Noncompact locally compact Hausdorff continuous mapping

[PsychonautQQ]

Self studying here :D...

Let X and Y be noncompact, locally compact hausdorff spaces and let f: X--->Y be a map between them; show that this map extends to a continuous map f* : X* ---> Y* iff f is proper, where X* and Y* are the one point compactifications of X and Y.

(A continuous map is said to be proper if the inverse image of each compact subset of Y is compact in X).

I have already shown that X* is a compact hausdorff space and that X is open and dense in X* and has the subspace topology.This one might take me awhile to work all the way through, I've meditated on the matter a bit but would appreciate some help getting my foot in the door.

 

[andrewkirk]

Since f* is an extension, the only choice we have to make is what point in Y* to map $\infty_X$ (the extra point that compactifies X) to. The most natural choice seems to be to map it to $\infty_Y$. I suggest investigating such a map to see if it is continuous. We will need to use the continuity and properness of f in the process. The only tricky bit will be trying to prove that the inverse images of open sets in $Y^*$ containing $\infty_Y$ are open.

If that doesn't work, we'll need to consider the case where $f^*(\infty_X)=y\neq \infty_Y$. My gut tells me we won't have to consider that case, but my gut is sometimes wrong.

 

[PsychonautQQ]

let $U$be an open neighborhood of $\infty_Y$. Then $Y^*-U$ will be compact in Y, call it $C_y$.

The inverse image of $C_y$ will be some compact set in X, call it $C_x$. We know this to be true because the map is proper.

Hmm... am i any closer to showing that the preimage of $U$ is open in $X^*$? I will think about how I can bring the locally compact and Hausdorff properties into the solution.

So $C_x = X^*-R$ for some neighborhood of $R$ of $\infty_X$ in $X^*$?

So $Y^*-U$ is a compact set in $Y$ and therefore closed in $Y$. Perhaps we can use this to show that $U$ is open in $Y^*$ and likewise that $R$ is open in $X^*$ and that R is the preimage of U?
I did not use local compactness anywhere yet. There must be some flaws in my logic or holes to be filled in.

edit: so it turns out a continuous proper map between two locally compact Hausdorff spaces is a closed map. That will probably help. I suppose it's also worth noting that a compact subset of a Hausdorff space is closed.

Also... any tips?? :D

 

[andrewkirk]

Using the definition of openness in $Y^*$, we can write $U=U'\cup\{\infty_Y\}$, where $U'$ is open in $Y$. Using that and our definition of $f^*$, that should enable us to say something more specific about the inverse image of $C_y$.

 

[PsychonautQQ]

Man, I'm certain there is something obvious that I'm not seeing... But why can we write $U=U'\cup\{\infty_Y\}$ where $U'$ is open in $Y$?

 

[andrewkirk]

Man, I'm certain there is something obvious that I'm not seeing... But why can we write $U=U'\cup\{\infty_Y\}$ where $U'$ is open in $Y$?

In the one-point compactification, any open set containing the additional point $\infty_Y$ must be of that form. See this definition.

 

[PsychonautQQ]

That seems to be under
Alexandroff one-point compactification,

that does not seem to be the same as the one point compactification i am using, where the open sets of $X^*$ are the sets that are open in $X$ or sets in V in $X^*$ such that $X^* - V$ is compact in X.

edit: are these definitions equivalent or am I mistaken somewhere in my definition or understanding?

 

[andrewkirk]

They are the same. Since $X$ is Hausdorff, the compactness of $X^*-V$ implies it is closed in X. So that $X-(X^*-V)=U'=V-\infty_X$ is open in $X$.

 

[PsychonautQQ]

They are the same. Since $X$ is Hausdorff, the compactness of $X^*-V$ implies it is closed in X. So that $X-(X^*-V)=U'=V-\infty_Y$ is open in $X$.

Oh, wow, i am a bit slow sometimes :-/

 

[PsychonautQQ]

They are the same. Since $X$ is Hausdorff, the compactness of $X^*-V$ implies it is closed in X. So that $X-(X^*-V)=U'=V-\infty_Y$ is open in $X$.

$X - (X^* - V) = V - \infty_Y$

Can you show me why this is ?

 

[andrewkirk]

Use subscript 'c' to denote complement in $X^*$. Then, using de Morgan's laws repeatedly:

\begin{align*}
\\X-(X^*-V)
&=X \cap ((X\cup\{\infty_X\} )\cap V^c)^c
\\&=X \cap ((X \cup\{\infty_X\} )^c \cup V)
\\&=X \cap ((X^c \cap\{\infty_X\}^c ) \cup V)
\\&=X \cap ((\{\infty_X\} \cap\{\infty_X\}^c ) \cup V)
\\&=X \cap (\emptyset \cup V)
\\&=X \cap V
\\&=V - \{\infty_X\}
%\\&=X \cap (X^c \cup \{\infty_X\}\cup V)
%\\&=X \cap V
%\\&= X-\\
\end{align*}

That $\infty_Y$ in post 8 should have been $\infty_X$. I've gone back and corrected it.

 

[PsychonautQQ]

So $preimage of f^*(C_y) = preimage of f^*(Y^* - (U' union {\infty_Y)})$

Hmm, i still don't see what i am supposed to see.

 

[andrewkirk]

What we need to prove is that, for U open in Y*, $f^{*-1}(U)$ is open in X*. For U open in Y* we know (from Hausdorff) that $Y^*-U$ is compact in Y* so that (by properness) $f^{*-1}(Y^*-U)=f^{*-1}(Y^*)-f^{*-1}(U)$ is compact in X*.

Can you reason from there to the sought conclusion, using what we know about compactness and the Haussdorf property?

 

[PsychonautQQ]

What we need to prove is that, for U open in Y*, $f^{*-1}(U)$ is open in X*. For U open in Y* we know (from Hausdorff) that $Y^*-U$ is compact in Y* so that (by properness) $f^{*-1}(Y^*-U)=f^{*-1}(Y^*)-f^{*-1}(U)$ is compact in X*.

Can you reason from there to the sought conclusion, using what we know about compactness and the Haussdorf property?

Oh, so it takes the complement of an open set, which is compact, to another compact set that is therefore closed, so it's complete, the preimage of U, must be open.

something like that? Thanks for being patient with me btw :D

 

[andrewkirk]

Yes that's it.