I am tyring to understand a passage in Mumford's red book (beginning of Theorem 3, II. 5). Let $X = \operatorname{Spec}R$ and let $Y \subseteq X$ be a closed scheme, $f: Y \to X$ is the inclusion. Let $Q$ be the $O_X$-ideal defining $Y$, $A = \Gamma(X, Q)$. He states "First of all, note that $f$ factors through $\operatorname{Spec}R/A$ . For, $f^*: R \to \Gamma(Y, O_Y)$ factors through $R/A$ by definition of $A$."
What is not clear to me is:
What is meant by the "inclusion map"? What does mean for the map of sheaves?
It seems that he is assuming that the map of the global sections $f^*: R \to \Gamma(Y, O_Y)$ is surjective. How do we know this is surjective? I assume that the map of sheaves is surjective, but my understanding is that this does not always mean the map at the open sets are not necessarily surjective.
Thank you.
1) For your inclusion map, go back and re-examine the definition of closed subscheme of $X$; by definition it is a subset $Y$ of $X$ equipped with a sheaf of rings $O_Y$ and a surjective sheaf morphism $\pi:O_X\to O_Y$. So the inclusion map is the pair given by the "topological" inclusion $Y\hookrightarrow X$ and the morphism $\pi$. Right now we do not know what exactly the map of sheaves looks like, but the theorem we are proving is supposed to be proving that locally $\pi$ "looks like" surjections $R\to R/I$.
2) He is not assuming $f^*$ is surjective as far as I can tell. By definition, $$A=\Gamma(X,Q)=\ker(\Gamma(X,O_X)\to\Gamma(Y,O_Y))=\ker(R\overset{f^*}\to\Gamma(Y,O_Y)),$$
and then it follows (a simple ring theory fact, sometimes called the "universal property of quotients") that the homomorphism $R/A\to\Gamma(Y,O_Y)$ given by sending $r+A\mapsto f^*(r)$ a well-defined homomorphism (and, again, this does not use surjectivity). In other words $f^*$ factors through $R/A$.
I hope this addresses your questions; let me know if not.