Why are projective morphisms closed? It is a well-known fact that if $X$ is a projective curve and $p \in X$ a smooth point, then any rational map $X \to Y$, $Y$ a projective variety, extends to a rational map $X \to Y$ regular at $p$. This is proposition I.6.8 in Hartshorne (in the case of $X$ an abstract non-singular curve), for example. However, the two proofs I have seen both assume that it suffices to consider the case $Y = \mathbb{P}^n$. As I understand it, this is because morphisms of projective varieties are proper, and in particular the image is closed. Where I can find a proof of this, in the case of varieties only? I found a proof here by Akhil Mathew, but I got lost when he started talking about base change.
 A: A) If you want to extend your rational map $f: X\to Y$ defined outside of $p$ across $p$, it is true that you can  asssume that $Y=\mathbb P^n$ by embedding $Y$ into $\mathbb P^n$ as a closed set. This however has nothing to do with the completeness of $\mathbb P^n$ but follows from topology. Indeed, if $f$ sends $X\setminus \{p\}$ into $Y$, it will send the  point $p$, which is in the closure of
$X\setminus \{p\}$, into the closure of $f(X\setminus \{p\})$ and so in particular into $Y$.
B) Still  completeness of $\mathbb P^n$ is a fundamental result that is proved  here.
This is a hand-out from a course given in 2009 by Mike Artin: I can't think of anybody more competent than him but I don't know if he wrote the note himself or if it was scribed by a student. Anyway,  here is a link to the whole course, with many hand-outs, notes and exercises.
C) Addendum Not only don't we need completeness of $\mathbb P^n$ but the extension result is quite intuitive. In the complex case we would say that locally around $p=0$ the rational map can be written $z\mapsto F(z)=(f_0(z):...:f_n(z))$ where each $f_i(z)$ is identically zero or meromorphic of the form $f_i(z)=z^{n_i}u_i(z)$ for some $i\in \mathbb Z$ and $u_i(0)\neq 0$. If $n_0$ , say, is the smallest of the $n_i$'s ($n_0 \leq n_i$)   then by multiplying all the homogeneous coordinates  of $F(z)$ by $f_0 ^{-1}(z)=z^{-n_0} u_0^{-1}(z)$ we obtain $F(z)=(1:g_1(z):...:g_n(z))$ where the $g_i$'s are now holomorphic near $0$ so that  the conclusion follows: $F$ can clearly be extended regularly across $p=0$.
In the purely algebraic case we adapt this idea by replacing $z$ by a uniformizing parameter near $p$, the $u_i$'s by units in  $\mathcal O_p^\ast$ and the exponents $n_i$ of $z$ by the valuations of $f_i\in \mathcal O_p$,  which is a discrete valuation ring thanks to the assumed  regularity of $p$.
I have written this sketch because the technicalities of some proofs in the literature might obscure the actual naturality of this extension result.
A: Let $k$ be an algebraically closed field.
Let $X$ be a reduced separated scheme of finite type over $k$.
Let $X_0$ be the set of closed points of $X$.
By this question, $(X_0, \mathcal{O}_X|X_0)$ is a variety in the sense of Serre's FAC.
Conversely let $V$ be a variety over $k$ in Serre's sense.
By this question, there exists a reduced separated scheme $X$ of finite type over $k$ such that $V$ is identified with $(X_0, \mathcal{O}_X|X_0)$.
Such $X$ is essentially unique(i.e. they are all $k$-isomorphic).
Let $V$ be a projective variety.
Let $W$ be a variety in Serre's sense.
Let $p\colon V\times W \rightarrow W$ be the projection map.
We will prove that $p$ is a closed map.
There exists a projective scheme $X$ over $k$ such that $V$ is isomoprphic to $(X_0, \mathcal{O}_X|X_0)$.
Similiarly there exists a scheme $Y$ over $k$ such that $W$ is isomoprphic to $(Y_0, \mathcal{O}_Y|Y_0)$.
Let $\pi\colon X\times_k Y \rightarrow Y$ be the projection.
