# Quotient Topology?

1. Jun 27, 2006

### waht

I can't understand it. No matter how much I try, Can anyone explain it step by step, and give some examples.

How can it be applied to contruct different shapes?

2. Jun 27, 2006

### mathwonk

all quotient objetcs have the same property, namely they are defiend so that it is easy to construct maps out of them. A quotient group G/H is a group such that any homomorphism G-->K which maps H to the identity e in K, induces a unique homomorphism G/H-->K such that the diagram

G-->G/H-->K equals the given map G-->K.

Similarly, the quotient topology on a set of equivalence classes X/S of elements of X is the unique topology such that contiinuous maps X/S-->T for any space T, are exactly induced by those maps X-->T which are constant on equivalence classes of elements of X.

at least that should be right, is it? i recall that a set of equivalence classes of X/S is open iff their union is open in X. does that check out?

it seems then that if X-->Y is any surjective continuousm map, then there is a factorization X-->X/S-->Y where X/S is the same set as Y but with the quotient topology induced by the equivalence relation that two elements of X are equivalent iff they map to the same element of Y.

i.e. if X is a top. space and X-->Y is a surjection, the quotient topology on Y is the finest (richest) topology for which f is continuous.

3. Jun 28, 2006

### matt grime

The quotient topology is not used to construct different spaces. You can completely ignore the topology part if you want to know how to construct different spaces from old ones and just look at quotient spaces.

Example

S^1 is the unit circle thought of as embedded in the plane. Define a relation on S^1 by x~-x, what is the quotient space? It is S^1 again. How to see this? I would pick one element in each equivalence class and think what happens as you move around these elements in S^1. In this case we have one equivalence class for each point in the upper semicircle, and as you walk round the semicircle to the left and move into the lower semicircle you hit the same equivalence class as at the start of the semicircle, i.e. the point (1,0) is identified with (-1,0). In any case we can see we have another circle.

Now, take R^2 and the lattice Z^2 in R^2. Define x~y if x-y isin the lattice, i.e. if the non-integer coordinates of x and y are the same. Pick a 'fundamental domain' again of a natural choice of equivalence classes in the plane, the points (a,b) with 0<=a,b<1 will do. What happens as we move out of this fundamental domain? We come back in from the other side - (a,b)~(a+1,b) - so we're wrapping the plane up into a torus - you can think of this by creating a cylinder first from making the identification of all the x coordinates that differ by an integer, and then doing the same for the y coordinates.

It really is a hell of a lot easier if we can draw pictures.

The simplest example would be R to S^1: define a relation on R by x~y if x-y is an integer, this is wrapping up the line into a circle. It is easier if you just think of the map from R to S^1 by t--> exp(2*t*i*pi).

4. Jun 28, 2006

### matt grime

Here are some easier ones to imagine (for small n)

Let S^n be the n-sphere, ie. S^0 is two points, S^1 is the circle, S^2 is the usual sphere and so on. S^n is the unit length vectors in R^{n+1}.

Given S^n, let I be the unit interval [0,1] and consider IxS^n. Identify all the points (t,x) when t=0 and identify all points when t=1. The resulting space is homeomphic to S^{n+1}.

Example. S^0 is two points, so IxS^0 is two copies of I, one for each element of the two element set S^0, so it looks like

| |
Now, we pinc together all the bits at the top to get

/\

and then we pinch together the bottom

/\
\/

so it really is a circle

O

What is IxS^1? It is a cyclinder

o
| |
o

And pinch together all the top end, and also the bottom - you can imagine squeexing the ends together at the end of a tube... and that's the same (upto homeomorphism) as a 2-sphere.

5. Jun 28, 2006

### waht

Ok, so IxS^n is homemorphic to S^{n+1}

In your example you map two I's (for n = 0) to S^{0+1} or a circle.

Is this mapping considered a quotient topology?

6. Jun 28, 2006

### matt grime

No, IxS^n is not homeomorphic to S^{n+1}. Try rereading what I wrote and try to draw the examples on paper. IxS^0 is just two copies of I, so definitely not homeomorphic to S^1. I have to construct a quotient space from IxS^n to get S^{n+1}

A mapping is not a topology. I've not mentioned topologies at all in this. Forget topologies, just think about the spaces. When your'e happy with constructing the quotient spaces then we can move on to this idea:

If X is a topological space and X/~ is some quotient space, then X/~ is a topological space in a natural way, via the quotient topology and it is the 'finest' topology (i.e. the one with most open sets) making the quotient map q:X--> X/~ continuous.

Last edited: Jun 28, 2006
7. Jun 28, 2006

### waht

I understand how you can make a circle with 2 I's, simply join the endpoints. Likewise, the cylinder example is simple. But I'm a little confused about the pinching process?

"I have to construct a quotient space from IxS^n to get S^{n+1}"

So basically is it your goal here to make IxS^n homemorphic to S^2{n+1}

Is the quotient space a bridge between these sets that makes them homemorphic?

Last edited: Jun 28, 2006
8. Jun 29, 2006

### matt grime

In some sense, I suppose it is 'the bridge'. The quotient space of IxS^n above is (in the appropriate sense) S^{n+1}.

When you join the end points of the two Is you are quotienting out by an equivalence relation (albeit not a very interesting one) and forming the quotient space which is the set of equivalence classes.

And at no point am I making IxS^n homeomorphic to S^{n+1}. I cannot do that; they are not homeomorphic. I am making a quotient space from IxS^n that is homeomorphic to S^{n+1}.

Have you looked at the other examples I gave of making S^1 as a quotient of the real line, or the torus as a quotient of the plane?

9. Jun 29, 2006

### waht

"I am making a quotient space from IxS^n that is homeomorphic to S^{n+1}"

I see, but during the process of joining the end points, are you actually forming a map that maps the endpoints (in or to) a quotient space?

So for a torus take the endpoints of the cylinder IxS^1 and join them to form a circle.

You could take the boundary of a circle and collapse it in a single point. That would make a sphere.

10. Jun 29, 2006

### matt grime

The end points of a cylinder are circles, we don't join anything there. But yes, we collapse them to a single point - in the quotient space they are all identified.

If we are identifying points in X under a relation ~ to form the space X/~ it is common to use q to denote the map that sends x in x to [x] in X/~. [x] is the equivalence class of x. Perhaps that helps.