In the Wikipedia page, a complete variety is defined to be a variety $X$ such that for any variety $Y$, the projection map $X \times Y \to Y$ is a closed map. Asking whether this map is closed is related to the natural question of whether the image of a subvariety is still a subvariety. The page also mentions that this is analogous to a characterization of compact topological spaces.
Theorem. Let $X$ be a topological space. Then $X$ is compact if and only if for any topological space $Y$, the projection map $p: X \times Y \to Y$ is a closed map, where $X \times Y$ has the product topology.
Proof. ($\implies$) Assume $X$ is compact, $Z \subseteq X \times Y$, and $y \in Y \backslash p(Z)$. Since $p$ is the projection we know that $X \times \{y\} \subseteq (X \times Y) \backslash p^{-1}(p(Z)) \subseteq (X \times Y) \backslash Z$. By the closeness of $Z$, for any $x \in X$ we can find $U_x$ open in $X$ and $V_x$ open in $Y$ such that $(x,y) \in U_x \times V_x$ and $(U_x \times V_x) \cap Z = \varnothing$. Now $X \times \{y\}$ is compact, so finitely many $U_{x_1} \times V_{x_1}, \dots, U_{x_n} \times V_{x_n}$ cover $X \times \{y\}$. Then $V = \bigcap_{i = 1}^n V_{x_n}$ is open in $Y$. Claim: $V \cap p(Z) = \varnothing$. Suppose not, say $y' \in V \cap p(Z)$. Then $(x', y') \in Z$ for some $x' \in X$. Since $U_{x_1}, \dots, U_{x_n}$ cover $X$, $x' \in U_{x_i}$ for some $i$. But then $(x', y') \in U_{x_i} \times V \subseteq U_{x_i} \times V_{x_i}$, a contradiction to $(U_{x_i} \times V_{x_i}) \cap Z = \varnothing$. Therefore we have found an open neighborhood $V$ of $y$ in $Y$ disjoint from $p(Z)$.
($\impliedby$) This direction is less trivial because of the construction of $Y$. Here I refer to Theorem 3.1.16 in Engelking's General Topology or Lemma 4.3 in Martin Escardó's note.
Let $\mathcal{U} = \{U_\alpha\}$ be an open cover of $X$ and $\widetilde{\mathcal{U}}$ be the set of all finite unions of sets in $\mathcal{U}$. To use the closed projection property, let $Y$ be the space whose points are open subsets of $X$, and define a subset $V \subseteq Y$ to be open if and only if (1) $V$ is upward-closed, i.e. if $U \in V$ and $U' \in Y$ satisfies $U \subseteq U'$ then $U' \in V$, and (2) if $X \in V$ then $V \cap \widetilde{\mathcal{U}} \neq \varnothing$. This defines a topology on $Y$, where the only nontrivial part is prove property (2) for the intersection $V_1 \cap V_2$ for $V_1, V_2$ satisfying (1) and (2): Suppose $X \in V_1 \cap V_2$ so we have $U_1 \in V_1 \cap \widetilde{\mathcal{U}}$ and $U_2 \in V_2 \cap \widetilde{\mathcal{U}}$. By the definition of $\widetilde{\mathcal{U}}$, we have $U_1 \cup U_2 \in \widetilde{\mathcal{U}}$. By property (1) of $U_1$ and $U_2$, we have $U_1 \cup U_2 \in V_1 \cap V_2$, so property (2) is proved. As a notation, for $U \in \widetilde{\mathcal{U}}$, let $V(U)$ be the set of all open subsets of $X$ containing $U$. Then $V(U)$ is clearly open in $Y$.
