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Suppose we have a space $X$, which has zero cohomology (except in degree zero). Does he neccesarly have zero homology (except in degree zero)?

If not, what if $X$ is a manifold?

Universal Coefficient theorem gives me, that $Hom(H_*(X),\mathbb{Z})$ vanishes, but this isn't enough to conclude, that $H_*(X)$ vanishes, right?

Tina
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  • The universal coefficient theorem gives you a short exact sequence with $H^*(X)$ in the middle. What happens to other terms? – Ayman Hourieh Dec 09 '13 at 20:21
  • These are $Ext(H_{-1}(X),\mathbb{Z})$ and $Hom(H_{}(X),\mathbb{Z})$ and if $H^*(X)$ is zero, of course they are zero too. – Tina Dec 09 '13 at 20:23
  • But why should that mean, that $H_*(X)$ is zero? – Tina Dec 09 '13 at 20:24

1 Answers1

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$\newcommand{\Z}{\mathbb{Z}}$We can assume $X$ path-connected, by restricting to each path-component.

Then yes, vanishing cohomology does imply vanishing homology. In Hatcher's Algebraic topology, proposition 3F.12, it is proven that $\tilde{H}^*(X) = 0$ implies that $\tilde{H}_n(X)$ is finitely generated for all $n$, because otherwise $H^*(X)$ would be uncountable.

Now the universal coefficient theorem tells you that for every $n > 0$, $$\begin{align} \hom(H_n(X), \Z) & = 0 & \mathrm{Ext}(H_n(X), \Z) & = 0 \end{align}$$

But since $H_n(X)$ is a finitely generated abelian group, you can write it as $$H_n(X) = \Z^r \oplus \Z/p_1^{\alpha_1} \oplus \dots \oplus \Z/p_s^{\alpha_s}$$

  • The first equation gives you $\hom(H_n(X), \Z) \simeq \Z^r = 0 \Rightarrow r = 0$;
  • The second equation gives you $\mathrm{Ext}(H_n(X), \Z) \simeq \Z/p_1^{\alpha_1} \oplus \dots \oplus \Z/p_s^{\alpha_s} = 0 \Rightarrow s = 0$.

Therefore $H_n(X) = 0$.

Najib Idrissi
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