# resolution of singularities and a projection formula

Let $Y$ be a normal surface and let $p:X\longrightarrow Y$ be a resolution of singularities.

Let $f$ be a rational function on $Y$.

Do we have that $p_\ast$div $(d(f\circ p)) =$ div $df$ as cycles?

I'm guessing this is some kind of projection formula. I'm interested in how general this formula holds (if it holds at all). That is, could I replace $Y$ by any integral normal scheme and $p:X\longrightarrow Y$ by a surjective birational proper morphism with X regular?

-

Since $Y$ is normal, div is only determined outside a set of codim 2. We may also assume that outside this set of codimension 2, $p$ is an isomorphism. We call this open set $U$.
Of course, $p_{*}$ of any divisor on $X$, is also determined on $U$ (which is both an open subset of $X$ and $Y$). In general, $p_{*}$ of a divisor on a map of surfaces (or any varieties), simply throws away any components which are contracted to non-divisors. So at this point, $p_{*}$ of the divisor is determined on an $U \subseteq Y$. The divisor on $X$ is determined on $U$ also.
EDIT: I should point out however that $p_* (O_X(D)) \neq O_Y(p_* D)$ for divisors on $X$. This can happen even in the context you are considering. It depends on the singularities of the varieties in quesiton.