Since $X$ is proper over $k$(Hartshorne, Theorem 4.9, Chapter II, or the theorem below), $\pi$ is a closed morphism.
Since $(X\times_k Y)_0 = X_0 \times Y_0$ as a set, $p$ is a closed map by this question.
Let $f\colon V \rightarrow W$ be a morphism.
Let $\Gamma = \{(x, f(x)) \in V \times W|\ x \in V\}$.
Let $\Delta = \{(y, y) \in W \times W|\ y \in W\}$.
Consider the map $f \times id_W\colon V\times W \rightarrow W \times W$.
Then $\Gamma = (f \times id_W)^{-1}(\Delta)$.
Since $\Delta$ is closed in $W\times W$, $\Gamma$ is closed in $V\times W$.
Hence $f(V) = p(\Gamma)$ is closed in $W$.
Since any closed subset of $V$ is a projective variety, $f$ is a closed map as desired.
Theorem
The projective scheme $\mathbf{P}_{\mathbf{Z}}^n$ is proper over $\mathbf{Z}$.
Proof(EGA II, theorem 5.5.3):
We need to show that the projection $\mathbf{P}_{\mathbf{Z}}^n \times Y \rightarrow Y$ is a closed morphism for any scheme $Y$ over $\mathbf{Z}$.
Since the problem is local on $Y$, we may assume that $Y$ is affine.
Let $Y = Spec(A)$.
Then $\mathbf{P}_{\mathbf{Z}}^n \times Y = \mathbf{P}_A^n = Proj(A[X_0,\dots,X_n))$.
Let $Z$ be a closed subset of $Proj(A[X_0,\dots,X_n))$.
Then $Z$ is isomorphic to $Proj(A[X_0,\dots,X_n]/I)$, where $I$ is a homogeneous ideal.
Let $S = A[X_0,\dots,X_n]/I$.
Let $f\colon Proj(S) \rightarrow Y$ be the canonical homomorphism.
It suffices to prove that $f(Proj(S))$ is closed in $Y$.
Let $y \in Y$.
$f^{-1}(y) = Proj(S)\times_Y Spec (k(y)) = Proj(S\otimes_A k(y))$.
Hence $f^{-1}(y) = \emptyset$ if and only if $S_n\otimes_A k(y) = 0$ for all sufficiently large $n$.Since $S_n$ is an $A$-module of finite type, $S_n\otimes_A \mathcal{O}_y$ is an $\mathcal{O}_y$-module of finite type. Hence $S_n\otimes_A k(y) = 0$ is equivalent to $S_n\otimes_A \mathcal{O}_y = 0$ by Nakayama's lemma. This is equivalent to that $\mathfrak{p}_y$ does not contain $\mathfrak{a}_n$, where $\mathfrak{p}_y$ is the prime ideal corresponding to $y$ and $\mathfrak{a}_n$ is the annihilator of $A$-module $S_n$.
Since $S_n = S_{n}S_1$, $\mathfrak{a}_n \subset \mathfrak{a}_{n+1}$.
Let $\mathfrak{a} = \bigcup_n \mathfrak{a}_n$.
Then $f(Proj(S)) = V(\mathfrak{a})$.
QED
A: We use the theory of specializations developed in Weil's Foundations of Algebraic Geometry, Chapter II. The main result of the theory of specializations(Lemma 7) below can be proved by "valuation theory + the axiom of choice". However, our proof(which is essentially the same as Weil's) does not use the axiom of choice.
Notation
Let $X$ be a topological space.
Let $E$ be a subset of $X$.
We denote by $cl(E)$ the closure of $E$ in $X$.
Lemma 1
Let $X$ be a topological space.
Let $x \in X$.
Then $cl(\{x\})$ is irreducible.
Proof:
Clear.
Definition 1
Let $X$ be a topological space.
Suppose there exists $x \in X$ such that $X = cl(\{x\})$.
We say $x$ is a generic point of $X$.
Definition 2
Let $X$ be a topological space.
