Chow's Lemma Immersion

Question

I have a question about a step in the proof of Chow's lemma introduced in https://en.wikipedia.org/wiki/Chow%27s_lemma:

After having defined $\phi: U \to P = P_1 \times_S ... \times_S P_n$ induced by given $\phi_i: U \to P_i$.

Futhermore we have the inclusion $i:U \hookrightarrow X$ (=open immersion)

Two questions:

1.: Why is the morphism $\psi:U \to X \times_S P$ induced by $\phi$ and $i$ a immersion?

2.:Why showing $f^{-1}(U) = \psi(U)$ means that $\psi(U)$ is closed in $U \times_S P$?

Answer 1

1. $\psi:U \to X \times_S P$ is immersion because it factors as follows: we have the morphisms $\Gamma:U\hookrightarrow U\times_S P$, $\tilde{i}:U\times_S P \hookrightarrow X\times_S P$. Notice that $\Gamma$ is the graph morphism of the defined morphism $\phi:U\to P$, since $P$ is seperated over $S$, $\Gamma$ is closed embedding. While, as a base change of $i:U\hookrightarrow X$, $\tilde{i}$ is still open immersion. Finally, clearly $\psi=\tilde{i}\circ \Gamma$, thus $\psi$ is an immersion.

2. showing $f^{-1}(U) = \psi(U)$ is aimed at clarifying $f$ becomes isomorphic on the reverse image of $U$. While that $\psi(U)$ is closed in $U \times_S P$ follows directly from the graph morphism argument.