Let $Z = \{(x, U) \in X \times Y \,|\, x \not \in U\}$. Then $Z$ is a closed subset of $X \times Y$: Pick $(x, U) \in (X \times Y) \backslash Z$, so $x \in U$. Also we have $x \in U'$ for some $U' \in \widetilde{\mathcal{U}}$. Then $(U \cap U') \times V(U \cap U')$ is an open neighborhood of $X \times Y$ containing $(x, U)$, so $Z$ is closed. By the closed projection property, $p(Z)$ is closed in $Y$, so $Y \backslash p(Z) = \{U \in Y \,|\, (X \times \{U\}) \cap Z = \varnothing\}$ is open in $Y$. We see that $U \in Y \backslash p(Z)$ if and only if for any $x \in X$, $x \in U$, i.e. $X \subseteq U$. Then $X \in Y \backslash p(Z)$, so by property (2) we know that $(Y \backslash p(Z)) \cap \widetilde{\mathcal{U}} \neq \varnothing$. Then we have some $X \subseteq U$ for some finite union $U$ of sets in $\mathcal{U}$. Therefore $X$ is a compact topological space. ∎
We need to replace compactness by completeness for varieties because any variety is a noetherian topological space, hence compact.
Resuming to varieties, I want to compare the definition of complete varieties given by Hartshorne's Algebraic Geometry and the definition above. Hartshorne gave the following definitions in Sections II.3-II.4:
(i) An (abstract) variety $X$ is an integral separated scheme of finite type over an algebraically closed field $k$, i.e. $X$ is an integral scheme and we have a separated morphism $X \to \operatorname{Spec} k$ of schemes of finite type.
(ii) A morphism $f: X \to Y$ of schemes is closed iff it is closed as a map of topological spaces.
(iii) A morphism $f: X \to Y$ of schemes is universally closed iff for any morphism $Y' \to Y$ of schemes, the induced projection morphism $f': X \times_Y Y' \to Y'$ is also closed. (Here $X \times_Y Y'$ is the fibred product of $X$ and $Y'$ over $Y$, and the process $X \mapsto X \times_Y Y'$ is called a base extension by $Y' \to Y$.)
(iv) A morphism $f: X \to Y$ of schemes is proper iff it is separated, of finite type, and universally closed.
(v) An (abstract) variety $X$ is complete iff the morphism $X \to \operatorname{Spec} k$ is proper.
Comparing definitions (i), (iv) and (v), we see a lot of redundancy. In fact, the requirement of $X$ being complete says only that the morphism $X \to \operatorname{Spec} k$ is universally closed, that is, for any schemes $Y$ over $k$, the induced morphism $X \times_{\operatorname{Spec} k} Y \to Y$ is also closed. This agree with the definition in the Wikipedia page, if we are talking about abstract varieties.
For a (classical) quasi-projective variety $V$ over an algebraically closed field $k$, as defined in Chapter 1 of Hartshorne's book, we have the associated abstract variety $t(V)$ given by Propositions II.2.6 and II.4.10 We can recover the topological space $V$ by taking the set of closed points of $\operatorname{sp}(t(V))$. Similarly, we can relate the products of quasi-projective varieties over $k$ to fibred products of abstract varieties over $k$.
For convenience, we should denote the fibred product symbol $\times_{\operatorname{Spec} k}$ by $\times_k$.
Proposition. Let $X$ and $Y$ be schemes locally of finite type over $k$. Then there is an 1-1 correspondence between the set of pairs of closed points in the Cartesian product $X \times Y$ and the set of closed points of the fibred product $X \times_k Y$. Furthermore, for quasi-projective varieties $V$ and $W$, there is a homeomorphism between the product $V \times W$ and the subspace of closed points of the fibred product $t(V) \times_k t(W)$.
Proof. Recall that $\operatorname{Spec} k$ is a 1-point space whose structure sheaf is $k$. Recall also that if $X$ is locally of finite type over $k$, $x \in X$ is a closed point and $U$ is an open neighborhood of $x$ such that $(U, \mathscr{O}_X|_U)$ is an affine scheme, $(\varphi, \varphi^\#): (U, \mathscr{O}_X|_U) \to \operatorname{Spec} A$ being the isomorphism of locally ringed space with $A \simeq \mathscr{O}_X(U)$ a finitely generated $k$-algebra, then $x$ corresponds to a maximal ideal of $A$. Such a maximal ideal corresponds to a ring homomorphism $\phi_1: A \to k$ (because the quotient ring is a finite extension of $k$, so it has to be $k$).