Let $x, y \in X$.
Suppose $y \in cl(\{x\})$.
Then we say $y$ is a specialization of $x$ and write $x \rightarrow y$.
We fix an algebraically closed field $\Omega$ which has infinite trancendence dimension over the prime subfield.
Let $k$ be a subfield of $\Omega$ such that tr.dim $\Omega/k = \infty$.
Let $E$ be a subset of a polynomial ring $k[X_1,\dots,X_n]$.
We denote by $V(E)$ the common zeros of $E$ in $\Omega^n$.
It is easy to see that by taking subsets of the form $V(E)$ as closed sets, we can define a topology on $\Omega^n$.
We call this topology $k$-topology.
A closed(resp. open) subset of $\Omega^n$ with respect to $k$-topology is called $k$-closed(resp. $k$-open) subset.
Let $E$ be a subset of $\Omega^n$.
We denote by $I(E)$ the set $\{f \in k[X_1,\dots,X_n] |\ f(x) = 0$ for all $x \in E\}$.
$I(E)$ is an ideal of $k[X_1,\dots,X_n]$.
It is easy to see that $cl(E) = V(I(E))$.
In particular, $cl(\{x\}) = V(I(\{x\}))$ for $x \in \Omega^n$.
Let $x, y \in \Omega^n$.
If $y$ is a specialization of $x$ with respect $k$-topology, we write $x \rightarrow_k y$ or simply $x \rightarrow y$ if there is no ambiguity.
Hence $x \rightarrow_k y$ if and only if $f(y) =0$ for any $f \in k[X_1,\dots,X_n]$ such that $f(x) = 0$.
Lemma 2
Let $V$ be a $k$-irreducible subset of $\Omega^n$.
Then $V$ has a generic point(this is one of the reasons why we need tr.dim $\Omega/k = \infty$).
Proof:
Let $\mathfrak{p} = I(V)$.
$\mathfrak{p}$ is a prime ideal.
Let $K$ be the field of fractions of $k[X_1,\dots,X_n]/\mathfrak{p}$.
Since tr.dim $K/k \le n$, there exists an $k$-embedding $\sigma\colon K \rightarrow \Omega$.
Let $\bar x_i$ be the image of $x_i$ by the canonical homomorphism $k[X_1,\dots,X_n] \rightarrow 
k[X_1,\dots,X_n]/\mathfrak{p}$.
Let $y = (\sigma(\bar x_1),\dots \sigma(\bar x_n))$.
Then $\mathfrak{p} = I(\{y\})$.
Hence $cl(\{y\}) = V(\mathfrak{p}) = V$.
QED
We consider the set $\Omega_{\infty} = \Omega \cup \{\infty\}$, where $\infty$ is an element which does not belong to $\Omega$.
Let $\rho\colon \Omega_{\infty} \rightarrow \Omega_{\infty}$ be the map defined as follows.
$\rho(x) = 1/x$ if $x \in \Omega - \{0\}$.
$\rho(0) = \infty$.
$\rho(\infty) = 0$.
Clearly $\rho$ is a bijection.
Let $I = \{1,\dots,n\}$.
For $i \in I$, let $\rho_i\colon \Omega_{\infty}^n \rightarrow \Omega_{\infty}^n$ be the map defined by $\rho_i(x_1,\dots,x_i,\dots,x_n) = (x_1,\dots, \rho(x_i), \dots,x_n)$.
Clearly $\rho_i$ is a bijection and $\rho_i\rho_j = \rho_j\rho_i$ for any $i, j \in I$.
Let $G$ be the subgroup of the symmetric group on $\Omega_{\infty}^n$ generated by the set $\{\rho_1,\dots,\rho_n\}$.
Let $J$ be a subset of $I$.
We define $\rho_J = \prod_{i\in J} \rho_i$
Clearly every element of $G$ can be uniquely written as $\rho_J$ for some $J \subset I$.
Hence $|G| = 2^n$.
Lemma 3
Let $x, x' \in \Omega^n$.
Let $\sigma \in G$.
