Factorization as morphism between schemes vs. as morphism between underlying sets

Question

I'm confused about an argument used by Mohan in his answer of this question.

We have two algebraic curves $C $ and $C'$ (algebraic curve = complete and nonsingular over an algebraically closed field) and it is supposed that $C\times \mathbb{P}^1$ is birational to $C' \times \mathbb{P}^1$. For open $U\subset C\times \mathbb{P}^1$ and $U' \subset C' \times \mathbb{P}^1$ with $U \cong U'$ we hape a composition map $U\to U'\to C'$.

As Mohan correctly observed for every $p \in C$ the induced morphism $U\cap (\{p\}\times\mathbb{P}^1)\to C'$ is indeed constant. From this observation he deduced that then the map $U\to U'\to C'$ factors through rational map $C \to C'$ as scheme morphism. Why we get such observation almost immediately? On set theoretic level such factorization is obvious. Why the factorization works also as morphism of schemes?

Answer 1

Let $f:X\to Y$ be a morphism of schemes. The scheme-theoretic image $Z$ is the smallest closed subscheme of $Y$ through which $f$ factors, and if $f$ is quasi-compact or $X$ is reduced, we have that the underlying topological space of $Z$ is the closure of the set-theoretic image of $f$ (see Stacks, Vakil theorem 8.3.4, other questions on MSE, etc). Further, if $X$ is reduced, the scheme-theoretic image is just the reduced induced structure on $\overline{f(X)}$.

In the linked argument, $C,C'$ are reduced, so the above paragraph applies and the set theoretic factorization immediately implies the scheme-theoretic factorization. Here's how: pick $q\in\Bbb P^1$ so that $C\times q$ has nonempty intersection with $U$; call this intersection $V$. Next, replace $U$ by $(V\times \Bbb P^1)\cap U$ and consider the map $f:U\to V\times C'$ given by the projection on the first coordinate and the map $U\to C'$ on the second coordinate. By the computation that $U\cap (p\times\Bbb P^1) \to C'$ is constant, this map lands in the graph of $g:(V\times q)\to C'$ inside $V\times C'$. As the scheme-theoretic image of $f$ is just the closure of the set-theoretic image equipped with the reduced induced structure, we have that we can upgrade the set-theoretic containment of $f\subset \Gamma_g$ to a scheme-theoretic containment. Since $\Gamma_g\cong V$ because $C'$ is separated, this gives that $U\to C'$ factors through $V$ and hence we have a dominant rational map $C\dashrightarrow C'$ as Mohan claimed.

Answer 2

I didn't look at Mohan's answer very closely, but hopefully these kind of observations will resolve your objections.

Those schemes are, in particular, algebraic varieties over algebraically closed fields, and therefore can be identified with their closed points. The closed points of the fiber product $C \times \mathbb P^1$ over the field can be literally identified with the set-theoretic cartesian product of $C$ and $\mathbb P^1$, a morphism of schemes between algebraic varieties (identified with their closed points) is literally the same thing as a map between the underlying sets satisfiying certain properties, etc. etc.

Other intuitive properties apply, for example if $f: X \rightarrow Y$ is a morphism of varieties between two algebraic varieties, $U \subset X$ and $V \subset Y$ are locally closed sets with $f(U) \subset V$, then $U$ and $V$ have the natural structure of varieties in their own right (as schemes we make them reduced to not allow nilpotents), and $f|_U$ is a morphism of varieties $U \rightarrow V$.