On Thu, Oct 16, 2003 at 02:51:14PM -0700, Eugene Salamin wrote:
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
On Thu, Oct 16, 2003 at 12:27:44PM -0700, Eugene Salamin wrote:
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
(There is some confusion in terminology, between the concepts of "trivial fundamental group" and "homotopy equivalent to S^n". Among topologists, "simply connected" always means the former, as far as I know. The two conditions are equivalent for 3-manifolds.)
What does "homotopy equivalent to S^n" mean?
There are a number of ways to say it. Here's one:
A compact n-dimensional manifold M is "simply connected" if every map of a circle into M can be contracted to a point.
It is "homotopy equivalent to S^n" if in addition, every map of a k-sphere, 1<=k<=n, can be contracted to a point.
But the identity map of the n-sphere into itself is not contractable.
Yes, there's a typo above: I should have written 1<=k<n.
Can we say that two compact n-dimensional manifolds M1 and M2 are homotopy equivalent if for all k, 1<=k<=n the homotopy groups pi[k](M1) and pi[k](M2) are isomorphic?
No, that's not true. What is true is that if there is a map between M1 and M2 which induces the identity on all homotopy groups (not just in this range) then M1 and M2 are homotopy equivalent. (Alternatively, if the M1 and M2 are simply connected and there is a map that induces an isomorphism on homology, then M1 and M2 are homotopy equivalent. Since homology groups H^k(M) vanish for k > n, this is close to your statement above.) (The case of spheres is special, since there is some special behaviour when the homotopy/homology groups are 0.)
(There is a more general notion of homotopy equivalence, which is a little harder to state.)
In fact, it's not so hard to state: M1 and M2 are homotopy equivalent if there are maps f: M1 -> M2 g: M2 -> M1 so that fg and gf can both be deformed to get the identity map. (Apologies for the excessive seriousness of this math...) Peace, Dylan