Now let $y \in Y$ be a closed point of $Y$ and $(V, \mathscr{O}_Y|_V) \simeq \operatorname{Spec} B$ be an open affine neighborhood of $y$, with $B$ a finitely generated $k$-algebra, and $\phi_2: B \to k$ be the corresponding ring homomorphism. From the construction of the fibred product $X \times_k Y$ we see that $U \times_k V \simeq \operatorname{Spec} (A \otimes_k B)$ is an open affine subscheme of $X \times_k Y$. Inside $U \times_k V$ we have a closed point corresponding to the ring homomorphism $\phi_1 \otimes \phi_2: A \otimes_k B \to k$. Therefore from a pair of closed points $(x, y) \in X \times Y$, we have associated a closed point in $U \times_k V$.
Conversely, if we have a closed point in $U \times_k V$ with the corresponding ring homomorphism $\phi: A \otimes_k B \to k$, then composed with the natural map we get a pair of ring homomorphisms $\phi_1: A \to A \otimes_k B \to k$ and $\phi_2: A \to A \otimes_k B \to k$, which again corresponds to a pair of closed points $(x, y) \in X \times Y$. Clearly these two procedures are inverse to each other, hence we have a 1-1 correspondence, proving the first assertion. We can often write the corresponding closed point in $U \times_k V$ as $(x,y)$.
For quasi-projective varieties $V$ and $W$, the underlying set of the product $V \times W$ is the same as the Cartesian product, but the topology is defined using the Serge embedding, as in Exercise I.3.16. On the other hand, Proposition II.2.6 showed that the product $V \times W$ is homeomorphic to the set of closed points of $t(V \times W)$. Since $V \times W$ is the product in the category of varieties, $t(V) \times_k t(W)$ is the product in the category of schemes over $k$, and the fully faithful functor $t$ preserves products, we see that $t(V) \times_k t(W) \simeq t(V \times W)$ naturally as schemes over $k$ (This is Exercise II.3.23). Then we have the homeomorphism from the product $V \times W$ to the set of closed points of $t(V \times W)$, then to the set of closed points $t(V) \times_k t(W)$. ∎
Proposition. Let $V$ and $W$ be quasi-projective varieties over $k$. Then a morphism $f: V \to W$ of varieties is closed if and only if the associated morphism $t(f): t(V) \to t(W)$ is closed.
Proof. ($\impliedby$) This is trivial if we consider $V$ as a topological subspace of $t(V)$.
($\implies$) On the set level, the points of $t(V)$ are irreducible closed subsets of $V$. A continuous maps certainly sends irreducible subsets to irreducible subsets, and the closure of irreducible subsets are also irreducible, so a continuous map $f: V \to W$ determines the map $t(f): t(V) \to t(W)$ by $t(f)(Y) = \overline{f(Y)}$ if $Y$ is an irreducible closed subset of $V$.
The closed subsets of $t(V)$ are subsets of the form $t(Y)$, where $Y$ is a closed subset of $V$. If $t(Y)$ is a closed subset in $t(V)$, then $Y$ is closed in $V$, so by assumption $Z = f(Y)$ is closed in $W$. We need to show that $t(Z)$ is the image of $t(Y)$ under $t(f)$. If $Y'$ is an irreducible closed subset of $Y$, then $f(Y') \subseteq Z$ so $t(f)(Y') = f(Y') \in t(Z)$. This shows that $t(f)$ maps $t(Y)$ into $t(Z)$. To show that it is surjective, I will use noetherian induction on the noetherian topological space $W$. (See Exercise II.3.16.) Let $\mathscr{P}$ be the property of closed subsets $Z$ that $t(f): t(Y \cap f^{-1}(Z)) \to t(Z)$ is surjective. Then it is vacuously true when $Z = \varnothing$. Now assume $\mathscr{P}$ holds for all proper closed subsets $Z'$ of $Z$. Let $Z'$ be an irreducible closed subset of $Z$. If $Z' = Z$, then $Y \in t(Y)$ and $t(f)(Y) = f(Y) = Z$. If $Z' \neq Z$, then by the inductive hypothesis, $t(f): t(Y \cap f^{-1}(Z')) \to t(Z')$ is surjective. Since $Z' \in t(Z')$, we have some irreducible closed subset $Y'$ of $Y \cap f^{-1}(Z')$ such that $t(f)(Y') = f(Y') = Z'$, and $Y' \in t(Y)$. This shows that $t(f): t(Y) \to t(Z)$ is surjective. Hence $t(f)$ maps $t(Y)$ to the closed subset $t(Z)$. ∎
Using $t(V) \times_k t(W) \simeq t(V \times W)$ again, we obtain the following corollary.
