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Let $X$ be a (connected) closed $n$-manifold and $G=\pi_1(X)$ be the fundamental group of $X$. There is a classifying map $f: X \rightarrow K(G, 1)$ which induces an isomorphism on $\pi_1$. I would like know when the map $f_*: H_n(X, \mathbb{Z}) \rightarrow H_n(K(G,1), \mathbb{Z})$ is injective, or even when $f_*: H_n(X, \mathbb{Q}) \rightarrow H_n(K(G,1), \mathbb{Q})$ is injective. (Here $K(G, 1)$ is not assumed to be a manifold.)

For example, if $X$ is simply connected, the induced map $f_*$ is the zero map on $H_n$ and when $X$ is the n-torus $T^n$, the induced map $f_*$ is an isomorphism. Is there any condition on $\pi_1(X)$ that will imply injectivity of $f_*$ on $H_n$?

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    $\begingroup$ $f_{\ast}$ 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 $\widetilde{X} \to X \to B \pi_1(X)$. $\endgroup$
    – Qiaochu Yuan
    Commented Feb 25, 2016 at 18:14
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    $\begingroup$ 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. $\endgroup$
    – Jens Reinhold
    Commented Feb 26, 2016 at 3:32
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    $\begingroup$ 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 $\endgroup$
    – Igor Belegradek
    Commented Mar 1, 2016 at 21:59
  • $\begingroup$ Thank you, Igor. Dranishnikov's paper that you referred to is very helpful to me. $\endgroup$
    – awivil
    Commented Mar 1, 2016 at 23:44

2 Answers 2

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There can not be such a condition.

For any finitely presented group $G$, we can find a closed 4-manifold $N$ with fundamental group $G$. In dimension $n \geq 6$, we can now take $M = N \times S^{n-4}$, a $n$-manifold with fundamental group $G$. Then the classifying map $f\colon M \to BG$ of the universal cover $\tilde{M} \to M$ factors through $N$, hence $$f_{\ast}\colon \ H_n(M;\mathbb Z) \to H_n(BG;\mathbb Z)$$ is zero.

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Here is another source of counterexamples. Take $N$ as in Jens Reinhold's answer. Then let $M$ be the connected sum $M=N\#\mathbb C P^2$. It has the same fundamental group $G$ as $N$, and the classifying map $M\to BG$ simply collapses $\mathbb C P^2$ to a point. But $H_2(M)\cong H_2(N)\oplus H_2(\mathbb C P^2)\cong H_2(N)\oplus\mathbb Z$, and the extra $\mathbb Z$ maps to $0\in H_2(BG)$. Examples like this exist in all dimensions $\ge 4$.

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