What do topologically stable maps mean?

[marlon]

Hi

I have this question on homotopy groups: Spacial infinity in two dimensional space is a unit circle S1 (topologically). I understand that. Now, in physics one can prove that fields will exhibit an equation (expressed by the map phy --> v =constant) that also represents a unit circle. Now, a map between two unit circles (S1 to S1) is topologically stable because this map cannot be transformed into a map where S1 is mapped onto a point of S1. Can someone explain that last sentence, please.

Now, a homotopy group counts the number of topologically inequivalent maps. Can some one give me an example of those kinds of maps. Does this mean that you cannot go from one map to the other without breaking it apart ?

A map phi--->v*exp(imx) (where x is the angle and m an integer) represents a map that wraps one circle around the other m times. Does anyone know what this means and how can i envision this. This is about solitons and instantons

thanks in advance

regards
marlon

 

[marlon]

PS these questions come from QFT in a Nutshell by A.Zee, page 283

marlon

 

[Hurkyl]

Now, a map between two unit circles (S1 to S1) is topologically stable because this map cannot be transformed into a map where S1 is mapped onto a point of S1. Can someone explain that last sentence, please.

It sounds like they're talking about contractability. Here's an example:

An example of a map from S1 to R^2 is one that maps S1 homeomorphically to a circle centered at the origin. Let f be one such map.

We can smoothly transform f into a constant map as follows: define g(x, t) = (1 - t) f(x) where t ranges over [0, 1]. We have that g(x, 0) = f(x) and g(x, 1) = (0, 0). So, we would say that this map is contractible.

If, however, we mapped into R^2 - (0, 0), then f would not be contractible. (because it "contains" a hole)

A less trivial example would be a map from S^1 into a circle that loops around a torus. (Either meaning to around is fine: they're inequivalent, I think)


Disclaimer: I'm not too familiar with homotopy, and I'm borrowing heavily from the other things I know about topology. I may have the terminology all wrong for this.

A map phi--->v*exp(imx) (where x is the angle and m an integer) represents a map that wraps one circle around the other m times. Does anyone know what this means and how can i envision this.

Think of how the functions z^m behave on the complex plane: it's essentially the same. (it is the same thing when you embed S^1 as the unit circle)

 

[matt grime]

The maps of the circle to some space are the same as the maps from [0,1] to the space with the images of 0 and 1 agreeing. Think of it as inserting a bit of length of elastic into the space and joining up the ends. Two of these loops are topologically the same if they can be deformed continuously into one another. INSIDE the space. Normally we fix the point where 0 and 1 are sent to, it is called the base point.

In the case when we're mapping into the circle we can loop the elastic once, twice, three times, etc round the circle. All these are different. We can send the elastic to a single point, or we can wrap it round in the opposite direction, ie minus once, -2, or -3 times. Thus these maps are the same as the abstract group of integers (i'm omitting details, obviously).

If you're used to winding numbers from, say, complex analysis, that is exactly what we're measuring here - the winding number of something round the origin.

"Now, a map between two unit circles (S1 to S1) is topologically stable because this map cannot be transformed into a map where S1 is mapped onto a point of S1. Can someone explain that last sentence, please."

I can't explain that since it is wrong. Though I admit they are using some odd terminology (bloody applied mathematicians). I think there must need to be some qualifications about what kind of map they're talking about.

Anything else?

 

[marlon]

The maps of the circle to some space are the same as the maps from [0,1] to the space with the images of 0 and 1 agreeing. Think of it as inserting a bit of length of elastic into the space and joining up the ends. Two of these loops are topologically the same if they can be deformed continuously into one another. INSIDE the space. Normally we fix the point where 0 and 1 are sent to, it is called the base point.

Hi matt, thanks for the reply. However, i must admit that i do not understand this. I have read previous explanations like this but i really don't get it. What does that mean : the images of 0 and 1 agreeing... What about this elastic that you insert into the space ? What is this elastic and what is the space ? I really don't get it.

