Verifying S1 in Quotient Topology of R with x~x+1

[andlook]

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.

 

[andlook]

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?

 

[andlook]

for anyone that needs more on this I found extrapolation given by M grime third post down on thread

https://www.physicsforums.com/showthread.php?t=124804


At least I found this helpful

 

[Fredrik]

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 \pi^{-1}(U) of open sets U in R, where \pi:\mathbb R\rightarrow\mathbb R/\sim is the function that takes a number to its equivalence class: \pi(x)=[x].

 

[andlook]

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?

 

[Fredrik]

What I'm suggesting is that you define a topology on R/~ by saying that the function \pi is continuous. Then you you show that your f is continuous with respect to that that topology on R/~.