Quotient geographies and also equivalence courses

I'm presently researching the idea of a quotient geography. The one point I'm having problem with understanding is what we are in fact doing to the factors as we are recognizing them. Claim we have a $[0,1] \times [0,1]$ in $E^2$ (euclidean $2$ - room) with the subspace geography and also we dividing $X$ right into:

  1. The set $\{(0,0),(1,0),(0,1),(1,1)\}$ of 4 edge factors
  2. collections of sets of factors $(x,0),(x,1)$ where $0<x<1$ ;
  3. collections of sets of factor $(0,y), (1,y )$ where $(0,y), (1,y)$ where $0<y<1$ ;
  4. establishes containing a solitary factor $(x,y)$ where $0<x<1$ and also $0<y<1.$

after that our quotient room will certainly be the torus, yet this will certainly set - talking be: $\{0,0),(1,0),(0,1),(1,1)\}$, collections of sets of factors as defined over for $(x,0),(x,1),$ collections of sets of factors as defined over for $(y,0),(y,1),$ and also $(x,y)$ where $0<x<1, 0<y<1$.

I ask yourself whether there is an any kind of all-natural means to see just how to "position" the factors? This instance is certainly kinda very easy, yet simply offered this set, exists any kind of means to identify just how it looks geometrically? If we would certainly have claim the set containing all factors from a dealt with range from a set factor, we would certainly recognize it as a circle, exists anything comparable below?

And also currently, for the last inquiry below: Allow $B^n$ represent the device round in $n$ - dimensional euclidean room, and also allow $S^{n-1}$ represent its border. Take into consideration the dividing of $B^n$ which has as participants:

  1. the set $S^{n-1}$ ;
  2. the specific factors of $B^n-S^{n-1}$.

OK, when we currently recognize $S^{n-1}$ it falls down right into one factor, right? Specifically what does it collapse right into and also where do we "placed it"?

Hope you can lose some light on this.

2019-05-18 22:52:51
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Answers: 3

Trying to position factors in geography is high-risk. The geometric instinct it offers can be really valuable, yet can additionally be really deceptive. Officially, a geography is specified by the open collections no matter any kind of capacity to visualize them. See "Counterexamples in Topology" as an examples of the troubles you encounter below.

Falling down factors can transform the form of a room substantially. I assume your details instance of falling down the border of $B^n$ offers you $S^n$ as you have a portable n - dimensional manifold without border in which every loop is contractible to a factor, yet I agree to be dealt with.

2019-05-21 09:03:14

There are 2 points you can do with Topology: (1) attempt to visualize what is taking place with ease, and also (2) write it down carefully. Both serve, essential actions.

As an example, if I recognize your first instance, you have the device square $[0,1]^2 \subset \mathbb{R}^2$ with the subspace geography. Allow me write the works with of its factors $(\theta, z) \in [0,1]^2$, for emotional factors that will certainly emerge soon.

On the instinctive side, what you are doing is flexing your square $[0,1]^2$ as if sides $\theta = 0$ and also $\theta = 1$ come to be more detailed and also more detailed ... You placed some adhesive on them, stick them ... Ok: and also what do you "see" currently? - A cyndrical tube, right?

(Except for the reality that in the real world you can not stick simply 2 sectors - or it is actually hard -, you can attempt to do this trying out a paper and also some adhesive.)

On the strenuous side, you might do the following: you recoup that parametrization of the cyndrical tube you have actually possibly seen in some geometry or indispensable calculus training course

$$ \varphi: [0,1]^2 \longrightarrow \mathbb{R}^3 \ , \qquad \varphi (\theta , z) = (\cos (2\pi\theta ), \sin (2\pi\theta ), z) \ . $$

This $\varphi$ is clearly continual (" clearly" suggests: "a person educated you this in some calculus training course a year earlier") and also surjective, if we limit the codomain to the cyndrical tube

$$ C = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 = 1 \ , \ z \in [0,1] \right\} \ . $$

That is, we are aiming to $\varphi$ as a map $\varphi : [0,1]^2 \longrightarrow C$. Notification that additionally in this manner it is continual with the subspace geographies on both sides.

Currently $\varphi $ is virtually injective also, with the exception of those factors $(0,z)$ and also $(1,z)$ that have the very same photo

$$ \varphi (0,z) = (1,0,z) = \varphi (1,z) \ , \qquad \text{for all}\qquad z \in [0,1] \ . $$

In order to repair this absence of injectivity, you claim: "Ok, allow is recognize (" recognize" is the fussy, proper, means to claim "adhesive" in Topology) each $(0,z)$ with its equivalent $(1,z)$ for all $z$." So, you are specifying an equivalence relationship $\sim$ amongst the factors of the square $[0,1]^2$, the one created by:

$$ (0,z) \sim (1,z) \qquad \text{for all} \qquad z \in [0,1] \ . $$

You call $X = [0,1]^2/\sim$ the room gotten from the square $[0,1]^2$ after gluing (sorry, "recognizing") each $(0,z)$ with its equivalent $(1,z)$.

