# Thread: Algebraic Topology - Quotient Spaces

1. ## Algebraic Topology - Quotient Spaces

In explaining quotient spaces Martin Crossley (Essential Topology) uses the closed disk $D^2$ in the plane $\mathbb{R}^2$ (see attachment showing Crossley's section on Quotient Spaces)

He states on page 78 (see attachment) that $D^2 - \partial D^2$ is homeomorphic with $\mathbb{R}^2$ , by

$( r cos ( \theta ), r sin(\theta) ) \longleftrightarrow ( tan ( \frac {\pi r }{2} ) cos ( \theta ) , tan ( \frac {\pi r }{2} ) sin ( \theta ) )$

Can anyone help me show explicitly and formally that this is a homeomorphism?

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Crossley then goes on to state that $\mathbb{R}^2$ is homeomorphic with $S^2 - \{ (0,0,1) \}$ by stereographic projection. (see attachement page 78)

I also need help to explicitly and formally prove that this is a homeomorphism.

Be really grateful for help on either one or both of these problems.

Peter

2. ## Re: Algebraic Topology - Quotient Spaces

to show a map is a homeomorphism, one needs to do 3 things:

1. show it is bijective (and thus has an inverse map)
2. show it is continuous
3. show its inverse is continuous

to show the map that sends the open disk D2 - ∂D2 is bijective, suppose that

tan(πr1/2)cos(θ1) = tan(πr2/2)cos(θ2);
tan(πr1/2)sin(θ1) = tan(πr2/2)sin(θ2)

since the two sides of these two equations represent the same point in R2, the two points have the same norm, so we have:

√[(tan(πr1/2)cos(θ1))2 + (tan(πr1/2)sin(θ1))2] = √[(tan(πr2/2)cos(θ2))2 + (tan(πr1/2)sin(θ2))2]

and simplifying both sides (using cos21) + sin21) = cos22) + sin22) = 1):

tan(πr1/2) = tan(πr2/2).

since tan(x) is 1-1 on the interval [0,π/2) (remember r is in [0,1)) taking arctangents of both sides, we get:

πr1/2 = πr1/2, and thus r1 = r2.

(technically we should restrict r to (0,1) and map the point (0,0) separately, but this is a minor matter).

thus cos(θ1) = cos(θ2) and sin(θ1) = sin(θ2).

from the former equation, we see that θ2 = ±θ1, and the second equation then forces us to choose θ2 = θ1 (since sin is an odd function).

this shows our map is injective. to show it is surjective, let (x,y) be any point in R2.

let r = (2/π)arctan(√(x2+y2))

if x > 0, y > 0, let θ = arctan(y/x) <---we'll ignore the niceties of the difference of definition and assume (x,y) is in the first quadrant.
if x < 0, y > 0, let θ = arctan(y/x) + π <---you can verify it for the other 3, if you wish. it's boring.
if x < 0, y < 0, let θ = arctan(y/x) + π
if x > 0, y < 0, let θ = arctan(y/x) + 2π
if x = 0, y ≥ 0 let θ = π/2
if x = 0, y < 0 let θ = 3π/2

then (rcos(θ),rsin(θ)) maps to:

([tan(π((2/π)arctan(√(x2+y2)))/2]cos(arctan(y/x)), [tan(π((2/π)arctan(√(x2+y2)))/2]sin(arctan(y/x))

= (tan(arctan(√(x2+y2))cos(arctan(y/x)), tan(arctan(√(x2+y2))sin(arctan(y/x)))

= (√(x2+y2)(x/√(x2+y2)), (√(x2+y2)(y/√(x2+y2)) = (x,y),

so our map is surjective, and thus it is bijective.

it should be clear that this is a continuous map, and that its inverse, given by:

(rcos(θ),rsin(θ)) → ((2/π)arctan(r)cos(θ), (2/π)arctan(r)sin(θ)) is likewise continuous.

3. ## Re: Algebraic Topology - Quotient Spaces

Thanks for a most helpful post!!

I will go through this carefully now

Will now have the confidence to push on into algebraic topology

Peter

4. ## Re: Algebraic Topology - Quotient Spaces

Hi Deveno,

You write

"to show the map that sends the open disk D2 - ∂D2 is bijective, suppose that

tan(πr1/2)cos(θ1) = tan(πr2/2)cos(θ2);
tan(πr1/2)sin(θ1) = tan(πr2/2)sin(θ2)

since the two sides of these two equations represent the same point in R2, the two points have the same norm"

I am struggling to get a good picture of these two equations and to see why they represent the same point in R2. Can you please help?

I would like to get a good physical (geometrical) picture of $tan ( \pi r/2 )$ and $tan ( \pi r )$ but am finding it difficult. Again, can you help?

Peter

EDIT - PROBLEM SOLVED - Just realised that the equations deal with r1 and r2? Sorry - stupidly (for some odd reason) I was reading them as r x 1 and r x 2

So to answer my own question: Why do the two equations represent the same point?

Looking at the equation:

$tan \frac{\pi r_1}{2} cos {\theta}_1 = tan \frac{\pi r_2}{2} cos {\theta}_2$ .... (1)

we see that LHS equals x-co-ord in the range of mapping (2) below and RHS equals x co-ord also!!

$(r cos \theta, r sin \theta ) \longrightarrow ( tan \frac{\pi r}{2} cos \theta, tan \frac{\pi r}{2} sin \theta )$

Similar reasoning with the second equation means the y co-ords arte the same

So we are dealing with the same point.

Is my reasoning correct?

Peter

5. ## Re: Algebraic Topology - Quotient Spaces

Hi Deveno,

"this shows our map is injective. to show it is surjective, let (x,y) be any point in R2."

You then say'

"Let r = $\frac{2}{\pi} arctan \surd ( x^2 + y^2 )$"

But this means

$tan \frac{\pi r}{2} = \surd ( x^2 + y^2 ) = r$

This does not seem right - can you clarify?

(maybe I have made an error manipulating symbols?)

EDIT -- I am assuming that you are using r to be a symbol for the vector or line segment to the point (x,y) and hence assuming that $r^2 = ( x^2 + y^2 )$ but maybe r is not used in this sense?

6. ## Re: Algebraic Topology - Quotient Spaces

"r" represents (at that part of my post) the radial distance from the origin of a point in D2-∂D2.

yes, we want to end up with √(x2+y2), which is the radial distance of the point (x,y) from the origin.

there are two distances involved here. the point (the pre-image of (x,y)) in the unit open disk has r in [0,1). the point (x,y) is not so constrained, so it would be unwise to denote BOTH distances by "r" (perhaps "r" and "R" would be ok).