Let $y = \sigma(x)$, $y' = \sigma(x')$.
Suppose $y$ and $y'$ are contained in $\Omega^n$.
Then $x \rightarrow_k x'$ if and only if $y \rightarrow_k y'$.
Proof:
Without loss of generality, we can assume that $\sigma = \rho_J$, where $J = \{1,\dots,m\}$.
Suppose $x \rightarrow_k x'$.
It suffices to prove that $y \rightarrow_k y'$.
Note that none of $x_1,\dots,x_m$ and $x_1',\dots, x_m'$ are $0$, since $y$ and $y'$ are contained in $\Omega^n$.
Suppose $f(y) = 0$ for $f \in k[X_1,\dots,X_n]$.
Then $g = (X_1\dots X_m)^k f(1/X_1,\dots,1/X_m, X_{m+1},\dots,X_n) \in k[X_1,\dots,X_n]$ for some integer $k > 0$.
Since $f(y) = 0$, $g(x) = 0$.
Since $x \rightarrow_k x'$, $g(x') = 0$.
Hence $f(y') = 0$.
Hence $y \rightarrow y'$.
QED
Definition 3
Let $x, x' \in \Omega_{\infty}^n$.
Suppose there exists $\sigma \in G$ such that $\sigma(x), \sigma(x') \in \Omega^n$ and $\sigma(x) \rightarrow_k \sigma(x')$.
Then we say $x'$ is a specialization of $x$ over $k$ and write $x \rightarrow_k x'$.
By Lemma 3, this does not depend on the choice of $\sigma$.
Lemma 1 also shows that, if $x, x' \in \Omega^n$ and $x \rightarrow_k x'$ in the new definition, $x \rightarrow_k x'$ in the old definition(Definition 2).
Lemma 4
$0$ has only one specialization over $k$, namely $0$.
Hence $\infty$ has only one specialization over $k$, namely $\infty$.
Proof:
$I(\{0\}) = (X)$.
Hence $V((X)) = 0$.
Hence $cl(\{0\}) = \{0\}$.
QED
Lemma 5
Let $x =(x_1,\dots,x_n), x' = (x_1',\dots,x_n') \in \Omega_\infty^n$.
Suppose $x \rightarrow_k x'$.
If $x_i = \infty$ for some $i$, then $x_i' = \infty$.
Proof.
Since $x \rightarrow_k x'$, $x_i \rightarrow_k x_i'$.
Hence, by Lemma 4, $x_i' = \infty$.
QED
Lemma 6
Let $x, x' \in \Omega^n$, $y \in \Omega$.
Suppose $x \rightarrow_k x'$.
Then there exists $y' \in \Omega_{\infty}$ such that $(x, y) \rightarrow_k (x', y')$.
Proof:
If there exists $y'$ in an finite extension of $k(x')$ such that $(x, y) \rightarrow_k (x', y')$,
we are done.
Suppose there exist no $y'$ in any finite extensuon of $k(x')$ such that $(x, y) \rightarrow_k (x', y')$.
Then, by Lemma 4 of my answer to this question, $y \neq 0$ and $(x, 1/y) \rightarrow_k (x', 0)$.
Hence $(x, y) \rightarrow_k (x', \infty)$.
QED
Lemma 7(Fundamental theorem of the theory of specializations)
Let $x, x' \in \Omega_\infty^n$, $y \in \Omega_\infty^m$.
Suppose $x \rightarrow_k x'$.
Then there exists $y' \in \Omega_{\infty}^m$ such that $(x, y) \rightarrow_k (x', y')$.
Proof:
By using induction on $m$, it suffices to prove the lemma when $m = 1$.
Since $x \rightarrow_k x'$, there exists $\sigma \in G$ such that $\sigma(x), \sigma(x') \in \Omega^n$ and $\sigma(x) \rightarrow_k \sigma(x')$.
Since $(\sigma(x), \infty) \rightarrow_k (\sigma(x'), \infty)$, we may assume that $y \neq \infty$.
