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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.