Corollary. Let $V$ and $W$ be quasi-projective varieties over $k$. Then $V \times W \to W$ is closed if and only if the associated morphism $t(V) \times_k t(W) \to t(W)$ of schemes over $k$ is closed. ∎
Therefore, one sees that a quasi-projective variety $V$ is complete if and only if the associated abstract variety $t(V)$ is complete.
Proposition. If $X$ is a complete abstract variety and $Z$ is a closed subscheme of $X$, then $Z$ is proper over $k$. In particular, if $Z$ is an irreducible closed subset of $X$ and $Z$ has the reduced induced closed subscheme structure, then $Z$ is a complete abstract subvariety of $X$.
Proof. We have a closed immersion $f: Z \to X$ where $Z$ has the closed subscheme structure. By Collary II.4.8, $f$ is proper and $X \to \operatorname{Spec} k$ is proper by definition, so the composition $Z \to X \to \operatorname{Spec} k$ is proper. In case $Z$ is both reduced and irreducible, then $Z$ is also integral, so $Z$ is an abstract variety. ∎
Proposition. If $X$ is a complete abstract variety, then any morphism $f: X \to Y$ from $X$ to an abstract variety is closed.
Proof. Since $X$ is a noetherian topological space, any closed subset of $X$ is a finite union of irreducible closed subsets of $X$. Therefore it suffices to show that the image of an irreducible closed subset of $X$ is closed in $Y$.
Let $Z$ be an irreducible closed subset of $X$. Give $Z$ the reduced induced closed subscheme structure, so $Z$ is now a complete abstract variety. The identity morphism $Z \to Z$ and $f: Z \to Y$ induces the graph morphism $\Gamma_f: Z \to Z \times_k Y$ with $p_1 \circ \Gamma_f = \operatorname{id}_Z$ and $p_2 \circ \Gamma_f = f$, where $p_1, p_2$ are projections of the fibred product $Z \times_k Y$. The morphisms $f \circ p_1: Z \times Y \to Z$ and $p_2: Z \times_k Y \to Y$ induce $g: Z \times_k Y \to Y \times_k Y$ with $\pi_1 \circ g = f \circ p_1$ and $\pi_2 \circ g = p_2$, where $\pi_1, \pi_2$ are projections of the fibred product $Y \times_k Y$. Intuitively, we think of $\Gamma_f$ as $(\operatorname{id}_Z, f)$ and $g$ as $(f, \operatorname{id}_Y)$. We see that $f: Z \to Y$ can be factorized as
Proposition. Let $X$ be a quasi-projective abstract variety over an algebraically closed field $k$. Then $X$ is complete if and only if $X$ is projective.
Proof. ($\impliedby$) Since $X$ is of finite type over $k$, $X$ can be covered by finitely many open affine subchemes $(U_i, \mathscr{O}_X|_{U_i}) \simeq \operatorname{Spec} A_i$ where each $A_i$ is a finitely generated $k$-algebra, so $X$ is a noetherian scheme. Then this direction follows from Theorem II.4.9, applied to the morphism $X \to \operatorname{Spec} k$.
($\implies$) By the definition of being quasi-projective, the morphism $X \to \operatorname{Spec} k$ can be factorized into an open immersion $j: X \to X'$ to some scheme $X'$ and a projective morphism $i: X' \to \operatorname{Spec} k$. Then $X'$ is a noetherian scheme. Since $j(X)$ is irreducible, by replacing $X'$ by a closed subscheme $Y'$ whose underlying topological space is an irreducible component of $\operatorname{sp}(X')$, we can assume that $X'$ is irreducible. Since $X$ is complete, the morphism $j: X \to X'$ is closed by the last proposition. But if $j(X)$ is a proper open subset of $X'$, then $j(X')$ cannot be closed in $X'$ because $X'$ is irreducible, a contradiction. So $j(X) = X'$ and $j$ is an isomorphism between $X$ and $X'$, so $X$ is projective. ∎
Finally, let me mention several facts that may be my next post on the topic.
