Graph of morphism , reduced scheme.

Let $f:X \rightarrow Y$ be a morphism of S-schemes. Let us suppose that X is reduced and endow the image of the graph morphism $\Gamma_f:X \rightarrow X \times_S Y$ , call it A, with a reduced closed subscheme structure. Show that if Y is separated over S, then the projection $X \times_S Y \rightarrow X$ induces an isomorphism from A onto X.

The graph morphism $\Gamma_f : X \to X \times_S Y$ induces a morphism $g : X \to A$ which satisfies by definition $(p|_A) \circ g = 1_X$ and $(q|_A) \circ g = f$, where $p : X \times_S Y \to X$ and $q : X \times_S Y \to Y$ are the projections. Since $Y$ is separated over $S$, the graph morphism $\Gamma_f$ is in fact a closed immersion; this is easy, see for example (EGA I, (5.2.4)) (in the Springer edition). Hence there exists a closed subscheme $A' \subset X \times_S Y$ and isomorphism $g' : X \stackrel{\sim}{\to} A'$ such that $\Gamma_f$ factors through $g'$ and the inclusion, so $\Gamma_f = g \circ j = g' \circ j'$, where $j : A \hookrightarrow X \times_S Y$ and $j' : A' \hookrightarrow X \times_S Y$ are the inclusions. It follows that $p \circ j \circ g = p \circ j' \circ g'$. Since the former is equal to the identity $1_X$ on $X$, one has $p \circ j' \circ g' = 1_X$; since $g'$ is an isomorphism, $p$ is thus an isomorphism on $A'$. Considered as a map of the underlying topological spaces, $g : X \to A$ is clearly surjective (and hence a homeomorphism) since $A = g(X)$. Consider the homeomorphisms $u = g' \circ g^{-1} : A \to A'$ and $v = g \circ g'^{-1} : A' \to A$ of the underlying spaces of $A$ and $A$'; since $j \circ v = j \circ g \circ g'^{-1} = j' \circ g' \circ g'^{-1} = j'$ and similarly $j' \circ u = j$, one obtains an equality $A = A'$ of the underlying spaces of $A$ and $A'$. Since $A'$ is isomorphic as a scheme to the reduced scheme $X$, and $A$ is by definition the unique reduced subscheme of $X \times_S Y$ having $A$ as its underlying space, it follows that $A$ and $A'$ are indeed the same subscheme of $X \times_S Y$. Finally, one sees that $p \circ j = p \circ j'$ is an isomorphism from $A = A'$ to $X$.
• By considering the morphism $v' : A' \to A$ which is the composite of $g'^{-1}$ and $g$, you can prove $j' = j \circ v$, where $j$ and $j'$ are the inclusions of $A$ and $A'$ into $X \times_S Y$, respectively. This shows $A' \subset A$. – user314 Feb 26 '13 at 17:19
• I believe it works! Thank you. However, couldn't one argue more directly by saying that A and X are both reduced closed subschemes of $(X \times_Y )_{red}$? – Heidar Svan Feb 27 '13 at 12:39