I'am reading Gortz, Wedhorn, Algebraic Geometry, p.403 Theorem 13.84
Here, the corresponding morphism $r$ in the statement means that the element of $Hom_S(X, \mathbb{P}(\mathcal{E}))$ corresponding from $\mathcal{H} := ker(f^{*}\mathcal{E}\to \mathcal{L})$ via the bijection in the page 381 of the book.
Anyway, I'm trying to understand the above underliend statement.
My strategy is as follows,
First, fix $s \in S$. Consider next commutative diagram
$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} X':=X \times_{S}k(s) \ \ & \ \ra {i_s^{'}} \ & X & \ra{r} \ & P:= \mathbb{P}(\mathcal{E}) \\ \da{f^{'}} & & \da{f} & & \swarrow{h} & & \\ \operatorname{Spec}k(s) & \ra{i_s} & S & \\ \end{array} $$ where $f^{'}$ is proper since the 'properness' is stable under base change.
Next, note that
$i_s^{'*}(\mathcal{L})$ is $f^{'}$ -ample ( by the proposition 13.64)
$i_s^{*}(\mathcal{E})$ is also quasi-coherent.
There is a surjective homomorphism $f^{'*}(i_s^{*}(\mathcal{E})) \to i_s^{'*}(\mathcal{L})$ (Since there is a surjective homomorphism $f^{*}\mathcal{E} \to \mathcal{L}$ and $i_s^{'*}$ is right exact, $i_s^{'*}(f^{*}\mathcal{E}) \to i_s^{'*}\mathcal{L}$ is also surjective and note that $i_s^{'*}(f^{*}\mathcal{E}) = f^{'*}(i_s^{*}\mathcal{E})$ by the above commutative diagram).
Now let $r_s : X^{'} \to \mathbb{P}(i_s^{*}(\mathcal{E})) = i_s^{*}(\mathbb{P}(\mathcal{E})) := \mathbb{P}(\mathcal{E}) \times_{S} k(s) $ be the corresponding morphism as in the Teorem 13.84( where the first equality is by his book p.215-(8.8.3) and the second equality is by the definition of the inverse image of scheme(his book p.105)). We have a commutative diagram ;
$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} X' \ \ & \ \ra {r_s} \ & \mathbb{P}(i_s^{*}\mathcal{E})=\mathbb{P}(\mathcal{E})\times_{S} k(s) \\ \da{f^{'}} & & \swarrow{h_s} & \\ \operatorname{Spec}k(s) \\ \end{array} $$
, where $i_s^{'*}(\mathcal{L})$ is $f^{'}$ -ample.
Since $s\in S$ is arbitrary, by the reduction assumption, we have that $r_s$ is finite for all $s \in S$. In particular, All $r_s$ is quasi-finite($\because$ Corollary 12.89) and has finite fibers.
Now, note that
- $ X' := X \times_S k(s) \cong f^{-1}(s)$ and $\mathbb{P}(i_s^{*}\mathcal{E})=\mathbb{P}(\mathcal{E})\times_{S} k(s) \cong h^{-1}(s)$ as topological spaces. So, we have $r_s : f^{-1}(s) \to h^{-1}(s)$ for all $s\in S$.
2-1) For all $s\in S$, $im(r|_{f^{-1}(s)}) \subseteq h^{-1}(s)$ ($\because f = h\circ r$)
2-2) For all $s\in S$, $r|_{f^{-1}(s)} = r_s$ ?
Assume that 2-2) is also true. Now let $p \in P:=\mathbb{P}(\mathcal{E}) = \bigcup_{s\in S}h^{-1}(s)$. Then $p\in h^{-1}(s_0)$ for some $s_0 \in S$. Consider $r_{s_0} : f^{-1}(s_0) \to h^{-1}(s_0)$. If the above 2-1), 2-2) are true, then we can show that $$r^{-1}(p) \subseteq r_{s_0}^{-1}(p)$$. So $r^{-1}(p)$ is finite set and we are done.
So in my strategy, key point is relation between $r$ and $r_s$ ; 2-2).
Is 2-2) true? Then how can we show that?
Can anyone help?
Thanks for reading.