"Now, a map between two unit circles (S1 to S1) is topologically stable because this map cannot be transformed into a map where S1 is mapped onto a point of S1. Can someone explain that last sentence, please."

I can't explain that since it is wrong. Though I admit they are using some odd terminology (bloody applied mathematicians). I think there must need to be some qualifications about what kind of map they're talking about.

Anything else?

Well the field configuration becomes phy --> |v|² at spatial infinity. This is a unit circle S1 and the spatial infinity itself is also a unit circle S1 (in the case of a two dimensional space) Hence the fact that the field MUST become a S1 circle at infinity leads to the map between S1 and S1...

marlon

 

[matt grime]

You don't follow the maths, and I don't follow the physics, so we're stuck: I've no clue as to what the phrase "spatial infinity" might begin to mean.

Let's start properly on the maths. Let X be a topological space, and let x be a distinguished point, eg X=S^1 and x=1.

A path is a continuous map (or function), f, from the interval I=[0,1] to X such f(0)=f(1)=x.

If you like, imagine a particle moving in the space X and at time t=0 it is at x, and at time t=1 it is back at x, inbetween it traces out a path in X.

Elastic is a metaphor for a trace of the path. Two paths are the same if there is a way to continuously deform one to the other, in the metaphor, move the two bits of elastic so that one covers the other exactly, in the technical terminology there is a homotopy between them.

Example X+S^1, x=1, define f(t)=1 for t in I, and g(t) = e(t) for 0<=t<=1/2, and e(1-t) for 1/2<=t<=1, where e(s) = exp{2.s.\pi}


so f would keep the particle fixed at 1, and g moves it around the upper half circle to -1 and then back again. g can be continuously deformed into f without moving outside the circle. Explicitly, define

h(t,s) = g(t) for 0<=t<=s/2, then it stays at g(s/2) for s/2<=t<1-s/2, and for 1-s/2<=t< 1 it is g(t) again, where s runs from 0 to 1, h is continuous and h(t,0) = f(t), h(t,1)=g(t). this is a homotopy.

You need to draw pictures and understand what is going on!

The map e(t), for 0<=t<=1 is not homotopic to f, it cannot be continuosly deformed to be f.

 

[marlon]

You don't follow the maths, and I don't follow the physics, so we're stuck: I've no clue as to what the phrase "spatial infinity" might begin to mean.

Let's start properly on the maths. Let X be a topological space, and let x be a distinguished point, eg X=S^1 and x=1.

A path is a continuous map (or function), f, from the interval I=[0,1] to X such f(0)=f(1)=x.

If you like, imagine a particle moving in the space X and at time t=0 it is at x, and at time t=1 it is back at x, inbetween it traces out a path in X.

Elastic is a metaphor for a trace of the path. Two paths are the same if there is a way to continuously deform one to the other, in the metaphor, move the two bits of elastic so that one covers the other exactly, in the technical terminology there is a homotopy between them.

Example X+S^1, x=1, define f(t)=1 for t in I, and g(t) = e(t) for 0<=t<=1/2, and e(1-t) for 1/2<=t<=1, where e(s) = exp{2.s.\pi}


so f would keep the particle fixed at 1, and g moves it around the upper half circle to -1 and then back again. g can be continuously deformed into f without moving outside the circle.

this is clear now, thanks matt.

Explicitly, define

h(t,s) = g(t) for 0<=t<=s/2, then it stays at g(s/2) for s/2<=t<1-s/2, and for 1-s/2<=t< 1 it is g(t) again, where s runs from 0 to 1, h is continuous and h(t,0) = f(t), h(t,1)=g(t). this is a homotopy.

Howdo you come to this ? You say that the h(t,s)=g(t) for 1-s/2<t<1. But let us take s=1 then we have that 1/2<t<1 and then the h is NOT equal to g(t) because this is only the case for 0<t<1/2, given the definition of h(t,s)

What am i doing wrong here ?

