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Quotient topology

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Homework Statement



verify that R, the reals, quotiented by the equivalence relation x~x+1 is S^1

Homework Equations





The Attempt at a Solution



All i can think of is to draw a unit square and identify sides like the torus, but this would be using IxI, a subset of R^2, and gives a cylinder at best... Obviously I am missing the point, so any help would be great.
 

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  • #2
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Ok thought about this some more.

What I have done above is actually on the right path, if the horizontals of the square are identified we get a cylinder, if the vertical height is shrunk to nothing, so we are working in just R, then we get a circle. Then the result follows. I think the logic I present is correct but is there a more mathematically concise way to present this?
 
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  • #4
Fredrik
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Sounds like you should try to write down a function from from R/~ into S1 (or from S1 into R/~) and then try to prove that it's a homeomorphism. You mentioned the "quotient topology" in the title. I take it that means that the open sets on R/~ are defined to be the preimages [itex]\pi^{-1}(U)[/itex] of open sets U in R, where [itex]\pi:\mathbb R\rightarrow\mathbb R/\sim[/itex] is the function that takes a number to its equivalence class: [itex]\pi(x)=[x][/itex].
 
  • #5
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Ok so the quotient R/~ = [0,1) for the relation x~x+1?

Then defining the map f:[0,1)---S^1 via

f(x)=exp(2*Pi*x*i)

for x in [0,1).

Yes I am working on the definition that open sets in the preimage are open defines continuity and so give definition of homeomorphism.

I take by showing that S^1 is homeomorphic to unit interval this shows that it has the same topology as the real line?
 
  • #6
Fredrik
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What I'm suggesting is that you define a topology on R/~ by saying that the function [itex]\pi[/itex] is continuous. Then you you show that your f is continuous with respect to that that topology on R/~.
 

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