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Mar 1, 2016 at 23:49 vote accept awivil
Mar 1, 2016 at 23:44 comment added awivil Thank you, Igor. Dranishnikov's paper that you referred to is very helpful to me.
Mar 1, 2016 at 21:59 comment added Igor Belegradek If the rational fundamental class of $X$ survives under the classifying map to $K(\pi_1(X),1)$, the manifold is called rationally essential. Gromov and more recently Dranishnikov studied the interplay between being rationally essential and the the so called "macroscopic dimension". See Dranishnikov's papers at arxiv: arxiv.org/find/grp_math/1/au:+dranishnikov/0/1/0/all/0/1
Mar 1, 2016 at 21:36 answer added Sebastian Goette timeline score: 2
Mar 1, 2016 at 21:17 answer added Jens Reinhold timeline score: 6
Feb 26, 2016 at 3:32 comment added Jens Reinhold It definitely also fails for any manifold of the form $X = N \times S^1$ with $N$ simply-connected, just because $B\mathbb Z = S^1$ has trivial homology above degree 1.
Feb 25, 2016 at 18:14 comment added Qiaochu Yuan $f_{\ast}$ is an isomorphism for the torus because the torus is already the classifying space of its fundamental group. And note that if $X$ has any nontrivial rational homology then the rational map can't be injective if $G$ is finite (say for $X = \mathbb{RP}^3$). In general, look at the Serre spectral sequence of the fibration $\widetilde{X} \to X \to B \pi_1(X)$.
Feb 25, 2016 at 17:50 review First posts
Feb 25, 2016 at 18:16
Feb 25, 2016 at 17:49 history asked awivil CC BY-SA 3.0