I was reading the paper A Characterization of Rational Singularities by Professor Kovács.
The main theorem is stated as follows:
THEOREM 1. Let $\phi: Y \rightarrow X$ be a morphism of varieties over $\mathbb{C}$, and let $\rho:\mathcal{O}_X \to R \phi_* \mathcal{O}_Y$ be the associated natural morphism. Assume that $Y$ has rational singularities and there exists a morphism (in the derived category of $\mathcal{O}_X$-modules) $\rho^{\prime}: R \phi_* \mathcal{O}_Y \rightarrow \mathcal{O}_X$ such that $\rho^{\prime} \circ \rho$ is a quasi-isomorphism of $\mathcal{O}_X$ with itself. Then $X$ has only rational singularities.
The proof goes as follows, first, we take a resolution on both $X$ and $Y$ (denote it $\pi:\tilde{X}\to X$ and $\sigma:\tilde{Y}\to Y$) which makes the diagram commute (i.e. there exist a morphism $\psi: \tilde{Y}\to \tilde{X}$ with $\phi\circ \sigma = \pi \circ \psi$) thus it will induce the following diagram in the derived category:
$$\require{AMScd}
\begin{CD}
\mathcal{O}_X @>{\rho}>> R\phi_*\mathcal{O}_Y;\\
@VVV @VVV \\
R\pi_*\mathcal{O}_{\tilde{X}} @>{\gamma}>> R\phi_*R\sigma_*\mathcal{O}_{\tilde{Y}};
\end{CD}$$
where $\alpha:\mathcal{O}_X\to R\pi_*\mathcal{O}_{\tilde{X}}$ and $\beta:R\phi_*\mathcal{O}_Y\to R\phi_*R\sigma_*\mathcal{O}_{\tilde{Y}}$, by the assumption of the main theorem, we have $\left(\rho^{\prime} \circ \beta^{-1} \circ \gamma\right) \circ \alpha$ is a quasi-isomorphism of $\mathcal{O}_X$ with itself, then the paper state that thus we may assume $\phi$ is the resolution of singularity.
And finally we can apply the standard result to prove $X$ is Cohen-Macaulay and the direct image of canonical sheaf is canonical sheaf.(using some vanishing theorem and quasi isomorphism conditions)
The question is why the quasi isomorphism $\left(\rho^{\prime} \circ \beta^{-1} \circ \gamma\right) \circ \alpha$ implies $\phi$ be a resolution?
By the definition of resolution, $Y$ must be nonsingular, the assumption in the theorem assumes only $Y$ being rational. The quasi-isomorphism condition only implies something on the cohomology level, why does it imply smoothness?
Best Answer
I think he means that instead of the data $(\phi \colon Y \to X, \rho)$ you can consider the data $(\pi \colon \tilde{X} \to X, \rho' \circ \beta^{-1} \circ \gamma)$. The new data satisfy the same assumptions as the old data do, but this time $\pi$ is a resolution.