# Homeomorphism between a cone in a compact space and this compactification

Let $X$ be a compact Hausdorff space. Show that the cone in $X$ is homeomorphic to the compactification of $X \times [0,1)$. If $A$ is closed in $X$, show that $X/A$ is homeomorphic to the compactification of $X\setminus A$.

I don't know even how to begin to solve this question, it's seems so hard, anyone could help me, please.

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To the one point compactification, you mean; there are many compactifcations! – Mariano Suárez-Alvarez Sep 30 '12 at 2:15
Try to define a map $\mathrm{Cone}(X)\to \mathrm{Compatification}(X\times[0,1))$ There are not many natural ways to do this... in fact, there is pretty much only one—and that one will do the trick. Similarly for the other problem: there is exactly one reasonable map you can define, and it is the homeo you are looking for. When you have the correct maps figured out, you will need to prove they are homeos, of course: for that, you will need to use the fact that your space is compact and Hausdorff, because for that sort of... – Mariano Suárez-Alvarez Sep 30 '12 at 2:16
...spaces there is a simple way of checking if a map is an homeo. – Mariano Suárez-Alvarez Sep 30 '12 at 2:21
yes, I think the one point compactification is better in this case. Speaking of the map, I know that, the problem is that, I don't know how to find this map, I new on this topic (compactifications). – user42912 Sep 30 '12 at 2:22
It is not that «the one point compactification is better in this case»: it is the only one which will make the statements true! – Mariano Suárez-Alvarez Oct 1 '12 at 0:43

Here’s a start on the second problem; you should be able to use some of the ideas to help you with the first, as well.

Suppose that $X$ is a compact Hausdorff space, and $A\subseteq X$ is closed. Let $Y=X\setminus A$, and let $Y^*=Y\cup\{p\}$ be the one-point compactification of $Y$. Finally, let $Z=X/A$, let $q:X\to Z$ be the quotient map, and let $a\in Z$ be the point corresponding to $A$ in $X$. You want to prove that $Y^*$ is homeomorphic to $Z$.

The first step is figure out what the homeomorphism should be. $Z=q[Y]\cup\{a\}$, where $q[Y]$ is a homeomorphic copy of $Y$, and $Y^*=Y\cup\{p\}$, where $Y$ is a homeomorphic ‘copy’ of $Y$ sitting inside $Y^*$. Thus, each of the spaces $Z$ and $Y^*$ consists of a copy of $Y$ together with one extra point. This suggests that we should try the function

$$h:Y^*\to Z:y\mapsto\begin{cases} q(y),&\text{if }y\in Y\\ a,&\text{if }y=p\;, \end{cases}$$

which sends each point of $Y$ to its copy in $Z$ and sends the extra point $p$ in $Y^*$ to the extra point $a$ in $Z$. Now you just have to check that this $h$ really is a homeomorphism.

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@user42912, to prove continuity, simply show that the preimage of an open set is open! – Mariano Suárez-Alvarez Oct 1 '12 at 0:42
@user42912: As Mariano said, just show that if $U\subseteq Z$ is open, then $g^{-1}[U]$ is open in $Y^*$. There are just two cases: $a\in U$, and $a\notin U$. If $a\notin U$, then $h^{-1}[U]=q^{-1}[U]$, and you know that $q$ is continuous. If $a\in U$, use what you know about the quotient map $q$ and what you know about the nbhds of $p$ in the one-point compactification. – Brian M. Scott Oct 1 '12 at 5:40
@user42912: It’s the same thing: $Y$ is open in $X$, so any open cover in $Y$ of a subset of $Y$ is also an open cover in $X$ of that same subset of $Y$. It’s compact in $X$ if and only if it’s compact in $Y$. – Brian M. Scott Oct 4 '12 at 1:53
@user42912: Yes, you can; just pick the right set for $A$. – Brian M. Scott Oct 6 '12 at 0:56
@user42912: Yes, that’s the right $A$. – Brian M. Scott Oct 6 '12 at 20:04