The map e(t), for 0<=t<=1 is not homotopic to f, it cannot be continuosly deformed to be f.

Sorry, i don't see how ?

regards
marlon and thanks again for helping me out

 

[matt grime]

Put s=1, then

h(t,1) = g(t) for 0<=t<=1/2

h(1/2,1) = g(1/2) for 1/2<=t<=1/2

h(t,1) =g(t) for 1/2<=t<=1

so h(t,1)= g(t) for all t in I.

Nothing wrong there.

The picture you should have in you mind is that at time 0 we stay put, then at time s we go an s'th of the way round the upper half circle, stay at the s'th of the way round and then come back, and so on until we reach at time s=1 the loop g given by going half way round and then turning around.



The proof that looping once round the circle is not homotopic (continuously deformable) to the constant map is complicated, but the result is obvious: you have a strip of elastic looped round the circle fixed at either end to the point one. How are you going to shrink it so that it is concentrated at 1 WITHOUT breaking the elastic or moving it outside the circle (which isn't allowed). If you think about it, it is the same as taking the punctured plane, ie one missing the origin. Wrap a loop round the (missing) origin. Sliding and stretching the loop in the plane only, try and shrink it down without passing over the hole at the origin. Cannot be done. The actual mathematical proof is long, about 600 words, or a page in a densely written textbook. You don't want to see it here, honestly. Go buy any decent book on algebraic topology, and it will prove it directly (the long way) or indirectly by creating a lot of machinery. However, if the loop were contractible, then it would follow that the integral of 1/z round the unit circle would be zero, but it isn't, is it?

 

[marlon]

Thanks matt

again a crystal clear explanation

regards
marlon

 

[mathwonk]

the fact that a map from a circle to a circle, whose winding numer, i.e. homotopy class, is non zero, cannot be contracted to a point, is just the fact that if you take a piece of string, wind it around a newspaper some non zero number of times, and then tie it tight, it won't just fall off. I.e. it would have to either pass through ther newspaper, or slide off the end to come off.

neither of those operations is allowed in your homotopy.

 

[mathwonk]

by the way here is the first topic in homotopy theory: why the identity map of the circle to itslef in the plane, is not homotopic to a constant map, as a map into the opunvtured plane, i.e. the unit circle cannot be contracted to a point without passing through the origin.

Problem: prove that a continuous map from the unit disc to the plane, that sends the unit circle identically onto itself, must send some point of the disc to the origin.

Hint: use polar cordinates. This can also be proved using calculus, if you give yourself a smooth, i.e. continuously differentiable map. Then it all follows from the existence of the angle form, "dtheta", and integration i.e. polar coordinates again.

this generalizes to 2 dimensions, the old intermediate value theorem, that a continuous map from the closed unit interval to the line, which sends the end points 0 and 1, to numbers of opposite sign, must send some point of the open unit interval to zero.

 

[mathwonk]

as to winding number or homotopy, think of your belt. if you wrap it one or two or three times around your waist then hook it, it does not just come off. But if you wrap it around one way then back the opposite way the same number of timnes, and then hook it, it could slide off through the belt loops, since the total winding number around your waist is zero.

 

[marlon]

nice mathwonk, thanks

marlon

 

[Hurkyl]

I've no clue as to what the phrase "spatial infinity" might begin to mean.

I think he means the "points at infinity" in a compactification of the plane. In particular, the one where equivalence classes of parallel rays form points. (Or, alternatively, the one analogous to adding the boundary to an open disk)

 

[mathwonk]

I. here is a short argument that a unit circle cannot be pulled away from the origin without passing through the origin. the unit circle has winding number 1 around the origin, and any small enough deformation of the circle will not change that. hence the concept of winding number, which is defined for any deformation of the circle not passing through the origin, is a continuous function.

but if you pulled the circle completely away from the origin in a continuous motion, the winding number would have become zero. that is impossible. a continuous integer valued function on a time interval cannot take values zero and one.


