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.
[Math] Graph of morphism , reduced scheme.
algebraic-geometryschemes
Related Solutions
In general, separated morphisms are stable under composition and under base change. This follows from the definitions, no valuative criterion is necessary here. You can also find this in every detailed introduction to schemes. This implies (c). Open immersions are seperated (in fact, every monomorphism is separated). This implies (a). Affine morphisms are separated. This gives the one direction of (b). Now assume that $X$ is separated over $R$, i.e. $X \to X \times_R X$ is a closed immersion. Then for every pair of open affines $U,V \subseteq X$ we have that $U \times_R V$ is affine, hence the preimage $U \cap V$ is affine and $\Gamma(U \times_R V) = \Gamma(U) \otimes_R \Gamma(V) \to \Gamma(U \cap V)$ is surjective. But this implies that also $\Gamma(U) \otimes_\mathbb{Z} \Gamma(V) \to \Gamma(U \cap V)$ is surjective. From this we see that $X$ is separated over $\mathbb{Z}$. Actually a more general statement is true: If $f \circ g$ is separated, then $g$ is separated (the standard proof factors $g$ over its graph).
You are correct that if $f : X \to Y$ is proper over some base $S$ then it has the property you described. If $Z \subset Y$ is a closed subscheme proper over $S$ then $f^{-1}(Z) \subset X$ is a closed subscheme proper over $S$.
Because $f^{-1}(Z)$ has $f^{-1}(Z)$ has a natural subscheme structure as $f^{-1}(Z) = Z \times_Y X$ which is not always reduced, it is probably better to require that $f^{-1}(Z)$ be proper with this standard subscheme structure.
However, it's also true if we give $f^{-1}(Z)$ the reduced subscheme structure because this is a closed subscheme of $f^{-1}(Z)$ with its standard structure and closed subschemes of proper schemes are proper.
Now let me prove the claim. The map $f^{-1}(Z) \to Z$ is proper because it is the base change of $f : X \to Y$. Therefore the composition $f^{-1}(Z) \to Z \to S$ is proper showing that $f^{-1}(Z)$ is proper over $S$.
However, I do not believe this property is enough to characterize proper morphisms even for varieties over an algebraically closed field. The problem is ``there aren't enough proper closed subvarieties''. For example, $\mathbb{A}^1 \setminus \{ 0 \} \to \mathbb{A}^1$ is not proper but the only proper closed subvarities of $\mathbb{A}^1$ are points whose preimages are also points.
In fact, we can make examples where $f : X \to Y$ is a closed surjective map of varieties over $\mathbb{C}$ with your property but not proper. For example, let $Y$ be the affine nodal curve $y^2 = x^2(x-1)$ and $X = \mathbb{A}^1 \setminus \{ i \}$. The normalization map $\nu : \tilde{Y} \to Y$ where $\tilde{Y} = \mathbb{A}^1$ sends $t \mapsto (t^2 + 1, t(t^2 + 1))$ so take $X \to \tilde{Y} \to Y$ i.e. we remove one of the two preimages of the node. Then $f$ is closed and surjective (even bijective!) pulling back proper closed subvarities (points) to proper closed subvarities. However, $f$ is not proper. To see this, consider $\tilde{f} : X \times \tilde{Y} \to Y \times \tilde{Y}$ and the closed set $$\Delta = \{ (x,x) \mid x \in X \} \subset X \times \tilde{Y} $$ however $\tilde{f}(\Delta) = \{ (f(x), x) \mid x \in X \} \subset Y \times \tilde{Y}$ is the graph of $f$ minus one point and thus not closed.
However, for smooth varieties over $\mathbb{C}$, we can view an algebraic map $f : X \to Y$ as a map of complex manifolds $f^{\mathrm{an}} : X^{\mathrm{an}} \to Y^{\mathrm{an}}$ viewing $X, Y$ as complex manifolds with the analytic topology. Then it is a comforting fact that $f$ is proper iff $f^{\mathrm{an}}$ is proper in the usual topology sense.
Best Answer
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 $X$.