# Why the following quasi isomorphism implies the morphism to be a resolution (a step in the paper A characterization of rational singularities)

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?

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.