My question arises from proposition 8.3 in the algebraic geometry text by Shigeru Iitaka:
Let $f:W\rightarrow S$ be a surjective morphism between nonsingular projective surfaces. Let $E$ be a divisor on $W$ such that $f(E) =p$ is a point. Then for any divisor $D$ on $S$, one has their intersection number on $W$: $(f^*D, E)_W=0$.
Here is the proof the author gave:
Basically we want to show that $(f^*D,E)_W =\operatorname{deg}(\mathcal{O}(f^*D)|_E)$ is zero. The author claims that $\mathcal{O}(f^*D)|_E \cong \mathcal{O}_E$ and it is proven as follows:
\begin{align*} \mathcal{O}(f^*D)|_E & = (f^* (\mathcal{O}(D))|_E \\ & = f|_E^* (\mathcal{O}(D)|_p) \quad \quad - \star \\ & \cong \mathcal{O}_E \quad \quad -\star\star \end{align*}
Where $f|_E:E\rightarrow \{p\}$ denotes the restriction of $f$ to $E$. I am lost at the steps labelled ‘$\star$‘ and ‘$\star\star$’
Q1) What exactly does $\mathcal{O}(D)|_p$ mean in this context? I believe you need to restrict to a divisor on $S$. But $\{p\}$ is a point in $S$ and so restricting to just merely $’p’$ does not make any sense. Given that this restriction makes sense, then do we jump from the preceding step to $\star$?
Q2) Naturally, I am unable to understand how $\star\star$ is derived, i.e. why is the preceding line isomorphic to $\mathcal{O}_E$.
I believe if I understand how $\star$ is derived I will be able to work out the rest. Would greatly appreciate any help given! Thank you, and wishing you guys a happy new year. :)
Looking at $f^*(\mathcal{O}(D))|_E$, we're pulling back $\mathcal{O}(D)$ along the two maps $E\to W\to S$, and because this composition $E\to S$ factors through $\{p\}$, it will equal the pullback of $\mathcal{O}(D)$ along $E\to \{p\} \to S$, which is what's going on in the first starred step.
In the second starred step, we're just using the fact that $\mathcal{O}(D)|_p$ is a line bundle on a point, so it's trivial, and pulling it back to $E$ again gives the trivial line bundle.