Let $X$ be a normal variety with only one singularity at $x$ and $(X,x)$ is a canonical singularity i.e. $(X.x)$ satisfies $(i)$ and $(ii)$.

$(i)$ $(X,x)$ is a $\mathbb{Q}$ Goreinstein singularity.

$(ii)$ For any resolution $f:Y \rightarrow X$, we can write $K_Y=f^*K_X+\sum_{i=1}^r m_iE_i$ with $m_i\geqq 0$

How can I prove that for any resolution $f:Y\rightarrow X$ and $m\in > \mathbb{Z}_{\geqq 0}$, $f_*\mathcal{O}(mK_Y)\cong \mathcal{O}(mK_X)$?

I don't know even what is a morphism.

Motivation. I want to show the unique ness of "canonical resolution"

A resolution of $f:Y\rightarrow X$ is called canonical resolution if

(i) $Y$ has at worst canonical singularity.

(ii) $K_Y$ is $f$ ample.

My strategy is this.

Let $f_i:Y_i\rightarrow X$ be two canonical resolution.

Make two common resolution $g_1:X\rightarrow Y_1$ and $g_2:X\rightarrow Y_2$.

From (ii), we get $Y_i\cong$ Proj($\oplus_{m\geqq 0}{f_i}*\mathcal{O}(mK_{Y_i})$).

So if we can show ${g_i}_*\mathcal{O}(mK_{X})\cong \mathcal{O}(mK_{Y_i})$, we can show $Y_1\cong Y_2\cong $Proj($\oplus_{m\geqq 0}{g_1}_*{f_1}_*\mathcal{O}(mK_{Y_i})$)

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