Equivalence of topological compactifications 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.
 A: 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 $\beta X$. 
A: This question is years old but I thought I'd share another point of view about the relevance of having a commutative diagram $f\circ c_1 = c_2$, for people in the future could perhaps find it useful too.
Using your notation, imagine that $Y_1$ and $Y_2$ are actually supersets of $X$ and assume that $c_i:X\hookrightarrow Y_i$ is the inclusion for $i\in\{1,2\}$. Then the compactifications $(Y_1,c_1)$ and $(Y_2,c_2)$ are equivalent iff there is a homeomorphism $f:Y_1\to Y_2$ such that $f(x)=x$ for all $x\in X$.
This says that $X$ is embedded in the same way in both compactifications.
In the general case where $Y_i$ is not necessarily a superset of $X$, the previous notion still makes sense, because the embedding $c_1$ acts essentially as a "generalized inclusion". The main idea of a compactification is to think of a space as if it were compact so that one can take advantage of properties that compact spaces have, so it seems natural to ask when two ways of compactifying a space $X$ end up looking exactly the same from the perspective of $X$.
David Ullrich's answer in this post is very insightful of the essence of this concept and the interpretation that I tried to provide. In his example, not only are his two compactifications not equivalent $-$ they are not even comparable (w.r.t. the preorder mentioned in Simon_Peterson's answer).
A: 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. 