(1) The class of abstract varieties is larger than the class of quasi-projective abstract varieties, and there exists a non-projective complete abstract variety.
(2) On the other hand, Chow's Lemma states that for any complete abstract variety $X$, there exists a projective abstract variety $X'$ and a surjective birational morphism $X' \to X$.
Let $\mathcal{U} = \{U_\alpha\}$ be an open cover of $X$ and $\widetilde{\mathcal{U}}$ be the set of all finite unions of sets in $\mathcal{U}$. To use the closed projection property, let $Y$ be the space whose points are open subsets of $X$, and define a subset $V \subseteq Y$ to be open if and only if (1) $V$ is upward-closed, i.e. if $U \in V$ and $U' \in Y$ satisfies $U \subseteq U'$ then $U' \in V$, and (2) if $X \in V$ then $V \cap \widetilde{\mathcal{U}} \neq \varnothing$. This defines a topology on $Y$, where the only nontrivial part is prove property (2) for the intersection $V_1 \cap V_2$ for $V_1, V_2$ satisfying (1) and (2): Suppose $X \in V_1 \cap V_2$ so we have $U_1 \in V_1 \cap \widetilde{\mathcal{U}}$ and $U_2 \in V_2 \cap \widetilde{\mathcal{U}}$. By the definition of $\widetilde{\mathcal{U}}$, we have $U_1 \cup U_2 \in \widetilde{\mathcal{U}}$. By property (1) of $U_1$ and $U_2$, we have $U_1 \cup U_2 \in V_1 \cap V_2$, so property (2) is proved. As a notation, for $U \in \widetilde{\mathcal{U}}$, let $V(U)$ be the set of all open subsets of $X$ containing $U$. Then $V(U)$ is clearly open in $Y$.
Let $Z = \{(x, U) \in X \times Y \,|\, x \not \in U\}$. Then $Z$ is a closed subset of $X \times Y$: Pick $(x, U) \in (X \times Y) \backslash Z$, so $x \in U$. Also we have $x \in U'$ for some $U' \in \widetilde{\mathcal{U}}$. Then $(U \cap U') \times V(U \cap U')$ is an open neighborhood of $X \times Y$ containing $(x, U)$, so $Z$ is closed. By the closed projection property, $p(Z)$ is closed in $Y$, so $Y \backslash p(Z) = \{U \in Y \,|\, (X \times \{U\}) \cap Z = \varnothing\}$ is open in $Y$. We see that $U \in Y \backslash p(Z)$ if and only if for any $x \in X$, $x \in U$, i.e. $X \subseteq U$. Then $X \in Y \backslash p(Z)$, so by property (2) we know that $(Y \backslash p(Z)) \cap \widetilde{\mathcal{U}} \neq \varnothing$. Then we have some $X \subseteq U$ for some finite union $U$ of sets in $\mathcal{U}$. Therefore $X$ is a compact topological space. ∎
We need to replace compactness by completeness for varieties because any variety is a noetherian topological space, hence compact.
Resuming to varieties, I want to compare the definition of complete varieties given by Hartshorne's Algebraic Geometry and the definition above. Hartshorne gave the following definitions in Sections II.3-II.4:
(i) An (abstract) variety $X$ is an integral separated scheme of finite type over an algebraically closed field $k$, i.e. $X$ is an integral scheme and we have a separated morphism $X \to \operatorname{Spec} k$ of schemes of finite type.
(ii) A morphism $f: X \to Y$ of schemes is closed iff it is closed as a map of topological spaces.
(iii) A morphism $f: X \to Y$ of schemes is universally closed iff for any morphism $Y' \to Y$ of schemes, the induced projection morphism $f': X \times_Y Y' \to Y'$ is also closed. (Here $X \times_Y Y'$ is the fibred product of $X$ and $Y'$ over $Y$, and the process $X \mapsto X \times_Y Y'$ is called a base extension by $Y' \to Y$.)
