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I'm reading Koll'ar and Mori's book about singularity theory. They state the following lemma without proof:

Lemma 2.30. Let $f:Y\rightarrow X$ be a proper birational morphism between normal varieties. Let $\Delta_Y$ resp. $\Delta_X$ be $\mathbb Q$-divisors on $Y$ resp. $X$ such that $$K_Y + \Delta_Y\equiv f^*(K_X+\Delta_X)\quad \text{and}\quad f_*\Delta_Y = \Delta_X$$ Then for any divisor $F$ over $X$, we have $a(F,Y,\Delta_Y) = a(F,X,\Delta_X)$.

My questions are:

1. $F$ is a divisor over $X$ but it may not be a divisor over $Y$. Because when we define $F$, we need a model $g:Z\rightarrow X$ such that $F$ is a divisor on $Z$, but we don't know if $g$ factors through $f$. If $F$ is not over $Y$, how do we define $a(Y,\Delta_Y,F)?$

2. I know how to prove it when $F$ is also over $Y$. But if $F$ is not over $Y$, how do we prove it?

I also asked this question on MathStackExchange but receive no answers.

I'm reading Koll'ar and Mori's book about singularity theory. They state the following lemma without proof:

Lemma 2.30. Let $f:Y\rightarrow X$ be a proper birational morphism between normal varieties. Let $\Delta_Y$ resp. $\Delta_X$ be $\mathbb Q$-divisors on $Y$ resp. $X$ such that $$K_Y + \Delta_Y\equiv f^*(K_X+\Delta_X)\quad \text{and}\quad f_*\Delta_Y = \Delta_X$$ Then for any divisor $F$ over $X$, we have $a(F,Y,\Delta_Y) = a(F,X,\Delta_X)$.

My questions are:

1. $F$ is a divisor over $X$ but it may not be a divisor over $Y$. Because when we define $F$, we need a model $g:Z\rightarrow X$ such that $F$ is a divisor on $Z$, but we don't know if $g$ factors through $f$. If $F$ is not over $Y$, how do we define $a(Y,\Delta_Y,F)?$

2. I know how to prove it when $F$ is also over $Y$. But if $F$ is not over $Y$, how do we prove it?

I also asked this question on mathstackexchange but receive no answers.

I'm reading Koll'ar and Mori's book about singularity theory. They state the following lemma without proof:

Lemma 2.30. Let $f:Y\rightarrow X$ be a proper birational morphism between normal varieties. Let $\Delta_Y$ resp. $\Delta_X$ be $\mathbb Q$-divisors on $Y$ resp. $X$ such that $$K_Y + \Delta_Y\equiv f^*(K_X+\Delta_X)\quad and\quad f_*\Delta_Y = \Delta_X$$ Then for any divisor $F$ over $X$, we have $a(F,Y,\Delta_Y) = a(F,X,\Delta_X)$.

My questions are:

1. $F$ is a divisor over $X$ but it may not be a divisor over $Y$. Because when we define $F$, we need a model $g:Z\rightarrow X$ such that $F$ is a divisor on $Z$, but we don't know if $g$ factors through $f$. If $F$ is not over $Y$, how do we define $a(Y,\Delta_Y,F)?$

2. I know how to prove it when $F$ is also over $Y$. But if $F$ is not over $Y$, how do we prove it?

I also asked this question on mathstackexchange but receive no answers.

I'm reading Koll'ar and Mori's book about singularity theory. They state the following lemma without proof:

Lemma 2.30. Let $f:Y\rightarrow X$ be a proper birational morphism between normal varieties. Let $\Delta_Y$ resp. $\Delta_X$ be $\mathbb Q$-divisors on $Y$ resp. $X$ such that $$K_Y + \Delta_Y\equiv f^*(K_X+\Delta_X)\quad \text{and}\quad f_*\Delta_Y = \Delta_X$$ Then for any divisor $F$ over $X$, we have $a(F,Y,\Delta_Y) = a(F,X,\Delta_X)$.

My questions are:

1. $F$ is a divisor over $X$ but it may not be a divisor over $Y$. Because when we define $F$, we need a model $g:Z\rightarrow X$ such that $F$ is a divisor on $Z$, but we don't know if $g$ factors through $f$. If $F$ is not over $Y$, how do we define $a(Y,\Delta_Y,F)?$

2. I know how to prove it when $F$ is also over $Y$. But if $F$ is not over $Y$, how do we prove it?