Allow is write $\widetilde{(\theta ,z)} \in X$ the equivalence courses generated by this equivalence relationship. Notification that, if $\theta \neq 0, 1$, you simply have $\widetilde{(\theta , z)} = \left\{ (\theta , z) \right\}$, yet $\widetilde{(0,z)} = \widetilde{(1,z)} = \left\{ (0,z), (1,z) \right\}$. That is, factors of $X$ are "the very same" as the factors of $[0,1]^2$, with the exception of those $(0,z)$ and also $(1,z)$ that are currently glued togehter (" recognized", yes).

In doing this, we additionally get an all-natural map (" estimate", "recognition")

$$ \pi : [0,1]^2 \longrightarrow X \ , \qquad \pi (\theta , z) = \widetilde{(\theta , z)} \ , $$

which is continual necessarily of the quotient geography on $X$.

Back to our $\varphi$: given that it recognizes the very same factors as $\sim$ does, $\varphi$ generates a well - specified map

$$ \widetilde{\varphi} : X \longrightarrow C \ , \qquad \widetilde{\varphi}\widetilde{(\theta , z)} = \varphi (\theta ,z) \ . $$

Now, given that we have actually glued the factors with the very same photo by $\varphi$, this new $\widetilde{\varphi}$ is bijective, yet it is additionally continual, as a result of the global building of the quotient geography and also the reality that $\widetilde{\varphi} \circ \pi = \varphi$. Right?

Last action. A continual bijective map does not require to be a homeomorphism, as you possibly recognize: the inverted $\widetilde{\varphi}^{-1}$ is not always continual.

What could we do? Well, you can attempt to list a specific formula for $\widetilde{\varphi}^{-1}$ and also examine its connection straight, yet this is tough and also excruciating, so no one does it.

Rather, everyone consider the adhering to spectacular, wonderful, unrivaled method (the most effective one you can acquire in primary Topology ; btw, "method" suggests "suggestion": you can confirm it, certainly):

" $X$ is portable, due to the fact that it is a ratio of a portable room (particularly, $[0,1]^2$). $C$ is Hausdorff, due to the fact that it is a subspace of a Hausdorff room (particularly, $\mathbb{R}^3$).

The GTET (Greatest Trick in Elementary Topology)

Now, I have a continual, bijective map $\widetilde{\varphi} : X \longrightarrow C$ in between a portable room and also a Hausdorff one. Therefore $\widetilde{\varphi}$ is a homeomorphism."

Isn't that superb?: -) (Don't neglect it the next time you require to confirm that some map is a homeomorphism: it will certainly conserve your life.)

So, you can additionally confirm carefully your first instinct: your room is a cyndrical tube, $X \cong C$.

2019-05-21 07:59:58

As for your 2nd instance, Ross Milikan is instinct is right. Allow is see just how we can get to the verdict that $B^n/S^{n-1} \cong S^n$ analyzing what takes place in reduced measurements.

For $n=1$ , the (shut) device round is the interval $B^1 = [-1,1]$ and also its border the "round" $S^0 = \left\{ -1, 1\right\}$. Currently, if we attempt to visualize what takes place to $B^1$ when we adhesive (" recognize") the factors $-1$ and also $1$, we see the sector $[-1,1]$ flexing, factors $-1$ and also $1$ coming close to one per various other ... Our adhesive carrier has currently place several of it on them, so when they fulfill, they stick and also we get ... an area! So $B^1 / S^0 \cong S^1$.

Ok, prior to mosting likely to greater measurements: just how can we confirm this? Well, it ends up that, with some adjustments, we currently recognize just how to do it. First, notification that $B^1 \cong [0,1]$ and also below $S^0 = \left\{ 0,1 \right\}$. So we resort once more to a parametrization ; this moment, of the area

$$ \varphi : [0,1] \longrightarrow \mathbb{R}^2 \ ,\qquad \varphi (\theta ) = (\cos (2\pi\theta) , \sin (2\pi \theta) ) $$

and also continue along the lines of the previous cyndrical tube instance: $\varphi$ recognizes the factors of the "round" $S^0$, $\varphi (0) = (1,0) = \varphi (1)$, so $\varphi$ generates a well - specified map

$$ \widetilde{\varphi} : B^1/S^0 \longrightarrow S^1 \ , \qquad \widetilde{\varphi}(\widetilde{\theta}) = \varphi (\theta) \ , $$

which is continual and also bijective. Currently make use of the GTET (Greatest Trick of Elementary Topology) you currently recognize in conclusion that $\widetilde{\varphi} $ is a homeomorphism $B^1/S^0 \cong S^1$.

