Compactness under homeomorphisms

[ehrenfest]

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

 

[morphism]

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.

 

[ehrenfest]

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?

 

[morphism]

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?

 

[ehrenfest]

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?

 

[morphism]

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) = ...

 

[ehrenfest]

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?

 

[morphism]

If it has an inverse it must be a bijection.

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

 

[ehrenfest]

I see. Thanks.