Let $X$ a compact Hausdorff space and let the cone \begin{align} CX := (X \, \times \, [0, 1]) \, / \, A \end{align} where $A:= X \times \{1\} \subset X \, \times \, [0, 1]$. Show that $CX$ is homeomorphic to the one-point compactification $Y^{+}$ of $Y$, where $Y:= X \times [0, 1)$ and $Y^{+}:= Y \overset{\cdot}\cup \{ p\} $.

I tried to follow the answer at the question on this topic (Homeomorphism between a cone in a compact space and this compactification). I have some trouble to prove that the function $h: Y^{+} \rightarrow CX$ definied by \begin{align}h(y) : = \begin{cases} f(y), \, \text{if} \, \, y \in Y \\ a, \, \text{if} \, \, y = p \end{cases} \end{align} is continuous in the case of the point $a\in U$, where $U \subset CX$ open set, $f: X \times [0, 1] \rightarrow CX= (X \times [0, 1])\, / \, A$ is the quotient map and $a \in CX$ is the point corresponding to $A$ in $X \times [0, 1]$.

I want to show that the preimage $h^{-1}(U) \subset Y^{+}$ is open. By definition I know that a set $V \subset Y^{+}$ is said to be open if and only if \begin{cases} V \subset X \, \, \text{open in} \, \, X, \text{if} \, p \notin V\\ V = Y^{+} \setminus C \, \, \text{for a compact set} \, \, C\subset X, \text{if} \, \, p \in V\end{cases} but I can't see how use it to prove the continuity of $h$ in the case where the point $a \in U$.

Any suggestions? Thanks in advance!

  • $\begingroup$ often it is easier to use the following: clearly, there is a bijection of sets between $CX$ and $Y^+$ (why?). Now show that it is a closed (or open) map and you are done (by closed/open map theorem) $\endgroup$ – Simonsays Mar 16 '18 at 18:30
  • $\begingroup$ @Simonsays Unfortunately I didn't see this theorem on lecture so I can't use it. Is there maybe another way to solve it? $\endgroup$ – userr777 Mar 17 '18 at 7:55
  • $\begingroup$ of cause, you can choose the laborious way by showing the existence of an inverse and its continuity. It might be even easier to prove the closed map theorem which you use most of the time in the future and then use it. $\endgroup$ – Simonsays Mar 17 '18 at 8:09

For an open set $U\subset CX$ containing $a=f(A)$, show that the preimage $h^{-1}(U)\subset Y^+$ is open.

Since $h(p)=a\in U$, $p\in h^{-1}(U)$. By the definition of the one-point compactification, it suffices to find a compact set $C\subset Y$ such that $h^{-1}(U)=Y^+\setminus C$.

We claim $C\equiv Y\setminus f^{-1}(U)=(X\times[0,1])\setminus f^{-1}(U)$. (The second equality comes from the assumption $h(p)=a\in U \Rightarrow f^{-1}(a)=X\times\{1\}\subset f^{-1}(U)$.)

  1. Since the quotient map $f$ is continuous, $f^{-1}(U)$ is open in $X\times[0,1]$ and so $C$ is closed. Furthermore, since $C$ is a closed subset of a compact space $X\times[0,1]$, $C$ is also compact.

  2. It remains to show that $h^{-1}(U)=Y^+\setminus C$. Note that $p\in h^{-1}(U)$ by assumption and $$ p\notin Y \Rightarrow p\notin Y\setminus f^{-1}(U)\equiv C \Rightarrow p\in Y^+\setminus C $$ Thus it suffices to show $h^{-1}(U)\setminus\{p\}=Y\setminus C=f^{-1}(U)$. But it is trivial since $h(x)=f(x)$ for all $x\in Y=X\times[0,1)$.


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.