II. to prove it by stokes theorem, note that the winding number, times 2pi, is merely the integral of the form dtheta, over the closed curve. now stokes theorem says that if you deform the circle, without passing through the origin, the integral of dtheta over the deformed circle stays the same. I.e. the deformation sweeps out an annulus in the space missing the origin, and by stokes theorem, the integral of d(dtheta) over the annulus, equals the difference of the integrals of dtheta over the two boundary curves of the annulus. but d(dtheta) = 0, so we are done.

the same stokes theorem argument says we cannot deform the circle to a point without passing therough the origin, i.e. that deformation would sweep out a disc, a shrinking family of circles, in the plane, missing the origin. Then by stokes, the integral of dtheta over the boundary of the disc, equals the integral of d(dtheta) = 0 over the disc. that is a contradiction since the original position of the boundary circle was around the unit circle, and polar coordinates gives that integral as 2pi.

as matt said one could also integrate dz/z and get 2pi i times winding number.


III. To prove it merely using polar coordinates, consider the polar coordinate map from the right half plane to the punctured plane, taking (r,t) to
(rcos(t), rsin(t)).

This takes the vertical segment at r = 1, and from t = 0 to t=2pi, onto the unit circle. i.e. as matt grime said, it takes the two end points to the same point of the plane, thus making a closed loop. now the whole point is to prove, using a little uniform continuity, that any deformation of the circle results fom a deformation of the segment.

i.e. given any continuous motion of the unit circle, not passing through the origin, there is a correponding motion of the segment described above, such that the moving segments map onto the moving closed curves.

then since the endpoints of the segment must always map to the same point of the plane, to give a closed image curve, the end points of the segments can never come together. i.e they always stay separated by a distance of 2pi vertically in the (r,t) plane.

since the segments can never be moved to become one point, the image curve can never become one point. so the unit circle cannot be contracted to a point, without passing through the origin.

so that's three "proofs" in only about a million words.

by the way since winding numbers are computed by path integrals, and path integrals are computed by parametrization, e.g. by polar coordinates, all three proofs are the same.

so here is a different proof due to mo hirsch. if there is a map from the disk to the punctured plane which restricts to the identity on the boiundary circle, then since the punctured plane retracts radially onto the unit circle, we can assume we have a map from the unit disc onto its own boundary circle, which restricts to the identity map from the boundary cirfcle to itself.

now consider a typical point of the circle and ask what its inverse image is under the map from the disc. since the circle is one dimensionala nd the disc two diemnsional, a general point of the circle will have inverse image a one dimensional object, i.e. a union of segments and circles.

but there is some segment since that point we are looking at does map to itself. but where is the other end of that segment? i.e. no other point of the circle maps to that point, since they all map to themselves.

\
but the other endpoint p cannot be inside the disc. i.e. locally, by the inverse function theorem the map looks like as projection near p, so there must be a whole curve through p collapsing to the given point on the circle.

this is a contradiction. that's 4.

 

[mathwonk]

when i was young and energetic, i taught this stuff in calculus of several variables, as an illustration of the power of stokes theorem.

an easy corollary is the fundamental theorem of algebra. another is the fact that there is no smooth family of non zero vectors on a 2 - sphere.

 

[mathwonk]

I have noticed that when I answer a question too thoroughly I get silence. But I never learn.

 

[Hurkyl]

What's wrong with that? Either you've answered all of the audience's questions, or they don't understand one word of it.

 

[mathwonk]

its like singing your best song into a canyon, and getting no echo.

and experience teaches that anything understood provokes questioons, so the troubling suspicion is they haven't understood a word of it.

but as I once argued, that's ok too, since one has provided food for future thought.

