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Let $X$ be a $\mathbb{Q}$-Gorenstein (isolated) singularity of dimension at least $2$ and $f:Y \to X$ be a resolution of singularities. In this case the canonical sheaf $K_X$ is not necessarily invertible, it is only reflexive.

Question. Is the pull-back $f^*K_X$ invertible? If not, can we say that $f^*K_X/\mathrm{tors}$ is invertible?

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I think that $f^*K_X$ is not invertible in general. For instance, take as $X$ a quotient surface singularity of type $\frac{1}{4}(1, \, 1)$. Then straightforward computations give $$K_Y=f^*K_X - \frac{1}{2}E,$$ where $E$ is the exceptional divisor. We infer that $$f^*K_X = K_Y + \frac{1}{2}E$$ is not an invertible sheaf on $Y$. In fact, the exceptional divisor $E$ is not $2$-divisible in $\operatorname{Pic}(Y)$. The way I see this is that $X$ is the singularity given by a cone over a rational normal curve of degree $4$ in $\mathbb{P}^4$, and $Y$ is the blow-up at the vertex. Then $E$ is a section of a (rational) fibration on $Y$, hence it cannot be divisible.

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  • $\begingroup$ Thanks for the answer. Can you also say something about the second question (i.e., pullback modulo torsion is invertible?) $\endgroup$
    – user45397
    Commented Oct 13, 2021 at 9:34
  • $\begingroup$ In my example, it seems to me that $f^*K_X$ is torsion-free, so the answer should be again no. Am I missing something? $\endgroup$ Commented Oct 13, 2021 at 11:58
  • $\begingroup$ I mean, a torsion sheaf is supported on a proper subvariety. But $2f^*K_X=2K_Y+E$ is a line bundle, hence its torsion part is zero. So also the torsion part of $f^*K_X$ is zero. $\endgroup$ Commented Oct 13, 2021 at 12:05
  • $\begingroup$ My problem is, I cannot imagine a torsion-free sheaf of rank 1 over a smooth variety which is not invertible but if we tensor it twice it becomes invertible. I am thinking using the natural short exact sequence that you get from a torsion-free sheaf to its reflexive hull with non-trivial cokernel. I cannot imagine this cokernel vanishing if we tensor it twice. I find the language used in the literatures a bit ambiguous. In particular, a lot of the times by $f^*K_X$ they mean actually the reflexive hull of this sheaf, which is of course invertible. $\endgroup$
    – user45397
    Commented Oct 14, 2021 at 21:16

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