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(This is a re-post of 1)

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?

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    $\begingroup$ It follows easily from the universal coefficient theorem that this is equivalent to the existence of a group $A$ such that $\operatorname{Ext}(A,\mathbb{Z})\cong\mathbb{Q}$ (as $\mathbb{Q}$ is indecomposable, and $\operatorname{Hom}(A,\mathbb{Z})\cong\mathbb{Q}$ is impossible). $\endgroup$
    – Eric Wofsey
    Commented Sep 23, 2015 at 17:47

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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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    $\begingroup$ That paper also states without proof that it follows from $V=L$ (and hence is consistent with GCH) that there is no such $G$. $\endgroup$
    – Eric Wofsey
    Commented Sep 23, 2015 at 17:54
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EDIT: If I also did more than 1 second worth of checking homological algebra books, I wouldn't immediately find a desired UCT for homology that involves Ext (instead of Tor). So there's a potential gap here, as Oscar points out in the comment:

This is known (and given be a 1 second google search). The assumption on $H^{n-1}(X)$ can be removed, and it has a cute proof:

Use the UCT for cohomology to see that $H^n$ has an Ext term from $H_{n-1}$, and use UCT for homology to see that $H_{n-1}$ has an Ext term from $H^n$, and then note that $Ext(\prod_\mathbb{Z}\mathbb{Q},\mathbb{Z})$ is uncountable (while $\mathbb{Q}$ is countable).

One Remark on the Realizability of Singular Cohomology Groups

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    $\begingroup$ What is the UCT for homology you (or the author of that paper) are using here? $\endgroup$
    – Oscar Randal-Williams
    Commented Sep 23, 2015 at 16:46
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    $\begingroup$ Universal Coefficient Theorem? $\endgroup$
    – Qfwfq
    Commented Sep 23, 2015 at 17:00
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    $\begingroup$ My question wasn't really a question, it was a polite way of saying: the "UCT for homology" which the author of that paper uses is wrong. Apply it to a countably infinite wedge of 2-spheres. $\endgroup$
    – Oscar Randal-Williams
    Commented Sep 23, 2015 at 17:38

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