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jk22
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that is what I am searching.
Careful where you step in : ). In general, this is a very difficult question. But have a look at http://www.math.psu.edu/katok_a/TOPOLOGY/Chapter2.pdf for instance.jk22 said:thanks, and where could i find the ways of calculating the homotopy groups ?
The two-dimensional case is always much simpler in such questions. Also I should have mentionned this: it is very easy to find a manifold that is homotopic but not homeomorphic to a given compact 2-manifold: it just won't be two-dimensional. (Can you see how to do that?)It is then only in dimension 3 that closed manifolds can be homotopic but not homeomorphic ?
jk22 said:thanks, and where could i find the ways of calculating the homotopy groups ?
It is then only in dimension 3 that closed manifolds can be homotopic but not homeomorphic ?
Take a deep breath, and start with www.math.cornell.edu/~hatcher/AT/ATch4.pdf for instance.jk22 said:where could i find the ways of calculating the homotopy groups ?
WWGD said:but trivially not homeomorphic by cardinality reasons alone.
Yes. That is correct.jk22 said:Yes the dimensions have to be the same but i thought R and R2 for example have the same cardinality ?
One thing to note is that the non homeomorphism of ## R^n ## and ## R^m ## is actually a nice early result of homotopy theory : once you remove one point from each, they are homotopic to the corresponding (n-1)-sphere, and the calculation of the lower* homotopy groups of spheres is one of the few easy ones you can make. If you are interested in the subject you should definitely look up this calculation which is found in pretty much any introduction to higher homotopy groups.lavinia said:R and R2 are not homeomorphic.
jk22 said:Thanks for the problems. I wanted to ask a question more : for pi1 we say that we take a loop and describe it by a continuous function f : [0;1]->Manifold such that f(0)=f(1)
But what if there is a knot in the loop we contract in the space ? Is a knot contractible to a point ?
Right!wabbit said:One thing to note is that the non homeomorphism (and non homotopy equivalence) of ## R^n ## and ## R^m ## is actually a nice early result of homotopy theory : once you remove one point from each, they hare homotopic to the corresponding (n-1)-sphere, and the calculation of the lower* homotopy groups of spheres is one of the few easy ones you can make. If you are interested in the subject you should definitely look up this calculation which is found in pretty much any introduction to higher homotopy groups.
* Of course this being homotopy, nothing is easy, which is why I had to add "lower" to this statement : the calculation of ## \pi_k(S^n) ## is easy only for ## k\leq n ## - but this turns out to be sufficient for the purpose at hand.
...except I had to remove the silly statement "non homotopy equivalence of ## R^n ## and ## R^m ## " from that post, but you have now captured my blooper for all posterity to contemplate : )lavinia said:Right!
Bloopers don't matter but I will be happy to delete if you want.wabbit said:...except I had to remove the silly statement "non homotopy equivalence of ## R^n ## and ## R^m ## " from that post, but you have now captured my blooper for all posterity to contemplate : )
I'll leave it to the opening poster to see why it's such a silly statement.
Not at all, I was just kidding, let this remain as a lesson in proof-reading before posting : )lavinia said:Bloopers don't matter but I will be happy to delete if you want.
wabbit said:Not at all, I was just kidding, let this remain as a lesson in proof-reading before posting : )
Oh dear ! I don't even remember what lens spaces are !lavinia said:BTW: Do you know the proof that the lens space L(7,1) andL(7,2) are homotopically equivalent but not homeomorphic?
This shows that homotopy is different from embedding and immersion.jk22 said:$$\mathbb R $$
\mathbb{R} seems to work
I'm completely new to homotopy theory but I do not understand how pi3(s2) can be not trivial since we cannot put a 3-sphere in a 2-sphere ?
Also, jk22, generally you'll find that such "natural, as expected" results do hold in homology theory, which is in a way "homotopy without all the complications", but not in homotopy theory : the homology groups turn out to be more or less the homotopy groups where you remove all cyclical elements, and you might say that "the devil is in the cycles" here.lavinia said:This shows that homotopy is different from embedding and immersion
wabbit said:If I understand what you mean by "it's ## \mathbb{Z}^3 ## in the plane", then no, it isn't. What is ## \pi_1(\mathbb{R}^n) ## ?
.
Right.jk22 said:This should be {0} since every loop is contractible to a point ?
OK, makes more sense. Still wrong, though : )By in the plane i meant a two holed plane.
Sounds right, going around either hole is a generator.jk22 said:So It is the free group F2 for the 2 holed plane ?