Hence the assertion follows from Lemma 6.
QED
Lemma 8(Weil, Foundations of algebraic geometry, Prop. 10, Sec. 3, Chap. II)
Suppose $x \rightarrow_k x'$, where $x, x' \in \Omega_\infty^n$
Let $y = (y_1,\dots,y_m) \in \Omega^m$.
Suppose $y \neq 0$.
Let $I = {1,\dots,m}$.
Then there exists $i_0 \in I$ such that $(x, z) \rightarrow_k (x, z')$, where $z = (y_1/y_{i_0},\dots,y_m/y_{i_0})$ and $z' \in \Omega^m$.
Proof:
Without loss of generality, we may assume that $y_i \neq 0$ for all $i \in I$.
Suppose no such $i_0$ exists.
Let $u_{i,j} = y_i/y_j$ for all $(i, j) \in I\times I$.
Let $u = (u_{i,j}) \in \Omega^{m^2}$.
By Lemma 7, there exists $v = (v_{i,j}) \in \Omega_{\infty}^{m^2}$ such that $(x, u) \rightarrow_k (x, v)$.
By the assumption, for each $j_0 \in I$ there exists $i \in I$ such that $v_{i,j_0} = \infty$.
Since $u_{j_0,i} = 1/u_{i,j_0}, v_{j_0,i} = 0$.
Hence we have a map $\phi\colon I \rightarrow I$ such that $v_{j, \phi(j)} = 0$.
Let $j_0 = 1$ and $j_{k+1} = \phi(j_k)$ for $k = 0,1,2,\dots$
Since $I$ is finite, there exists $p, r$ such that $j_{p+r} = j_p$.
Then $v_{j_p,j_{p+1}} = \cdots = v_{j_{p+r-1}, j_p} = 0$.
But this is a contradiction, because $u_{j_p,j_{p+1}}\cdots u_{j_{p+r-1}, j_p} = 1$.
QED
Let $\mathbf{P}^n(\Omega)$ be the $n$-projective space over $\Omega$.
Let $E$ be a set of homogeneous polynomials in $k[X_0,\dots,X_n]$.
We denote by $V_+(E)$ the common zeros of $E$ in $\mathbf{P}^n(\Omega)$.
It is easy to see that by taking subsets of the form $V_+(E)$ as closed sets, we can define a topology on $\mathbf{P}^n(\Omega)$.
We call this topology $k$-topology.
A closed(resp. open) subset of $\mathbf{P}^n(\Omega)$ with respect to $k$-topology is called $k$-closed(resp. $k$-open) subset.
It is easy to see that $\mathbf{P}^n(\Omega)$ is a $k$-variety defined in this question.
Lemma 9
Let $X$ be a topological space.
Let $U$ be an open subset of $X$.
Let $x, y \in U$.
Then $x \rightarrow y$ in $X$ if and only if $x \rightarrow y$ in $U$.
Proof:
Suppose $x \rightarrow y$ in $X$.
Let $V'$ be an open neighborhood of $y$ in $U$.
Then there exists an open subset $V$ of $X$ such that $V' = V \cap U$.
Since $x \in V$, $x \in V$.
Hence $x \in V'$.
Hence $x \rightarrow y$ in $U$.
Conversely suppose $x \rightarrow y$ in $U$.
Let $V$ be an open subset of $X$ such that $y \in V$.
Then $V \cap U$ is an open neighborhood of $y$ in $U$.
Hence $x \in V \cap U \subset V$.
Hence $x \rightarrow y$ in $X$.
QED
Lemma 10
Let $X = \mathbf{P}^n(\Omega)$.
We regard $X$ as a $k$-variety.
Let $Y$ be a $k$-variety.
Then $X\times Y$ is a $k$-variety as defined in this question.
Let $(x, y) \in X\times Y$.
Suppose $y \rightarrow y'$ in $Y$.
Then there exists $x' \in X$ such that $(x, y) \rightarrow (x', y')$ in $X\times Y$.
Proof.
