{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

245B_W09_hwk5 - 245B Winter 2009 Assignment 5 Notes and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
245B, Winter 2009, Assignment 5: Notes and selected model answers (Model solutions follow question numbers in bold .) Folland Chapter 4 28. a. We must verify that T satisfies the axioms of a topology. 1. We always have π - 1 ( ) = and π - 1 ( ˜ X ) = X , and both of these pre- images are open in X , so , ˜ X ∈ T ; 2. If U is any collection of members of T , then by definition { π - 1 ( U ) : U U} is a collection of open subsets of X . Therefore U ∈U π - 1 ( U ) is a union of open sets in X and so is itself open, but since π - 1 U = U ∈U π - 1 ( U ) this tells us that π - 1 ( U ) is open, and so U ∈ T ; 3. Exactly similarly, if U is a finite family of members of T then π - 1 U = U ∈U π - 1 ( U ) is now a finite intersection of open subsets of X , so is itself open, and so the finite intersection U also lies in T . b. Suppose that Y is another topological space and f : ˜ X Y a function. First note that if U ˜ X is open (i.e., it lies in the family T that we have shown above to be a topology) then by the definition of T we have that π - 1 ( U ) is open 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
in X . Phrased differently, this tells us that π
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}