 

[mathwonk]

here some exciting news on homotopy groups: poincare's conjecture is proved! I just heard this today, is this old news?

i.e. suppose we say any space where every point has a neighborhood looking like a disc, is a 2 manifold. and we say it is connected if every pair of points can be joined by an arc in the space. and we say it is simply connecte if every closed loop in the space contracts to a point in the space. then the only object with all three proeprties is a sphere.

I.e. the only connected, simply connecetd, 2 manifold, is a sphere.


now ask the same question in one dimension up. i.e. a 3 manmifold is a space where evry point has a neighborhood looking like an open ball, and connected and simply connected are as before. then poincare asked in 1904 whetehr every such space is a "3 -sphere", i.,e. homeomoirphioc to the set in 4 space satisfying the equation

x^2 + y^2 + z^2 + w^2 = 1.


the answer was unknown for almost 100 years, and now is said to be yes!

and the proof uses riemannian geometry and curvature, (although the question does not.)

 

[marlon]

all very well understood mathwonk...

marlon :)

 

[marlon]

I. here is a short argument that a unit circle cannot be pulled away from the origin without passing through the origin. the unit circle has winding number 1 around the origin, and any small enough deformation of the circle will not change that. hence the concept of winding number, which is defined for any deformation of the circle not passing through the origin, is a continuous function.

but if you pulled the circle completely away from the origin in a continuous motion, the winding number would have become zero. that is impossible. a continuous integer valued function on a time interval cannot take values zero and one.


II. to prove it by stokes theorem, note that the winding number, times 2pi, is merely the integral of the form dtheta, over the closed curve. now stokes theorem says that if you deform the circle, without passing through the origin, the integral of dtheta over the deformed circle stays the same. I.e. the deformation sweeps out an annulus in the space missing the origin, and by stokes theorem, the integral of d(dtheta) over the annulus, equals the difference of the integrals of dtheta over the two boundary curves of the annulus. but d(dtheta) = 0, so we are done.

got it

the same stokes theorem argument says we cannot deform the circle to a point without passing therough the origin, i.e. that deformation would sweep out a disc, a shrinking family of circles, in the plane, missing the origin. Then by stokes, the integral of dtheta over the boundary of the disc, equals the integral of d(dtheta) = 0 over the disc. that is a contradiction since the original position of the boundary circle was around the unit circle, and polar coordinates gives that integral as 2pi.

as matt said one could also integrate dz/z and get 2pi i times winding number.

If i am getting this, you mean by this that you cannot "take away" a circle of radius 2 (surrounding the unit circle) for example, whithout having to pass over the origin.
Is that correct ?
marlon

 

[mathwonk]

yes. a circle of radius 2, with center the origin, cannot be continuously moved to lie entirely inside say the circle of radius 5 with center the point (10,0), without passing through the origin.

 

[marlon]

thanks mathwonk

marlon

 

[mathwonk]

do you see how this implies the fundamental theorem of algebra?

 

[marlon]

do you see how this implies the fundamental theorem of algebra?

As i understand it, this theorem states that any polynome can be factored by using complex numbers. There is always a complex root.

The fact that the windingnumber is non-zero implies that the origin is always a part of the disc, right ? Now, this means that any function with winding number non-zero must cross the horizontal x-axis and it must have roots in those points. Is this correct ?

marlon

 

[mathwonk]

a non constant complex polynomial defines a continuous map from the complex plane to the complex plane which you want to prove must hit the origin.

by the winding number ideas above, if you can show that some circle in the plane is mapped so as to wind around the origin, then it follows that some point inside the circle must map onto the origin.

thus the proof reduces to showing that on a large enough circle, the winding number of the image under the polynom,ial, equals the winding number of the image under just the leading term, which is Z^n.

then one uses polar coordinates to actually compute this latter winding number to be 2pi d, where d is the degree of the polynomial.

 

[marlon]

if you can show that some circle in the plane is mapped so as to wind around the origin, then it follows that some point inside the circle must map onto the origin.

