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.



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> **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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Is the pull-back of canonical sheaf invertible (modulo torsion)?
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.


