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Let $X=|\Delta|$ be the geometric realization of an abstract simplicial complex $\Delta$. Let $k$ be a field. Assume that $X$ is path connected .

Consider the following two conditions:

(1) $\pi_i (X)=0, \forall 1\le i\le n$

(2) $\tilde H_i(X,k)=0, \forall 0\le i\le n$

I know that (1) implies (2) by Hurewicz theorem and Universal coefficient theorem. My question is : Does (2) imply (1) ? If not, then would some other vanishing result on (reduced) Homology imply (1) ?

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2 Answers 2

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The Poincare Homology sphere has $\tilde H_0$, $H_1$ and $H_2$ trivial, yet has non-trivial $\pi_1$. But the Abelianisation of $\pi_1$ is trivial. I suspect that "(2) implies (1)" is far too optimistic, even if you replace $\pi_1(X)$ by its Abelianisation in (1).

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  • $\begingroup$ Thanks for the interesting example ... is it a simplicial complex ? $\endgroup$
    – user
    Sep 12, 2019 at 22:47
  • $\begingroup$ Every smooth manifold can be triangulated. In the wikipedia page there's a link to a particularly economical triangulation. $\endgroup$ Sep 13, 2019 at 3:01
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If we use $\mathbb Z$-coefficients, then the Hurewicz theorem tells us that (2) implies (1) provided $X$ is simply connected. If it is not, then there are counterexamples (see Lord Shark the Unknown's answer).

If we use coefficients in a field $k$, then it is possible that (2) is satisfied, but $\tilde{H}_i(X) \ne 0$ for some $i \le n$. As an example take the space $X$ obtained by attaching to $S^2$ a cell $D^3$ via map $f : S^2 \to S^2$ of degree $2$. This space is simply connected and has $\pi_2(X) = \mathbb Z_2$. By the Hurewicz theorem $H_2(X) = \mathbb Z_2$ and by the universal coefficient theorem $H_2(X,\mathbb Q) = 0$. See An abelian group A is torsion iff A ⊗ Q = 0.

This example shows that (for $k = \mathbb Q$) even for simply connected $X$ (2) does not imply (1).

However, there exists a "rational Hurewicz theorem" which allows to get information about $\pi_i(X) \otimes \mathbb Q$. See https://en.wikipedia.org/wiki/Hurewicz_theorem.

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  • $\begingroup$ I thought Hurewicz theorem only says that vanishing of the homotopy groups imply vanishing of homology groups and NOT the other way around ... $\endgroup$
    – user
    Sep 13, 2019 at 9:35
  • $\begingroup$ The converse holds for simply connected spaces. See for example Theorem 7.5.5 in Spanier, Edwin H. Algebraic topology. Vol. 55. No. 1. Springer Science & Business Media, 1989. You can also derive it from the wikipedia-article. You can also use Hatcher Theorem 4.32 (prove the converse inductively). $\endgroup$
    – Paul Frost
    Sep 13, 2019 at 9:42

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