Here i don't follow, Why can't a point inside the circle wind around the origin. Or is this just because, if the point inside the circle could wind around the origin, you'd get another circle which is topologically equivalent to the first circle.

Please, come again with the statement on the fundamental theorem of algebra because i think i missed it.

marlon

 

[mathwonk]

how can a point wind around the origin? a loop can wind around a point. apoint cannot.

 

[marlon]

Well, that is indeed quite obvious, so i don't see how the fundamental theorem of algebra is implied ?

marlon

 

[mathwonk]

well where in my argument above did i say anything about a point winding around a point? that was your version.

 

[marlon]

But, if you say that a circle is mapped as to wind around the origin, then way can't a point wind around the origin by this map?

marlon

 

[marlon]

In all honesty, i really don't understand the use of all these elastic strings...

marlon

i never used them in QFT...

 

[mathwonk]

imagine the map defined by the po;ynomial z^n. it takes the unit circle, and wraps it around the origin n times.

thus every circle of radius less than 1, also wraps n times around the origin, unless one of them maps onto the origin.

now this same reasoning applies to a large circle.


take nay polynomial at all of degree n. applying it to a large circle, it behaves much like the simpl,e term z^n, i.e. it winds the large circle n times around the origin.

now our polynomial behaves differently from z^n on the interior of this large circle, but nonetheless, if it maps zero to the origin we are done, and if not then it maps zero to some point say at (10,0) away from the origin. then every small circle maps near that point (10,0) hence with winding number zero.


since the large circle mapped with winding number n, somewhere between the large circle and the small one, some point of the large disc must have crossed over the origin.


i call this the invisible butterfly net proof. i.e. if the boundary of your butterfly net winds around the butterfly, then the buterfly must be inside your net, even if you can't see the net.

But in fact the last one, which ahs radius zero maps onto the origin.


now suppose some other polynomial also maps the unti circle in the same way as z^n does, but behaves differently in the disc. well if the center is not mapped onto the origin, then the center is mapped to some point away from the origin. that means by continuity that any small enough circle is mapped clsoe to that point, hence also well away from the origin. i.e. with winding number zero.


but then some one of the circles between radius 1 and radius zero, must have croissed over the origin, i.e. some point mapped to the origin.

 

[mathwonk]

to answer your question, if T is a map from a circle to the plane that misses the origin, then the circle winds around the origin if the integral of dz/z around the image of the circle is not zero.

how do i define what it means for a point to wind around a point? the integral of dz/z taken over a point is never non zero.


you do not need to think of elastic strings, but it helps if you have done path integral. ??

 

[marlon]

no i got, thanks again...mathwonk

 

[Haelfix]

Marlon if you are interested in a great free online textbook on homotopy/homology/cohomology with lots of pictures (and in my opinion the best book bar none) try Allen Hatchers book.

http://www.math.cornell.edu/~hatcher/AT/ATpage.html

 

[marlon]

Thanks for the great link Haelfix

marlon

 

[mathwonk]

If like me, you find Hatcher's book tedious, I suggest the more elementary book of Artin and Braun, or the differential topology book of Milnor, Topology from the differentiable veiwpoint, or the more detailed rewrite by Guillemin and Pollack, or the nice book on algebraic topology by William Fulton.

But Hatcher's book is free.

Another book I like a lot, is Differential forms in algebraic topology, by Bott / Tu.

My friends who are professional algebraic topologists also praise Hatcher, but I fail to see why. and I even liked Spanier.

Of course I have not read Hatcher closely, nor taught from it. maybe it grows on you. What do you like about it, Haelfix?

another excellent introduction for beginners is Andrew Wallace's book on the fundamental group, or any book by Wallace. Vick's Homology Theory is also easy to get into.