Let $V$ be an affine open neighborhood of $y'$ in $Y$.
Since $y \rightarrow y'$, $y \in V$.
Hence, by Lemma 9, we may assume $Y$ is an affine $k$-variety.
Then the assertion follows from Lemma 8.
QED
Lemma 11
Let $X$ be an irrreducible topological space.
Let $U$ be a non-empty open subset of $X$.
Suppose $U$ has a generic point(Definition 1) $x$.
Then $x$ is a generic point of $X$.
Proof:
Let $y \in X$.
Let $V$ be an open neighborhood of $y$ in $X$.
Since $X$ is irreducible, $V \cap U = \emptyset$.
Hence $x \in V \cap U \subset V$.
Hence $x \rightarrow y$ in $X$.
QED
Lemma 12
Let $f\colon X \rightarrow Y$ be a continuous map of topological spaces.
Suppose $X$ is irreducible and has a generic point $x$.
Then $f(x)$ is a generic point of $cl(f(X))$.
Proof:
Let $y = f(x)$.
Since $cl(\{y\}) \subset cl(f(X))$, it suffices to prove that $cl(f(X)) \subset cl(\{y\})$.
Let $y' \in cl(f(X))$.
It suffices to prove that $y \rightarrow y'$.
Let $V$ be a neighborhood of $y'$ in $Y$.
There exists $x' \in X$ such that $f(x') \in V$.
Since $f$ is continuous, there exists a neighborhood $U$ of $x'$ such that $f(U) \subset V$.
Since $x \in U$, $y = f(x) \in V$.
Hence $y \rightarrow y'$ as desired.
QED
Lemma 13
Let $X = \mathbf{P}^n(\Omega)$.
Let $Y$ be a $k$-variety.
Then the projection map $\pi\colon X\times Y \rightarrow Y$ is a closed map.
Proof:
Let $Z$ be a closed subset of $X\times Y$.
It suffices to prove that $\pi(Z)$ is closed.
Since $X\times Y$ is a Noethrian space, $Z = Z_1\cup\cdots\cup Z_r$, where each $Z_i$ is a closed irreducible subset.
Then $\pi(Z) = \pi(Z_1)\cup\cdots\cup \pi(Z_r)$.
Hence $cl(\pi(Z)) = cl(\pi(Z_1)\cup\cdots\cup cl(\pi(Z_r))$.
Hence we may assume that $Z$ is irreducible.
By Lemma 2 and Lemma 11, $Z$ has a generic point $(x, y)$. 
By Lemma 11, $y$ is a generic point of $cl(\pi(Z))$.
Let $y' \in cl(\pi(Z))$.
Then $y \rightarrow y'$.
By Lemma 10, there exists $x' \in X$ such that $(x, y) \rightarrow (x', y')$ in $X\times Y$.
Since $(x',y') \in Z, y' \in \pi(Z)$.
Hence $cl(\pi(Z)) = \pi(Z)$.
QED
Lemma 14
Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties(see this question).
Let $x \in X$ and $y = f(x)$.
Let $y' \in Y$ be such that $cl(\{y\}) = cl(\{y'\})$.
Then there exists $x' \in X$ such that $cl({x}) = cl({x'})$ and $f(x') = y'$.
Proof:
Let $X'$ be the reduced scheme of finite type over $k$ corresponding to $X$ by this question.
Then the underlying set of $X$ can be idenitified with $Hom_k(Spec(\Omega), X')$.
Then the assertion follows from this question.
QED
Definition 4
Let $X$ be a $k$-variety.
Let $x \in X$.
We denote by $k(x)$ the residue field of the local ring $\mathcal{O}_x$.
We say $x$ is algebraic if $k(x)$ is algebraic over $k$.
Lemma 15
Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties.
Let $X_0$(resp. $Y_0$) be the set of algebraic points of $X$(resp. $Y$).
Then $f(X_0) \subset Y_0$.
Proof:
Let $x \in X_0$ and $y = f(x)$.
$f$ induces a $k$-homomorphism $\mathcal{O}_y \rightarrow \mathcal{O}_x$.
