Morphisms from complete connected schemes to affine schemes are constant

By [F1] choose a finite affine open cover $Z=U_1\cup \cdots \cup U_s$ with $U_j=\mathrm{Spec}{A}_j$ and $A_j$ a finitely generated $k$-algebra. Each $A_j$ is Noetherian by [F2], so each $U_j$ is a Noetherian topological space by [F3]. A descending chain of closed subsets of $Z$ restricts to descending chains in the finitely many $U_j$, which stabilize from some index on; the largest of the finitely many indices then stabilizes the chain in $Z$, because the $U_j$ cover $Z$. Hence the underlying space of $Z$ is Noetherian by [F4].

By [F5] the space $Z$ is a finite union of irreducible closed subsets, so $Z$ has finitely many irreducible components; let $Z_1,\dots ,Z_r$ be the distinct components, each viewed with its reduced induced closed subscheme structure as in [F6]. Each $Z_i$ is nonempty, reduced and irreducible, hence integral by [F6], and each is a closed subscheme of $Z$. Since $Z$ is complete, $Z\to \mathrm{Spec}k$ is proper by [F8]; the closed immersion $Z_i\hookrightarrow Z$ is proper by [F7], and the composite $Z_i\to \mathrm{Spec}k$ is proper by [F7] again. Thus every $Z_i$ is a nonempty proper integral finite-type $k$-scheme, and [F9] gives that $K_i:=\mathrm{\Gamma }(Z_i,{\mathcal{O}}_{Z_i})$ is a finite field extension of $k$.

Write $Y=\mathrm{Spec}B$ with $B$ a finitely generated $k$-algebra; by [F10] the morphism $f$ corresponds to the $k$-algebra map $\phi :B\to \mathrm{\Gamma }(Z,{\mathcal{O}}_Z)$. Fix $i$ and let ${\psi }_i:\mathrm{\Gamma }(Z,{\mathcal{O}}_Z)\to K_i$ be the restriction. The composite ${\phi }_i={\psi }_i\circ \phi$ is a $k$-algebra map from $B$ into the field $K_i$, so its image is a $k$-subalgebra of the finite-dimensional $k$-vector space $K_i$; it is a domain of finite dimension over $k$, hence a field, and its kernel ${\mathfrak{m}}_i$ is a maximal ideal of $B$. It follows that the restriction of $f$ to $Z_i$ factors through $\mathrm{Spec}(B\mathrm{/}{\mathfrak{m}}_i)=\{{\mathfrak{m}}_i\}\subseteq Y$ by [F10], that is, $f$ is constant on $Z_i$ with value $y_i$, the closed point corresponding to ${\mathfrak{m}}_i$ by [F11].

Suppose $Z_i\cap Z_j\mathrm{\ne }\mathrm{\varnothing }$ for some $i,j$. Choosing a point $p$ in the intersection, step 2.1 gives $y_i=f(p)=y_j$. Hence the images of two components that meet coincide. If the components could be split into two nonempty groups with no member of one meeting any member of the other, then the union of each group would be a nonempty closed subset of $Z$ — a finite union of the closed $Z_i$ — and the two unions would be disjoint and cover $Z$, contradicting connectedness of $Z$. Therefore the intersection graph of $Z_1,\dots ,Z_r$ is connected, and iterating the observation just made along a path of intersections shows $y_1=\cdots =y_r=:y$.

As the finitely many $Z_i$ cover $Z$ by [step 1.2], every point of $Z$ lies in some $Z_i$ and therefore has image $y$; thus $f(Z)=\{y\}$ with $y$ a closed point of $Y$ by [step 2.1]. This proves the first assertion.

For the final assertion take $Y=Z$ and $f={\mathrm{id}}_Z$, which is a $k$-morphism of affine finite-type $k$-schemes when $Z$ is affine. By [step 4.1] the identity map has image a single point, so the underlying space of $Z$ consists of one point. Then $\mathrm{\Gamma }(Z,{\mathcal{O}}_Z)$ is a reduced finite-type $k$-algebra whose spectrum is a single point, hence a field, and it is finite over $k$ by [F9] applied to $Z$, which is nonempty proper integral because it is complete, connected, reduced and a single point.