(iv) A morphism $f: X \to Y$ of schemes is proper iff it is separated, of finite type, and universally closed.
(v) An (abstract) variety $X$ is complete iff the morphism $X \to \operatorname{Spec} k$ is proper.
Comparing definitions (i), (iv) and (v), we see a lot of redundancy. In fact, the requirement of $X$ being complete says only that the morphism $X \to \operatorname{Spec} k$ is universally closed, that is, for any schemes $Y$ over $k$, the induced morphism $X \times_{\operatorname{Spec} k} Y \to Y$ is also closed. This agree with the definition in the Wikipedia page, if we are talking about abstract varieties.
For a (classical) quasi-projective variety $V$ over an algebraically closed field $k$, as defined in Chapter 1 of Hartshorne's book, we have the associated abstract variety $t(V)$ given by Propositions II.2.6 and II.4.10 We can recover the topological space $V$ by taking the set of closed points of $\operatorname{sp}(t(V))$. Similarly, we can relate the products of quasi-projective varieties over $k$ to fibred products of abstract varieties over $k$.
For convenience, we should denote the fibred product symbol $\times_{\operatorname{Spec} k}$ by $\times_k$.
Proposition. Let $X$ and $Y$ be schemes locally of finite type over $k$. Then there is an 1-1 correspondence between the set of pairs of closed points in the Cartesian product $X \times Y$ and the set of closed points of the fibred product $X \times_k Y$. Furthermore, for quasi-projective varieties $V$ and $W$, there is a homeomorphism between the product $V \times W$ and the subspace of closed points of the fibred product $t(V) \times_k t(W)$.
Proof. Recall that $\operatorname{Spec} k$ is a 1-point space whose structure sheaf is $k$. Recall also that if $X$ is locally of finite type over $k$, $x \in X$ is a closed point and $U$ is an open neighborhood of $x$ such that $(U, \mathscr{O}_X|_U)$ is an affine scheme, $(\varphi, \varphi^\#): (U, \mathscr{O}_X|_U) \to \operatorname{Spec} A$ being the isomorphism of locally ringed space with $A \simeq \mathscr{O}_X(U)$ a finitely generated $k$-algebra, then $x$ corresponds to a maximal ideal of $A$. Such a maximal ideal corresponds to a ring homomorphism $\phi_1: A \to k$ (because the quotient ring is a finite extension of $k$, so it has to be $k$).
Now let $y \in Y$ be a closed point of $Y$ and $(V, \mathscr{O}_Y|_V) \simeq \operatorname{Spec} B$ be an open affine neighborhood of $y$, with $B$ a finitely generated $k$-algebra, and $\phi_2: B \to k$ be the corresponding ring homomorphism. From the construction of the fibred product $X \times_k Y$ we see that $U \times_k V \simeq \operatorname{Spec} (A \otimes_k B)$ is an open affine subscheme of $X \times_k Y$. Inside $U \times_k V$ we have a closed point corresponding to the ring homomorphism $\phi_1 \otimes \phi_2: A \otimes_k B \to k$. Therefore from a pair of closed points $(x, y) \in X \times Y$, we have associated a closed point in $U \times_k V$.
Conversely, if we have a closed point in $U \times_k V$ with the corresponding ring homomorphism $\phi: A \otimes_k B \to k$, then composed with the natural map we get a pair of ring homomorphisms $\phi_1: A \to A \otimes_k B \to k$ and $\phi_2: A \to A \otimes_k B \to k$, which again corresponds to a pair of closed points $(x, y) \in X \times Y$. Clearly these two procedures are inverse to each other, hence we have a 1-1 correspondence, proving the first assertion. We can often write the corresponding closed point in $U \times_k V$ as $(x,y)$.
