Let $X$ be an algebraic variety (say, smooth and projective) over $\mathbb{C}$, and fix $$\alpha\in H^i(X^{\text{an}}, \mathbb{Q})$$
with $i>0$.



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> Does there always exist a variety $Y$ and a smooth proper morphism $f: Y\to X$ such that $f^\*\alpha$ is integral, i.e., it is in the image of $H^i(Y, \mathbb{Z})\to H^i(Y, \mathbb{Q})$?
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Examples: 


* If $i=1$, $\alpha$ gives a map $\pi\_1(X)\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}$. The kernel of this map defines a (finite) covering $Y$ of $X$. Pulling back to $Y$ makes $\alpha$ integral.
* If $i=2$, there is a map $\mathbb{Q}\to \mathcal{O}\_X\to\mathcal{O}\_X^\*$. Let $\gamma$ be the image of $\alpha$ under the induced map $H^2(X, \mathbb{Q})\to H^2(X, \mathcal{O}\_X^\*)$; this is an analytic Brauer class on $X$. I believe a result of Serre(?) tells us that it is also an etale-cohomological Brauer class. As $X$ is projective, Gabber tells us that $\gamma$ is in fact an honest Brauer class, so we may choose a Severi-Brauer variety $f: Y\to X$ with class $\gamma$. Then $f^\*(\gamma)=0$. If $f^\*\alpha$ is non-zero, it lives in $H^{1,1}(Y)\cap H^2(Y, \mathbb{Q})$. Let $n$ be such that $n\cdot f^\*\alpha$ is integral; then by the Lefschetz $(1,1)$-theorem, there is a line bundle $\mathcal{L}$ on $Y$ with first Chern class $n\cdot f^\*\alpha$. Let $\mathcal{Y}\_\mathcal{L}\to Y$ be the $\mu\_n$-gerbe of $n$-th roots of $\mathcal{L}$. This isn't quite a smooth projective variety over $X$, admittedly, but the pullback of $\alpha$ to $\mathcal{Y}\_\mathcal{L}$ is integral. In any case, I'm fine with $Y$ being a DM stack. (There is always some construction that works involving higher stacks in the analytic category, but this is not what I'm looking for.) I guess I'd prefer $Y$ to be an actual variety, so this argument reduces the $i=2$ case to the following special case:



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> Let $X$ be as above, and let $\mathcal{L}$ be a line bundle on $X$. Fix a positive integer $n$. Is there always some smooth proper $f: Y\to X$ so that $f^\*\mathcal{L}$ has an $n$-th root?
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* If $i>2$, I have no idea what to do.


Some more remarks:


* For varieties so that the cup product $H^1(X, \mathbb{Q})^{\otimes n}\to H^n(X, \mathbb{Q})$ is surjective (say, Abelian varities or curves of genus at least $1$), this is certainly possible. Namely, write $\alpha$ as a linear combination of things coming from $H^1$, and then make all of those integral.
* Qiaochu addresses the topological version. This is already difficult, I think, but it is slightly easier than the algebraic one. For example, I don't know how to make classes in $H^2(\mathbb{P}^1, \mathbb{Q})$ integral, but the Hopf map $S^3\to S^2$ makes every class in $H^2(S^2, \mathbb{Q})$ integral, since $H^2(S^3)=0$.
* To do the case $i=2$, it suffices to consider the case when $X=\mathbb{P}^n$. To see this, observe that my argument above (under Examples, $i=2$) reduces to the case where $\alpha\in H^{1,1}(X)\cap H^2(X, \mathbb{Q})$. Then use that ample classes generate $H^{1,1}(X)\cap H^2(X, \mathbb{Q})$.
* Jason Starr suggests that every smooth projective morphism $Y\to \mathbb{P}^1$ might have a section, which would give a negative answer for $X=\mathbb{P}^1, i=2$. Is that true? I certainly don't know a counterexample...



Making $\mathbb{Q}$-cohomology integral
As Jason Starr remarks in the comments, [this answer](https://mathoverflow.net/a/281793/6950) to a question of his implies the answer to both of my questions is "no." For the latter, one may take:


$X=\mathbb{P}^1, \mathcal{L}=\mathcal{O}(1)$


Then there is no smooth proper morphism $f: Y\to X$ such that $f^\*\mathcal{L}$ has a square root, for example. (Jason's question is about projective morphisms, but I think it's fairly easy to deduce the proper case.)


For the first question, $X=\mathbb{P}^1, i=2$ is a counterexample.


I would be curious to see a simpler proof of this result. I would also be curious to know the answer with the smoothness condition relaxed to flatness.


