Question about an inclusion of a closed scheme into an affine scheme 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.
 A: 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.