For quasi-projective varieties $V$ and $W$, the underlying set of the product $V \times W$ is the same as the Cartesian product, but the topology is defined using the Serge embedding, as in Exercise I.3.16. On the other hand, Proposition II.2.6 showed that the product $V \times W$ is homeomorphic to the set of closed points of $t(V \times W)$. Since $V \times W$ is the product in the category of varieties, $t(V) \times_k t(W)$ is the product in the category of schemes over $k$, and the fully faithful functor $t$ preserves products, we see that $t(V) \times_k t(W) \simeq t(V \times W)$ naturally as schemes over $k$ (This is Exercise II.3.23). Then we have the homeomorphism from the product $V \times W$ to the set of closed points of $t(V \times W)$, then to the set of closed points $t(V) \times_k t(W)$. ∎
Proposition. Let $V$ and $W$ be quasi-projective varieties over $k$. Then a morphism $f: V \to W$ of varieties is closed if and only if the associated morphism $t(f): t(V) \to t(W)$ is closed.
Proof. ($\impliedby$) This is trivial if we consider $V$ as a topological subspace of $t(V)$.
($\implies$) On the set level, the points of $t(V)$ are irreducible closed subsets of $V$. A continuous maps certainly sends irreducible subsets to irreducible subsets, and the closure of irreducible subsets are also irreducible, so a continuous map $f: V \to W$ determines the map $t(f): t(V) \to t(W)$ by $t(f)(Y) = \overline{f(Y)}$ if $Y$ is an irreducible closed subset of $V$.
The closed subsets of $t(V)$ are subsets of the form $t(Y)$, where $Y$ is a closed subset of $V$. If $t(Y)$ is a closed subset in $t(V)$, then $Y$ is closed in $V$, so by assumption $Z = f(Y)$ is closed in $W$. We need to show that $t(Z)$ is the image of $t(Y)$ under $t(f)$. If $Y'$ is an irreducible closed subset of $Y$, then $f(Y') \subseteq Z$ so $t(f)(Y') = f(Y') \in t(Z)$. This shows that $t(f)$ maps $t(Y)$ into $t(Z)$. To show that it is surjective, I will use noetherian induction on the noetherian topological space $W$. (See Exercise II.3.16.) Let $\mathscr{P}$ be the property of closed subsets $Z$ that $t(f): t(Y \cap f^{-1}(Z)) \to t(Z)$ is surjective. Then it is vacuously true when $Z = \varnothing$. Now assume $\mathscr{P}$ holds for all proper closed subsets $Z'$ of $Z$. Let $Z'$ be an irreducible closed subset of $Z$. If $Z' = Z$, then $Y \in t(Y)$ and $t(f)(Y) = f(Y) = Z$. If $Z' \neq Z$, then by the inductive hypothesis, $t(f): t(Y \cap f^{-1}(Z')) \to t(Z')$ is surjective. Since $Z' \in t(Z')$, we have some irreducible closed subset $Y'$ of $Y \cap f^{-1}(Z')$ such that $t(f)(Y') = f(Y') = Z'$, and $Y' \in t(Y)$. This shows that $t(f): t(Y) \to t(Z)$ is surjective. Hence $t(f)$ maps $t(Y)$ to the closed subset $t(Z)$. ∎
Using $t(V) \times_k t(W) \simeq t(V \times W)$ again, we obtain the following corollary.
Corollary. Let $V$ and $W$ be quasi-projective varieties over $k$. Then $V \times W \to W$ is closed if and only if the associated morphism $t(V) \times_k t(W) \to t(W)$ of schemes over $k$ is closed. ∎
Therefore, one sees that a quasi-projective variety $V$ is complete if and only if the associated abstract variety $t(V)$ is complete.
Proposition. If $X$ is a complete abstract variety and $Z$ is a closed subscheme of $X$, then $Z$ is proper over $k$. In particular, if $Z$ is an irreducible closed subset of $X$ and $Z$ has the reduced induced closed subscheme structure, then $Z$ is a complete abstract subvariety of $X$.
Proof. We have a closed immersion $f: Z \to X$ where $Z$ has the closed subscheme structure. By Collary II.4.8, $f$ is proper and $X \to \operatorname{Spec} k$ is proper by definition, so the composition $Z \to X \to \operatorname{Spec} k$ is proper. In case $Z$ is both reduced and irreducible, then $Z$ is also integral, so $Z$ is an abstract variety. ∎
Proposition. If $X$ is a complete abstract variety, then any morphism $f: X \to Y$ from $X$ to an abstract variety is closed.
