9
$\begingroup$

Stone-Čech compactification $\beta X$ of a completely regular space $X$ is defined by the following property: Let $X$ be a completely regular space. Let $i \colon X \hookrightarrow \beta X$ be an embedding into a compact Hausdorff space $\beta X$. Then for every continuous map $f\colon X \to K$ where $K$ is a compact Hausdorff space there exists a unique continuous map $\widehat f \colon \beta X \to K$ such that $\widehat f \circ i =f$. (In the other words, every continuous map $X\to K$ has a continuous extension $\beta X\to K$.)

commutative diagram

http://presheaf.com/?d=d4l86n4i40s4n18675m3rw6cye1p

It is known that if we require the above property to be true not for all compact Hausdorff spaces $K$ but only for $K=[0,1]$, i.e. for the unit interval, then we get an equivalent definition. (One possible argument to show this is based on the fact that every compact Hausdorff space embeds into a power of the unit interval.)

My question is:

Which (compact Hausdorff) spaces $K$ has similar property as unit interval, i.e., the property that if we require the universal property from the definition of Stone-Čech compactification to hold only for this space $K$, then we obtain an equivalent definition? Is complete characterization known?

Are these spaces precisely the generators of the reflective subcategory $\mathbf{CHaus}$ of the category $\mathbf{Top}$? Here $\mathbf{CHaus}$ denotes the category of compact Hausdorff spaces, $\mathbf{Top}$ is the category of topological spaces. By a generator of a reflective subcategory I mean a space such that its reflective hull is precisely this subcategory.

For example, this is true if we take the unit circle $S$, simply because $[0,1]$ is a closed subspace of $S$.

If we take $K=\{0,1\}$ with the discrete topology, then we cannot repeat the same argument as for unit interval. (Not every compact space, is a subspace of some power $\{0,1\}^a$.) So the above is probably not true for $K=\{0.1\}$

$\endgroup$
  • $\begingroup$ It seems rather unnatural to state your definition of the Stone-Cech compactification in terms of completely regular spaces $X$ with an embedding into $\beta X$, rather than in terms of arbitrary spaces $X$ with a map to $\beta X$. If you use the latter statement, you can just take $X=[0,1]$ in the first part of my answer, and you can give a rather formal argument that if $K$ is a test space, then any compact Hausdorff space embeds in a power of $K$. $\endgroup$ – Eric Wofsey Dec 16 '15 at 9:51
  • $\begingroup$ @EricWofsey The only reason I have added the condition is completely regular was that only completely regular space can have compactification (can be densely embedded in a compact Hausdorff space). (E.g., Engelking requires dense embedding in the definition of compactification and then defines $\beta X$ as the largest compactification. Thanks for your comment - it is good to know that $\beta X$ is studied also for other spaces (I.e., if we require only continuous map instead of embedding.) $\endgroup$ – Martin Sleziak Dec 16 '15 at 10:03
7
$\begingroup$

Such "test spaces" are exactly the compact Hausdorff spaces that contain a subspace homeomorphic to $[0,1]$. Clearly any such space is a test space; conversely, suppose $[0,1]$ does not embed in $K$. Then in fact every map $[0,1]\to K$ is constant, since any path-connected Hausdorff space is arc-connected. So taking $X$ to be any path-connected completely regular space, every map $X\to K$ is constant. It follows that any compactification of $X$ satisfies your universal property for $K$. Thus $K$ is not a test space.

As for your second question, yes, these are the generators of CHaus as a reflective subcategory. For any test space $K$ and any compact Hausdorff space $X$, there is an embedding $i:X\to K^S$ for some set $S$ (this follows from the fact that $[0,1]$ embeds in $K$). Moreover, this embedding can be realized as the equalizer of a pair of maps $K^S\rightrightarrows K^T$ for some $T$: it is the equalizer of its cokernel pair $K^S\rightrightarrows Y$, and $Y$ is again compact Hausdorff, so $Y$ embeds in $K^T$ for some $T$. Composing the cokernel pair with the inclusion $Y\to K^T$, we get a pair of maps $K^S\rightrightarrows K^T$ whose equalizer is $i$. Thus $X$ is generated by $K$ using limits. Conversely, if $K$ is not a test space, then it is totally path-disconnected, and then it is easy to see that the reflective subcategory generated by $K$ consists entirely of totally path-disconnected spaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.