Ok I just looked at hatchers book, at least the first few paghes, and remember why I dislike it. He introduces the mapping cylinder on page 2. Thats ridiculous. No one can appreciate that before seeing a sphere or a torus. and the retraction illustrations are ugly computer generated ones.

You see now how impatient I am as well. No doubt this is just a poor choice of how to frame an introduction, and in the book proper he does better, but so far this is not my idea of pedagogy.

if you want an ideal example of good writing / teaching, see the first two pages of milnor's morse theory. with a few simple minded but attractive pictures he actually conveys almost as much morse theory as most people need, in the first two pages of the introduction.

but probably it is unfair to compare anyone's work to milnor's. even serre comes in second in my opinion. and maybe no one else comes close.

i will look further, but perhaps someone will kindly point me to a recommended well written section of hatcher to examine more closely.


Ok, it is indeed chapter zero which is so off putting, and chapter one is quite reasonable. In my view he should simply have begun with chapter one.

 

[Haelfix]

Well one thing I really like about Hatcher's book is the homework problems. Some of them are damn hard, and if you are able to tackle those, then you pretty much are well off and understand the topic at a high lvl. I remember spending many hours on a few of them.

I will agree some of the discussion gets a little complex, for instance he starts talking about non standard spaces (Eilenberg Maclane spaces) in the appendix of chapter 1. Thats very subtle material, and confused the hell out of me the first time I took the course.

Otoh hand his chapter on homology is very complete. I remember finding the answer to several problems I had (I had this fuzzy misunderstanding relating CW complexes and simplicial complexes), the latter I knew well but couldn't find many places where they talk about the relationship.

 

[mathwonk]

yes as i look further, the book looks useful. he gives a lot of information in a few well chosen words, at least in places.

i do not know that much algebraic topology, and never read any book on it thoroughly.

i did learn some things from greenberg's nice book, which in turn is based on the lovely elementary book of artin and braun.

spivak's differential geometry book, volume 1, has a wonderful chapter on de rham cohomology which is quite readable.

i also worked out the basic techniques of de rham theory for myself when i was teaching calculus of differential forms in order to give applications of stokes theorem to geometry for my students. i still remember how excited i was when i realized that stokes theorem implied the unit circle was not homotopic to a point in the complement of the origin in the plane, and that this implied the fundamental theorem of algebra.

i also took a course from ron stern in which we read great notes of griffiths and morgan on homotopy theory and postnikov towers. the postnikov towers came in very handy years later for me when i was proving some results on the homotopy and cohomology of the moduli space of abelian varieties.

and i had a course from ed brown. brown was a master of homotopy theory and made the material seem very intuitive.

fortunately brown discussed such things as eilenberg maclane spaces, or K(<pi>,n) spaces with only one homotopy group. he built them up by attaching spheres to generate the group and then attaching cells to kill the relations.

afterwards he used these spaces to show cohomology is a representable functor in the homotopy category, a nice technique that was imitated later in algebraic geometry and deformation theory by mike schlessinger and others.

i have just looked at hatcher's discussion of this topic on page 448, and am disenchanted with his book in this section again. i.e. he uses more sophisticated language than necessary, no doubt seking greater efficiency. but i can hardly understand his statement, whereas brown himself made it seem like the most natural thing in the world.

i.e. i do not even know what an omega spectrum is, but i do know what brown theorem says and if i think hard enough to remember the idea, possibly also essentially how to prove it.

i like easy clear versions of things, not sophisticated, technical versions.

as brown put it, his theorem says that the cohomology functor on finite cw complexes, apparently rather techical and contrived, actually "occurs in nature!"

one topic i still feel ill at ease with is characteristic classes, e.g. chern classes, although of course the classifying space approach makes them technically simple to define: just embed your manifold and its tangent spaces in euclidean space and then this defines a clasifying amap from the manifold to a grassmannian, and pull bcak the standard cohomology classes from the grassmannian.

does hatcher explain these too? maybe in his later books on vector bundles.