Hence it induces a $k$-homomorphism $k(y) \rightarrow k(x)$.
Since $k(x)$ is algebraic over $k$, $k(y)$ is algebraic over $k$.
QED
Lemma 16
Let $x$ be a point of $k$-variety $X$.
Then there exists an algebraic point $x_0 \in X$ such that $x \rightarrow x_0$.
Proof:
By Lemma 9, we may assume that $X$ is affine.
Let $A$ be the affine coordinate ring of $X$.
Let $P = \{f \in A|\ f(x) = 0\}$.
Let $\mathfrak{m}$ be a maximal ideal such that $P \subset \mathfrak{m}$.
Let $K = A/\mathfrak{m}$.
Since $K$ is algebraic over $k$, there exists a $k$-embedding $\phi\colon K \rightarrow \Omega$.
$\phi$ induces a $k$-homomorphism $\alpha\colon A \rightarrow \Omega$.
$\alpha$ can be idenitified with a point $x_0 \in X$.
Clearly $x_0$ is algebraic.
Since $Ker(\alpha) = \mathfrak{m}$, $x \rightarrow x_0$.
QED
Lemma 17
Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties.
Let $X_0$(resp. $Y_0$) be the set of algebraic points of $X$(resp. $Y$).
By Lemma 15, $f$ induces a map $f_0\colon X_0 \rightarrow Y_0$.
Suppose $f$ is a closed map.
Then so is $f_0$.
Proof:
Let $Z_0$ be a closed subset of $X_0$.
There exists a closed subset $Z$ of $X$ such that $Z \cap X_0 = Z_0$.
It suffices to prove that $f(Z) \cap Y_0 = f(Z_0)$.
Clearly $f(Z_0) \subset f(Z) \cap Y_0$.
Let $y_0 \in f(Z) \cap Y_0$.
It suffices to prove that $y_0 \in f(Z_0)$.
There exists $z \in Z$ such that $f(z) = y_0$.
By Lemma 16, there exists $z_0 \in Z_0$ such that $z \rightarrow z_0$.
Then $y_0 = f(z) \rightarrow f(z_0)$.
Since $y_0$ is algebraic, $cl(\{y_0\}) = cl(\{f(z_0)\})$.
Hence, by Lemma 14, there exists $w_0 \in X$ such that $cl({z_0}) = cl({w_0})$ and $f(w_0) = y_0$.
Since $Z$ is closed, $w_0 \in Z$.
Since $z_0$ is algebraic, so is $w_0$.
Hence $w_0 \in Z_0$.
Hence $y_0 \in f(Z_0)$.
QED
Proposition
Let $k$ be an algebraically closed field.
Let $X = \mathbf{P}^n(k)$ be a projective $n$-space over $k$.
Let $Y$ be a variety over $k$ in the sense of Serre's FAC.
Then the porojection map $\pi\colon X\times Y \rightarrow Y$ is a closed map.
Proof:
Let $Z$ be a closed subset of $X\times Y$.
Let $(U_i)$ be a finite affine open cover of $Y$.
Since $(X \times U_i)$ is an affine open cover of $X\times Y$ and $\pi(Z) \cap U_i = \pi(Z\cap (X\times U_i))$, we may assume that $Y$ is affine.
Then $Y$ is identified with a Zariski closed subset of $k^m$.
There exists an ideal $I$ of $k[X_1,\dots,X_m]$ such that $Y$ is the set of common zeros of $I$ in $k^m$.
Let $Y'$ be the set of common zeros of $I$ in $\Omega^m$, where $\Omega$ is an algebraically closed field such that tr.dim $\Omega/k = \infty$.Since $Y'$ is a $k$-closed subset of $\Omega^m$, $Y'$ is an affine $k$-variety.
Let $X' = \mathbf{P}^n(\Omega)$.
Since $X\times Y$ is the set of algebraic points of a $k$-variety $X'\times Y'$, the assertion follows from Lemma 13 and Lemma 17.
QED
