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.


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 $\beta X$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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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