Nonconstant morphisms of proper curves are finite and surjective

By [F1] the schemes $C$ and $D$ are nonempty, integral, separated and of finite type over $k$, with $C$ proper over $k$; by [F3] the morphism $f$, being a $k$-morphism from a proper $k$-scheme to a separated $k$-scheme, is proper. Hence $f$ is of finite type, universally closed and closed by [F2], and its image $f(C)$ is closed and irreducible; it is nonempty because $C$ is nonempty.

The hypotheses of [F8] hold: $D$ is separated over $k$, $C$ is proper of dimension one over $k$, and the image of its sole one-dimensional irreducible component has more than one point. Therefore $f$ is finite. The cited lemma checks finite fibres over images of closed points; it does not treat the fibre over the generic point as a closed subset.

If $f(C)$ were a proper closed subset of $D$, then by [F4] it would be a finite set of closed points, hence discrete. A nonempty finite discrete irreducible space is a single point, contradicting nonconstancy. Thus $f(C)=D$ and $f$ is surjective.

Let $U=\mathrm{Spec}A$ be a nonempty affine open of $D$. By finiteness [F5], $f^{-1}(U)=\mathrm{Spec}B$ with $B$ finite as an $A$-module. Both rings are domains [F1]. Surjectivity implies $A\to B$ is injective: an element in its kernel lies in every prime of $A$, hence is zero. Set $L=\mathrm{Frac}(A)$ and $K=\mathrm{Frac}(B)$ [F6]. The localization $B{\otimes }_{A}L$ is a finite-dimensional domain over $L$, hence a field; since it contains $B$, it equals $K$. Thus $k(D)=L\hookrightarrow K=k(C)$ is a finite field extension.

Steps 1.2, 2.1 and 3.1 prove finiteness, surjectivity and the finite function-field embedding. The stated Choice premise is inherited from [F4] in step 2.1; [F8] itself states no Choice premise.