Until now, so excellent. What is next? Allow is have a look at $n=2$. The device round $B^2$ is currently a disk and also its border the area $S^1$. We placed some adhesive throughout $S^1$ and also we flex the disk and also at the very same time reduce the area making it diminish and also smaller sized till it is simply one factor. What do we get? A round!

So we might believe that $B^2 / S^1 \cong S^2$. In order to confirm it, we might attempt to duplicate our "aesthetic" homeomorphism with solutions. Allow is place our disk $B^2$ in $\mathbb{R}^3$ in the $xy$ - aircraft, focused at $(0,0,0)$:

$$ B^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 \leq 1 \ , \ z= 0 \right\} \ . $$

And we placed our round focused at $(0,0, 1)$:

$$ S^2 = \left\{ (x,y,z) \in \mathbb{R}^3 \ \vert \ x^2 + y^2 + (z-1)^2 = 1 \right\} \ . $$

So the South Pole is $(0,0,0)$ - the facility of the disk $B^2$ -, the North Pole is $(0,0,2)$ and also the equator hinges on the aircraft $z= 1$.

Our "aesthetic" homeomorphism flexes the disk over the round and also in doing this reduces its border till it is understood the North Pole of the round. So, allow is attempt to do the very same with a map like

$$ \varphi : B^2 \longrightarrow S^2 \ , \qquad \varphi (x,y) = (f(r)x , f(r)y, 2r) \ . $$

Here $r= + \sqrt{x^2+y^2}$. Allow is evaluate what this sort of map is doing:

  1. When going from $r = 0$ to $r=1$, we desire the factors $\varphi (x,y)$ rise the round, from elevation $z=0$ (South Pole) to $z=2$ (North Pole). So $z$ needs to climb at double rate as $r$ goes from the facility of the disk to its border. Therefore we placed that $2$.
  2. To make points less complicated, we desire the estimates of the factors $\varphi (x,y)$ onto the $xy$ - aircraft to hinge on the very same instructions as the factor of the disk $(x,y)$ they originate from. So this estimate needs to be something like $(\lambda x, \lambda y)$. At the very same time this lambda will certainly differ with the range $r$ to the facility of the disk so that $\varphi (x,y)$ continues to be on the round. So we attempt this $(f(r)x , f(r)y)$.

To locate which $f(r)$ we require, we enforce the problem that $\varphi (x,y) $ pushes the round:

$$ 1 = x^2 + y^2 + (z-1)^2 = f(r)^2 r^2 + (2r-1)^2 $$

And we get $f(r) = 2 \sqrt{\frac{1-r}{r}}$. So our map is

$$ \varphi (x,y) = \begin{cases} \left( 2 \sqrt{\frac{1-r}{r}}x , 2 \sqrt{\frac{1-r}{r}}y, 2r \right) & \text{if}\ (x,y) \neq (0,0) \\\ (0,0,0) & \text{if}\ (x,y) = (0,0) \ . \end{cases} $$

Exercise. Confirm that $\varphi$ is continual.

Currently, we continue a lot similarly as in the cyndrical tube instance. By building and construction, $\varphi$ is plainly surjective. Yet not injective: all the factors in the border $r=1$ of the disk most likely to the North Pole. So we quotient out this border and also, as in the past, $\varphi$ generates a bijective continual map

$$ \widetilde{\varphi} : B^2/S^1 \longrightarrow S^2 \ , \qquad \widetilde{\varphi}\widetilde{(x,y)} = \varphi (x,y) \ . $$

We use once more our nuclear tool GTET and also we are done.

All right, what around greater measurements? - We've obtained no geometric or aesthetic instinct there (at the very least, I've obtained none ; if you assume you can "see" what takes place in, claim, $\mathbb{R}^4$, congratulations, yet simply in instance do a browse through to your psychoanalyst and also inform him/her). Anyhow, solutions permit us to surpass our aesthetic instinct. We placed our round and also round inside $\mathbb{R}^{n+1}$ as prior to:

$$ B^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2 \leq 1 \ , \ x_{n+1} = 0 \right\} $$

and also

$$ S^n = \left\{ (x_1, \dots , x_n, x_{n+1}) \ \vert \ x_1^2 + \cdots + x_n^2 + (x_{n+1}-1)^2 = 1 \right\} \ . $$

And we attempt some map of the very same kind:

$$ \varphi : B^n \longrightarrow S^n \ , \qquad \varphi (x_1, \dots , x_n ) = (f(r)x_1, \dots , f(r)x_n, 2r) \ . $$


  1. Full the information.
  2. This is not the very same homeomorphism we formerly located for $n=1$. Write this basic formula for $n=1$.
2019-05-21 07:30:22