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?