that regard?

Consider again the diagram above, where $\Phi$ is defined via $\pi_g(\Phi(p))=\pi_{g\circ \phi}(p)$. Let $(Z,e)$ be a compactification of $X$, and define $\mathcal{A}_Z=\{f\circ e:f\in C(Z)\}$. This is a so-called completely regular subalgebra of $BC(X)$ (it is closed and contains the constant functions, as well as $\mathcal A_Z\cap C(X,I)$ separates points and closed sets).

If all functions in $f\in BC(X)$ are of the form $g\circ\phi$ for some $g\in C(Y)=BC(Y)$, in particular for $g\in \mathcal{G}$, then $p=q$ since they agree on all components $f\in \mathcal{F}=C(X,I)\subset BC(X)$.

@psie I think I understand now (thanks to those who helped!). We need to show $$\forall p,q\in I^{\mathcal{F}}, \Phi(p)=\Phi(q)\implies p=q.$$ But $\Phi(p)=\Phi(q)$ means all the components agree for all $g\in\mathcal{G}=C(X,I)$, i.e. $\pi_g(\Phi(p))=\pi_g(\Phi(q))$, or using the definition of $\Phi$, $\pi_{g\circ\phi}(p)=\pi_{g\circ\phi}(q)$ for all $g\in \mathcal{G}$.

Ok, thanks for the help though :) appreciate it

hmm, not really sure why you assumed $f(X)\subseteq [0,1]$, when $f\in BC(X)$

ok 👍

Ok.

So $C(Y)=BC(Y)$. Don't know if that helps.

We do assume here that $(Y,\phi)$ is a compactification of $X$, and therefore $\phi(X)$ dense in $Y$.

> Finally, if every $f\in BC(X)$ is of the form $g\circ\phi$ for some $g\in C(Y)$, then $\Phi$ is injective.