projection formula for birational map Let $X$ be a smooth projective variety of dimension $n$ , and $D$ be a divisor.  Suppose the linear system $|D|$ induce a birational map $$f: X -\to Y, $$ and let $H$ be the very ample line bundle such that $f^*H=D$.
I was wondering if the projection formula is still valid in this case? To be precise, I what to have the following two equations:
$$D^n = (f^*H)^n=H^n,$$ and $$E\cdot H^{n-1} = 0$$ where $E$ is an exceptional divisor of $f^{-1}$.
Moreover, I was wondering if $Y$ is a normal variety in the above setting? 
Any suggestion/comment are very welcome!!!
 A: I don't think this is true.  A simple example is that the self-intersection of a divisor changes when you take its strict transform under a flop.  In more detail:
Suppose that $\phi : X \dashrightarrow Y$ is a standard flop between smooth threefolds.  Here's what I mean: there is a rational curve $C \subset X$ with normal bundle $\mathcal O(-1) \oplus \mathcal O(-1)$, we blow it up to get $f : Z \to X$ with an exceptional divisor  $E$ isomorphic to $\mathbb P^1 \times \mathbb P^1$, which can be blown down along the other ruling via $g : Z \to Y$, contracting $E$ to a curve $C^+$.  The induced map $\phi : X \dashrightarrow Y$ is birational, with no exceptional divisors in either direction.
Take $H$ very ample on $Y$, with $D$ its pullback via $\phi$ to $X$ (equivalently, its strict transform).  This $D$ has base locus along the flopping curve, and its linear system gives the flop we want.
Now, $f^\ast D = g^\ast H + aE$, where $a = H \cdot C^+$ (which is greater than $0$, since $H$ is ample).  Take the top self-intersection of both sides.  On the left, we just have $D^3$.  On the right, we have $H^3 + 3 a \, (g^\ast H)^2 E + 3 a^2 \, (g^\ast H) E^2 + a^3 E^3 = H^3 -3a^2  \, (H \cdot C^+) + 2a^3 = H^3 - a^3$.  So $D^3 \neq H^3$.
A: Since you write $D=f^*H$ I think you are assuming that $|D|$ is base point free and $f$ is a morphism (otherwise you must blow up the base locus to achieve this).
In this case we have $D^n=deg(f) H^n$ where $deg(f)$ is the generic degree.
$E\cdot H^{n-1}=0$ is clear as we can choose generic members $H_i\in|H|$ so that $H_1\cap \ldots \cap H_{n-1}\cap f(E)=\emptyset $.
$Y$ is not necessarily normal (just pick a $Y$ singular cubic and $X$ its normalization and $|D|=f^*|O_Y(1)|$. However, if you consider $|tD|$ for $t\gg 0$, then the image is normal.  
