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- Thread starter madness
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- #2

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Why don't you think it's a bijection? Are you doubting it's injective or surjective? The map is indeed a bijection so it would be helpful if you could point out why you doubt this.However, this map doesn't even seem to be a bijection.

The two simplest approaches I can think of is:The proof simply says "by construction". Can anyone help?

1) Show that \sigma is a bijective first. Then show that it's open and continuous by the usual set-theoretic arguments involving the image and pre-image of open sets (or just basis elements).

2) Show that \sigma is continuous, and construct a continuous inverse.

Personally I prefer (2), and using a couple of facts about quotient topologies it's a pretty simple construction (construct a map [itex]S^1 \to S^1[/itex] that sends [itex]\cos(\pi t)+i\sin(\pi t) \mapsto \cos(2\pi t) + i\sin(2\pi t)[/itex] and then factor it through [itex]RP^1[/itex]).

- #3

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Well I'm having trouble seeing how the inverse maps to just 1 element on the circle. Taking the inverse projection of an element [z] in RP1 gives {z,-z} in S1. I can see conceptually that the change from 2pi to pi makes it 1 to 1 but still find it difficult to formalise mathematically.

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That would be the case if [itex]\sigma[/itex] sent x to [x], but it does not. To see why this doesn't give any problems suppose [itex]\sigma(\cos(2\pi t_1)+i\sin(2\pi t_1)) = \sigma(\cos(2\pi t_2) + i\sin(2\pi t_2))[/itex] with [itex]t_1,t_2 \in [0,1)[/itex] and show that you must get [itex]t_1=t_2[/itex].Well I'm having trouble seeing how the inverse maps to just 1 element on the circle. Taking the inverse projection of an element [z] in RP1 gives {z,-z} in S1.

To conceptually visualize why this works consider the map [itex]f : S^1 \to S^1[/itex] defined by [itex]f(\cos(2\pi t) + i \sin(2\pi t)) = \cos(\pi t) + i \sin (\pi t)[/itex]. This sends the point at an angle v to the point at an angle v/2. Thus the image of the whole sphere lies entirely in the first and second quadrant. If you think of it as a movement around the circle parameterized by t, then you can think of f as slowing down the movement speed to 50% so we only get half the way around so we never get two antipodal points, but we get a representative for every pair of antipodal points.

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- #6

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- Let [itex]g : X \to Y[/itex] be a continuous function with the property that g(x)=g(y) if x~y. Then there exists a unique continuous function [itex]g' : (X/\sim) \to Y[/itex] such that [itex]g' \circ p = g[/itex] where [itex]p : X \to X/\sim[/itex] is the projection map.

This property is the key to constructing the inverse. As you say it's kind of hard since you somehow have to invert the projection and then show that in the end we end up with a well-defined function. With this property this is taken care of automatically.

Let ~ be the antipodal equivalence relation. Now basically what you do is that you construct the inverse map you want, but for a moment ignore that we are working in [itex]RP^1[/itex] and instead consider working in [itex]S^1[/itex]. So you define [itex]\tau : S^1 \to S^1[/itex] by

[tex]\tau(\cos(\pi t) + i \sin(\pi t)) = \cos(2\pi t) + i\sin(2\pi t)[/tex]

You can now show that if x~y, then [itex]\tau(x)=\tau(y)[/itex] (HINT: x~y iff x and y differ by a multiple of pi, which in turns becomes a multiple of 2pi, which disappears).

Thus by the previous result we get an induced map [itex]\tau' : (S^1/\sim) \to S^1[/itex] and since [itex](S^1/\sim) = RP^1[/itex] you can check that [itex]\tau'[/itex] is the correct inverse.

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