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Why is equivalence of compactifications defined the way it is?

For two compactifications $(Y_1,c_1)$ and $(Y_2,c_2)$ of some space $X$, why isn't the existence of a homeomorphism $f$ between them enough - what's the usefulness of $f(c_1(x))=c_2(x)$ for all $x$ in $X$ ?

Intuitively, I would guess that you want to keep the elements from $X$ in some sort structure, so that you can keep going between respective compactifications, without changing what happens with the elements in $X$ (more precisely, their imbeddings), but that's all very vague and not very convincing.

Explanation/motivation for defining equivalence of compactifications the way it is defined, and an example of it's use, both ideally simple, would be appreciated.

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The main reason for defining the equivalence this way is to give an order structure not to the points in $X$, but to the different compactifications of $X$. You should start by defining an order on the compactifications by saying that one compactification $(Y_1,c_1)$ is less than or equal to another compactification $(Y_2,c_2)$ if and only if there is a continuous surjection $\phi:Y_2\rightarrow Y_1$ such that $c_2\circ \phi=c_1$. In this case the equivalent compactifications are the ones for which you have a homeomorphism. The reason behind wanting an order to the compactifications is to be able to then use Zorn's Lemma to obtain a maximal element in the family of compactifications of a space $X$, which is also known as the Stone-Čech compactification $\beta X$.

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  • $\begingroup$ I think there are also other ways of constructing Stone-Cech compactification though - embedding of all continuous functions into $[0,1]$ for example, and it can be shown that every compactification is equivalent to a quotient space of it. Is the way that uses an order nicer in some way? Also, I though Stone-Cech compactification is the largest, not just maximal, could you explain or prove that? $\endgroup$ – Jake1234 Dec 28 '15 at 22:05
  • $\begingroup$ What do you mean by 'largest'? It is 'largest' in terms of cardinality, but it is also 'largest' in terms of there being a continuous surjection from $\beta X$ to any other compactification of $X$ such that the triangle commutes. Also, there are other ways of constructing $\beta X$, but the main point of defining equivalence the way you did is a consequence of the definition of order on the compactifications, which leads naturally to defining $\beta X$ as the maximal element. $\endgroup$ – Simon_Peterson Dec 28 '15 at 22:10
  • $\begingroup$ IMO, this is a more elegant proof of the existence of $\beta X$. From your notation, I assume you are following Engelking, in which case you should be looking at Corollary 3.5.10. For a (perhaps) more detailed treatment of this material, please see Chapter 6 of these lecture notes Engelking seems to put the cart before the horse when first definint the equivalence and then the order; the lecture notes do it the other way around, which seems more natural, imo. $\endgroup$ – Simon_Peterson Dec 28 '15 at 22:29
  • $\begingroup$ Thanks for the answer and lecture notes. I meant largest, with respect to the order defined, as far as equivalence classes go. I think this is exactly the Corollary 3.5.10 you've mentioned. Could you explain where it is that Zorn's lemma is used in the approach in the lecture notes/Engelking ? Lastly, where would this approach (defining the order this way) fail, if we removed the need for $c_2\circ \phi=c_1$ ? $\endgroup$ – Jake1234 Dec 28 '15 at 23:22
  • $\begingroup$ Zorn's Lemma states that if every chain in a partially ordered set $S$ has an upper bound in $S$, then $S$ has a maximal element. Theorem 3.5.9 says that every non-empty family of compactifications of $X$ has a least upper bound. Then, by Zorn's Lemma, Corollary 3.5.10 concludes that there is a largest element in the family of compactifications of $X$. $\endgroup$ – Simon_Peterson Dec 29 '15 at 9:21
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If you have a homeomorphism $f:Y_1\to Y_2$, then there is another compactification $(Y_2,c_3)$ of $X$ such that $f(c_1(x))=c_3(x)$ for all $x\in X$ (let $c_3=f\circ c_1$). By your version of equivalence (homeomorphism alone), $(Y_2,c_2)=(Y_2,c_3)$. So we may as well include that extra condition the definition to begin with.

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  • $\begingroup$ I think you want $(Y_2,c_2)=(Y_2,c_3)$ in there :) Also, this still doesn't explain why you would care if the diagram commutes (i.e. the condition of compositions of functions being fulfilled). $\endgroup$ – Simon_Peterson Dec 28 '15 at 21:13

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