- #1
christoff
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Homework Statement
This question comes out of "Introduction to Topology" by Mendelson, from the section on Identification Topologies.
Let D be the closed unit disc in R^2, so that the boundary, S, is the unit circle. Let [tex] C=S\times [0,1], [/tex] and [tex]
A=S \times \{1\} \subset C. [/tex] Prove that C/A is homeomorphic to D.
Homework Equations
The Attempt at a Solution
I feel as though the map [tex]
p:C/A \rightarrow D \\
p(x,y,z)=(x,y) [/tex]
should define a nice homeomorphism. It has an obvious inverse, but even proving that the forward one is continuous is proving to be a problem for me. This may be for lack of experience working with the identification topology on C/A, or maybe I'm taking the wrong approach here.
p is continuous iff for every δ open in D, [tex] p^{-1}(δ):=ψ[/tex] is open in C/A. A subset of C/A is open iff in turn,
[tex]
f^{-1}(ψ)\subset C
[/tex]
is open, where
[tex]
f:C\rightarrow C/A
[/tex] is defined by
[tex]
f|_{C-A}=id|_{C-A}\\
f(A)=(0,0,1).
[/tex]
So for the moment, the problem is: prove that [tex]f^{-1}(ψ)[/tex] is open.
I'm open to more elegant approaches, if you can get me started. I feel as though my approach might be too mechanical, to be honest. There must be a nicer way of doing this.