# Question about a divisor and its image

Let $X$ be a projective normal variety, $f: X\rightarrow Y$ a proper birational morphism to normal variety $Y$ and let $D$ be a Cartier divisor on $X$.

Write $D=\Sigma D_k$, where the image by $f$ of each component of $D_k$ has codimension $k$ in $Y$.

Let $H\subset X$ be a general very ample divisor. Set $E:=D|_H$.

Question> Why is it that $E_1=(D_1)|_H+(D_2)|_H$?

Think about it this way: Say $\dim X=\dim Y=n$. Then a component $\Delta$ of $D$ is in $D_k$ if and only if the general fiber of $f\left|_{\Delta}\right.$ has dimension $\dim \Delta -(\dim Y-k) = (n-1)-(n-k)=k-1$.

Now if $H$ is ample, then it intersects every positive dimensional subscheme and hence the intersection of $H$ with any positive dimensional fiber has codimension $1$ in that fiber. And this applies to $f\left|_{\Delta}\right.$ as well.

So, when will a component of $E=D\left|_{H}\right.$ be in $E_1$? First notice that since $H$ is general, its intersection with any component of $D$ is irreducible and hence the components of $E$ are just the restrictions of the components of $D$.

So, if $\Delta$ is a component of $D$, then $\Delta\left|_{H}\right.$ will be included in $E_1$ if and only if the general fiber of $f\left|_{\Delta\cap H}\right.$ has dimension $0$.

This can happen two ways:

1. the general fiber of $f\left|_{\Delta}\right.$ has dimension $0$, i.e., $f\left|_{\Delta}\right.$ is also birational and then the same is true for the restrictionn of the map to a general very ample divisor, or
2. the general fiber of $f\left|_{\Delta}\right.$ has dimension $1$, i.e., $\Delta$ is in $D_2$, but then the intersection of $H$ with any (including the general) fiber of $f\left|_{\Delta}\right.$ has codimension $1$, which means it is of dimension $0$.

Voilà.