Proof. Since $X$ is a noetherian topological space, any closed subset of $X$ is a finite union of irreducible closed subsets of $X$. Therefore it suffices to show that the image of an irreducible closed subset of $X$ is closed in $Y$.
Let $Z$ be an irreducible closed subset of $X$. Give $Z$ the reduced induced closed subscheme structure, so $Z$ is now a complete abstract variety. The identity morphism $Z \to Z$ and $f: Z \to Y$ induces the graph morphism $\Gamma_f: Z \to Z \times_k Y$ with $p_1 \circ \Gamma_f = \operatorname{id}_Z$ and $p_2 \circ \Gamma_f = f$, where $p_1, p_2$ are projections of the fibred product $Z \times_k Y$. The morphisms $f \circ p_1: Z \times Y \to Z$ and $p_2: Z \times_k Y \to Y$ induce $g: Z \times_k Y \to Y \times_k Y$ with $\pi_1 \circ g = f \circ p_1$ and $\pi_2 \circ g = p_2$, where $\pi_1, \pi_2$ are projections of the fibred product $Y \times_k Y$. Intuitively, we think of $\Gamma_f$ as $(\operatorname{id}_Z, f)$ and $g$ as $(f, \operatorname{id}_Y)$. We see that $f: Z \to Y$ can be factorized as
$Z \overset{\Gamma_f}{\longrightarrow} Z \times_k Y \overset{g}{\longrightarrow} Y \times_k Y \overset{\pi_i}{\longrightarrow} Y$
where $\pi_i$ can be $\pi_1$ or $\pi_2$. Since $Y$ is separated over $k$, the diagonal morphism $\Delta: Y \to Y \times_k Y$ is a closed immersion, and its image $\Delta(Y)$ is closed in $Y \times_k Y$, so $g^{-1}(\Delta(Y))$ is closed in $Z \times_k Y$. We see that on the set level the image $f(Z) = p_2(\Gamma_f(Z)) = p_2(g^{-1}(\Delta(Y))$ is the image of a closed subset of $Z \times_k Y$ under the projection $p_2: Z \times_k Y \to Y$, so it is closed in $Y$ by the completeness of $Z$. ∎
The next proposition says that a quasi-projective variety $V$ is complete if and only if it is projective.
Proof. ($\impliedby$) Since $X$ is of finite type over $k$, $X$ can be covered by finitely many open affine subchemes $(U_i, \mathscr{O}_X|_{U_i}) \simeq \operatorname{Spec} A_i$ where each $A_i$ is a finitely generated $k$-algebra, so $X$ is a noetherian scheme. Then this direction follows from Theorem II.4.9, applied to the morphism $X \to \operatorname{Spec} k$.
($\implies$) By the definition of being quasi-projective, the morphism $X \to \operatorname{Spec} k$ can be factorized into an open immersion $j: X \to X'$ to some scheme $X'$ and a projective morphism $i: X' \to \operatorname{Spec} k$. Then $X'$ is a noetherian scheme. Since $j(X)$ is irreducible, by replacing $X'$ by a closed subscheme $Y'$ whose underlying topological space is an irreducible component of $\operatorname{sp}(X')$, we can assume that $X'$ is irreducible. Since $X$ is complete, the morphism $j: X \to X'$ is closed by the last proposition. But if $j(X)$ is a proper open subset of $X'$, then $j(X')$ cannot be closed in $X'$ because $X'$ is irreducible, a contradiction. So $j(X) = X'$ and $j$ is an isomorphism between $X$ and $X'$, so $X$ is projective. ∎
Finally, let me mention several facts that may be my next post on the topic.
(1) The class of abstract varieties is larger than the class of quasi-projective abstract varieties, and there exists a non-projective complete abstract variety.
(2) On the other hand, Chow's Lemma states that for any complete abstract variety $X$, there exists a projective abstract variety $X'$ and a surjective birational morphism $X' \to X$.
