Birational smooth proper curves are isomorphic

Let $\phi :C⇢D$ be a birational rational map and let $\alpha :k(D)\to k(C)$ be the pullback induced by a representative ${\phi }_U:U\to D$; the definition of birationality makes $\alpha$ an isomorphism of $k$-algebras [F2]. In particular $\alpha$ is injective, so under Choice [F5] and [F1] there is a unique dominant $k$-morphism $F:C\to D$ with $F^{\ast }=\alpha$, and $F$ represents the rational map $\phi$ because $F$ extends the representative and representatives of a rational map with the same generic pullback agree [F2, F4].

Similarly ${\alpha }^{-1}:k(C)\to k(D)$ is an injective $k$-algebra homomorphism, so it is the pullback of a unique dominant $k$-morphism $G:D\to C$ [F1].

The composites satisfy $(F\circ G{)}^{\ast }=G^{\ast }\circ F^{\ast }={\alpha }^{-1}\circ \alpha ={\mathrm{i}\mathrm{d}}_{k(D)}$ and $({\mathrm{i}\mathrm{d}}_D{)}^{\ast }={\mathrm{i}\mathrm{d}}_{k(D)}$; both $F\circ G$ and ${\mathrm{i}\mathrm{d}}_D$ are dominant $k$-morphisms $D\to D$ with the same pullback, so by the injectivity part of the bijection [F1] they are equal. Symmetrically $(G\circ F{)}^{\ast }={\mathrm{i}\mathrm{d}}_{k(C)}$ gives $G\circ F={\mathrm{i}\mathrm{d}}_C$. Hence $F$ is an isomorphism with inverse $G$, and it represents $\phi$.

If moreover $f:C\to D$ is a dominant $k$-morphism which is birational in the sense of birational morphisms of integral finite-type schemes, then its pullback $f^{\ast }:k(D)\to k(C)$ is an isomorphism by definition [F2], so step 3.1 applied to the rational map represented by $(C,f)$ produces an isomorphism representing it; as $f$ and that isomorphism are dominant morphisms with the same pullback, they are equal by [F1], so $f$ itself is an isomorphism.

Steps 3.1 and 4.1 prove both assertions; the only choice-theoretic inputs are the extension lemma [F4] and the function-field bijection [F1], both used under Choice [F5].