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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Sep 21, 2015 at 11:27 comment added David White I like the new title much better.
Sep 21, 2015 at 6:49 answer added Qiaochu Yuan timeline score: 6
Sep 21, 2015 at 6:21 answer added Neil Strickland timeline score: 7
Sep 21, 2015 at 5:09 answer added Greg Friedman timeline score: 7
Sep 18, 2015 at 15:49 answer added David C timeline score: 20
Sep 18, 2015 at 14:41 comment added Vitali Kapovitch @Najib I just mean that if two simply connected spaces have isomorphic minimal models then they are rationally homotopy equivalent. I view the isomorphism class of a minimal model as an algebraic invariant of space here.
Sep 18, 2015 at 14:34 answer added archipelago timeline score: 11
Sep 18, 2015 at 14:28 comment added Najib Idrissi @archipelago What I mean is that if a space $X$ is an Eilenberg-MacLane space of type $K(A,n)$ (and by this I mean $\pi_n(X) = A$, $\pi_{k \neq n}(X) = 0$), then it is weakly homotopy equivalent to all the other spaces of type $K(A,n)$. This is not a priori evident, and the analogue statement for homology isn't true (as homology spheres show), for example.
Sep 18, 2015 at 14:18 history made wiki Post Made Community Wiki by Todd Trimble
Sep 18, 2015 at 14:05 comment added Najib Idrissi @VitaliKapovitch Can you elaborate (about minimal models)? Do you mean the theorem "if two minimal models are quasi-isomorphic then they are isomorphic"?
Sep 18, 2015 at 13:53 history edited Najib Idrissi CC BY-SA 3.0
added 43 characters in body; edited title
Sep 18, 2015 at 13:51 comment added Vitali Kapovitch I agree with Dylan. what you really are asking is when some algebraic topology invariants are complete invariants for a given class of objects and it would be less confusing if you state it using that terminology. Minimal models in rational homotopy theory is one example.
Sep 18, 2015 at 12:00 history asked Najib Idrissi CC BY-SA 3.0