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In the paper "On the realizability of singular cohomology groups" by Kan and Whitehead, it is shown that there is no space $X$ and integer $n\geq 1$ such that $H^{n-1}(X)=0$ and $H^n(X)=\mathbb{Q}$ (cohomology with integral coefficients).

At the very end of the article there is a remark where it is stated that, at the time of writing (around 1960, I suppose), it was not known whether $\mathbb{Q}$ could be a (integral) singular cohomology group at all.

My question is: is this still not known?

Aldo Guzmán Sáenz
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    Maybe I'm missing something, but I always thought that any group can be easily realized as the fundamental group of some space, and the first homology group is just it's abelianization... – Peter Franek Sep 22 '15 at 05:32
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    @PeterFranek Cohomology, not homology. $\Bbb Q$ is not of the form $A = \text{Hom}(G,\Bbb Z)$ because $A$ cannot be a divisible group. (If $g \in G$ and $\varphi(g) = k \neq 0$, then $\varphi$ certainly can't be divisible by anything bigger than $k$.) –  Sep 22 '15 at 06:14
  • @MikeMiller Right, thank you. – Peter Franek Sep 22 '15 at 08:13
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    I think this could get a better answer (or at least a quicker answer) on mathoverflow. – Najib Idrissi Sep 22 '15 at 08:33
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    By the universal coefficient theorem we know $H^n(X;\Bbb Z) = \text{Hom}(H_n(X;\Bbb Z), \Bbb Z) \oplus \text{Ext}(H_{n-1}(X; \Bbb Z),\Bbb Z)$. $\Bbb Q$ is indecomposable so only one of these can be nonzero, and because $\Bbb Q$ is divisible, it cannot be of the form $\text{Hom}(A;\Bbb Z)$. Because $H_{n-1}(X;\Bbb Z)$ can be whatever we want it to be, this question is equivalent to asking if there is some group $A$ with $\text{Ext}(A,\Bbb Z) = \Bbb Q$. Hatcher's theorem 3F.12 gives a basic restriction: if this were true, then $\text{Hom}(A,\Bbb Z)$ would be uncountable. –  Sep 23 '15 at 23:26

1 Answers1

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This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Oscar Randal-Williams below.

This may depend on your axioms, see

S. Shelah "The consistency of Ext(G,Z)=Q", Israel J. Math. 39 (1981), no. 1-2, 74–82.

There it is shown that it is consistent with the generalised continuum hypothesis that there exists a group $G$ having $Ext(G, \mathbb{Z})=\mathbb{Q}$. Then a Moore space $M(G,n-1)$ has the required property.

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Michael Albanese
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