# locally compact and proper functions

• November 13th 2010, 11:55 PM
Sogan
locally compact and proper functions
Hello,

i try to solve following exercise:

Let f:X-> Y be an continuous immersion. X a Hausdorff-space and Y locally compact, i.e. Y is a Hausdorff-space with the property that every point y $\in$ Y has a nbh. U with closure(U) compact.
Show that f is proper iff f^(-1) (K) $\subset$ X is compact for all K $\subset$ Y compact.

"=>" This direction was not so difficult. I could solve it.

But the other direction "<=" i couldn't solve.

We have shown before, that "f is proper" is equivalent to "f is closed" and to "f is an embedding".

I try to show that f is closed, since i think this must be the most easy way.

Have you some help for me.

Thanks a lot
• November 14th 2010, 06:08 AM
Sogan
I have a solution, but i'm not sure whether it is correct:

"<="
We want to show f is proper <=> f is closed.

Let V be a closed subset of X.

f(V) is closed in Y <=> Y-f(V) is open in Y.

so let y $\in$ Y-f(V) a arbitrary point. We know Y is Hausdorff, hence for every point v $\in$ f(V) we have a open set $U_v$ of v and a open set U of y, s.t.
$U_v \bigcap$ U= $\emptyset$

(*)hence it exist a nbh U of y , s.t. U $\bigcap$ f(V) = $\emptyset$

hence f(V) is closed.

I'm not sure whether this prove is correct.
I mean, i havnÄt use locally compactness of Y. So i think this can't be correct.
the part with (*) above isn't correct, or? how can i involve locally compactness in that point, such that my proof is correct?

Regards
• November 14th 2010, 08:05 AM
Drexel28
Quote:

Originally Posted by Sogan
I have a solution, but i'm not sure whether it is correct:

"<="
We want to show f is proper <=> f is closed.

Let V be a closed subset of X.

f(V) is closed in Y <=> Y-f(V) is open in Y.

so let y $\in$ Y-f(V) a arbitrary point. We know Y is Hausdorff, hence for every point v $\in$ f(V) we have a open set $U_v$ of v and a open set U of y, s.t.
$U_v \bigcap$ U= $\emptyset$

(*)hence it exist a nbh U of y , s.t. U $\bigcap$ f(V) = $\emptyset$

hence f(V) is closed.

I'm not sure whether this prove is correct.
I mean, i havnÄt use locally compactness of Y. So i think this can't be correct.
the part with (*) above isn't correct, or? how can i involve locally compactness in that point, such that my proof is correct?

Regards

Yeah, it's not so clear to me why you can do what you said you can. But how about this:

I'm a little unsure what you're question actually is. Is it that $f$ is proper implies that $f$ is closed?
• November 14th 2010, 08:32 AM
Sogan
Hello,

No thants not my question. I want to show following:

if f^(-1)(K) is compact for every compact K => f is proper.

This statement is what i want to prove
• November 14th 2010, 08:36 AM
Drexel28
Quote:

Originally Posted by Sogan
Hello,

No thants not my question. I want to show following:

if f^(-1)(K) is compact for every compact K => f is proper.

This statement is what i want to prove

Isn't that the definition of a proper map?
• November 14th 2010, 08:47 AM
Sogan
Oh ok. we have a different definition of proper map:

f is proper iff f x id_z : X x Z->Y x Z is closed.

f in C(X,Y)

but both definitions are equivalent. That is what we should show in this exercise.
• November 14th 2010, 08:50 AM
Drexel28
Quote:

Originally Posted by Sogan
Oh ok. we have a different definition of proper map:

f is proper iff f x id_z : X x Z->Y x Z is closed.

f in C(X,Y)

but both definitions are equivalent. That is what we should show in this exercise.

What is $Z$? Any space?
• November 14th 2010, 08:51 AM
Sogan
yes any topological space
• November 14th 2010, 08:57 AM
Drexel28
Quote:

Originally Posted by Sogan
yes any topological space

Ok, one last question. Are we assuming that $X,Y$ are Hausdorff and $Y$ locally compact?
• November 14th 2010, 08:58 AM
Sogan
Yes you are right. X,Y are Hausdorff and Y is locally compact