Compactness under homeomorphisms

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Homework Help Overview

The discussion revolves around the preservation of compactness under homeomorphisms in topology. The original poster seeks to understand how a homeomorphism, defined as a continuous function with a continuous inverse, maintains the property of compactness when mapping between sets or spaces.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of a homeomorphism on open covers and question the nature of continuity and compactness in this context. There is a focus on understanding why the pre-image of an open cover remains an open cover and the necessity of injectivity in homeomorphisms.

Discussion Status

The discussion is active, with participants raising questions about the definitions and properties of homeomorphisms, particularly regarding their bijective nature. Some guidance has been provided regarding the implications of continuity and the relationship between open covers and their pre-images.

Contextual Notes

There is uncertainty regarding the definitions of injectivity and surjectivity in the context of homeomorphisms, as well as the implications of these properties on the preservation of compactness.

ehrenfest
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Homework Statement


I want to show that homeomorphism preserve compactness on a set or a space. The definition of a homeomorphism is a continuous function with a continuous inverse.
The definition of a continuous function is a function such that the pre-image of an open set is open.

Let f: X to Y be continuous. Let X be compact
So, if you have an open cover in X then you have a finite subcover. But if you have an open cover in Y, then you could map it to X but how would you know that it is still an open cover in X?



Homework Equations





The Attempt at a Solution

 
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Suppose f:X->Y is a homeomorphism. If {U_i} is an open cover of Y, then {f^-1(U_i)} is an open cover of X.
 
morphism said:
Suppose f:X->Y is a homeomorphism. If {U_i} is an open cover of Y, then {f^-1(U_i)} is an open cover of X.

That is exactly the statement form of my question. Could you please explain why that is true?
 
We know that f is continuous and that each U_i is open. It follows that each f^-1(U_i) is open.

If Y = \cup U_i, can you guess what \cup f^-1(U_i) should be? Can you prove it?
 
Say, that there an x in X s.t. x is not in \cup f^-1(U_i). Then take a nbhd of x N_x. f(N_x) must be in \cup U_i.

OK. So, it comes down to whether the function is injective. Can we assumed it is? That is not part of the definition, is it?
 
A homeomorphism is a bijection, so yes it's injective... But what does that have to do with anything?

Look, Y = \cup U_i. So X = f^-1(Y) = f^-1(\cup U_i) = ...
 
Last edited:
OK. I guess the problem is more fundamental. I do not understand why a homeomorphism is a bijection. How would one show from the definition I wrote in the first post that it is surjective? That it is injective?
 
If it has an inverse it must be a bijection. :smile:

By the way, the f^-1's I used above were used to indicate taking the preimage.
 
I see. Thanks.
 

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