I would like to prove (or find a counterexample to) the following statement:


Let $X$ be a complex analytic scheme and let $\pi: Y \to X$ be its universal cover. Let $F$ be a coherent sheaf on $X$ and suppose that $L\otimes F\cong F $ for all $L\in \operatorname{Pic}^0(X)$. Then there exists a sheaf $G$ on $Y$, essentially unique up to deck transformations, such that $F\cong \pi\_\*G$. **Edit: In light of Jason's counterexample, I will additionally assume that $\pi\_1(X)$ is Abelian.**


Edit: I am willing to add the following strong assumptions on $X$ (although I'm not sure they are needed). Namely that $X$ is compact and that the normalizations of all irreducible components of $X$ are simply-connected.


Remarks:


(1) I'm also (maybe even more) interested in the algebraic setting and would happy to see the correct formulation of the above statement in the algebraic category.


(2) I stated the above in the generality in which I think it might be true, but the situation I am actually interested is far more specific, so if the above is false, I would be interested in additional hypothesis (on either $F$ or on $X$) such that the statement would be true.


Example:


A simple example is where $X$ is rational curve with one node (e.g. a nodal cubic curve). Then $Y$ is an infinite chain of $\mathbb{P}^1$s meeting in nodes. A torsion free, rank 1, degree 0 sheaf on $X$ is either a degree 0 line bundle (which does not push forward from $Y$ but fails the assumption) or is the push forward of the structure sheaf of one of the components of $Y$. 


In general, I do not want to assume that $X$ is either reduced or irreducible. 



Criteria for a coherent sheaf pushing forward from the universal cover
Let $E$ be the elliptic curve. Let $E\_1$ and $E\_2$ be two different double covers of $E$, with $E = E\_1 /x\_1$ and $E=E\_2/x\_2$ for two-torsion points $x\_1,x\_2$. 


Let $M$ be the minimal resolution of singularities of $ E\_1 \times E\_2 /\langle (a,b) \to (-a,-b), (a,b) \to (a+ x\_1, x\_2-b) \rangle $.


Then $M$ is an Enriques surface, with fundamental group $\mathbb Z/2$. We can embed $E$ into $M$ either horizontally or vertically. In the horizontal case the inverse image in the double cover is $E\_1$, and in the verticle case the inverse image is $E\_2$.


By combining these, we get an embedding of $E$ into $M \times M$, whose inverse image in the universal cover is $E$, mapping to $E$ by the multiplication by two map. 


On the other hand the Picard group of $M \times M$ is $\mathbb Z/2 \times \mathbb Z/2$ (as long as it is defined using rational Chern classes, and not integral Chern classes, for which the statement is trivially false as $\mathcal O\_X$ on $M$ is a counterxample). Pulled back to $E$, these line bundles are all the two-torsion vector bundles.


So the question becomes:



> 
> Is any coherent sheaf on $E$ which is isomorphic to its own tensor product with any two-torsion line bundle equal to a pushforward for $E$ under the multiplication-by-two map?
> 
> 
> 


Then the answer is "no", because indecomposable vector bundles on $E$ are classified by their determinant, so every indecomposable rank two bundle has this property, but none is a pushforward under multiplication by two as that would force it to have rank a multiple of four.


However, it may still be true if the fundamental group is cyclic.


