Morphisms and the induced maps on local rings

Because \(\phi\) is a morphism, \(f\circ\phi\) is regular on the open neighborhood \(\phi^{-1}(U)\) of \(P\). If two representatives \(f\) on \(U\) and \(g\) on \(V\) determine the same germ at \(Q\), then they agree on some open neighborhood \(W\subseteq U\cap V\) of \(Q\). Their pullbacks therefore agree on the neighborhood \(\phi^{-1}(W)\) of \(P\), so the displayed rule is independent of the representative. It plainly preserves sums, products, and \(1\).

The maximal ideal of \(\mco_{Q,Y}\) consists of germs vanishing at \(Q\), and similarly at \(P\). Since \[ (f\circ\phi)(P)=f(Q), \] we have \[ (\phi_P^*)^{-1}(\mathfrak m_{P,X})=\mathfrak m_{Q,Y}. \] Thus \(\phi_P^*\) is a local homomorphism.

Let \(\psi=\phi^{-1}\) be the inverse morphism. The underlying continuous maps \(\phi\) and \(\psi\) are inverse to one another, hence \(\phi\) is a homeomorphism. Pullback of germs is functorial under composition, so \[ \psi_Q^*\circ\phi_P^*=\operatorname{id}_{\mco_{Q,Y}}, \qquad \phi_P^*\circ\psi_Q^*=\operatorname{id}_{\mco_{P,X}}. \] Hence \(\phi_P^*\) is an isomorphism with inverse \(\psi_Q^*\).

It is enough to check locally that pullback by \(\psi\) preserves regular functions. Let \(Q\in Y\), put \(P=\psi(Q)\), and let \(f\) be regular on an open neighborhood \(U\) of \(P\). Because \(\phi_P^*\) is surjective, there is a germ \([g]_Q\in\mco_{Q,Y}\) with \[ \phi_P^*([g]_Q)=[f]_P. \] Choose \(g\) regular on an open neighborhood \(V\) of \(Q\). Equality of these germs means that, after shrinking to an open neighborhood \[ U'\subseteq U\cap\phi^{-1}(V) \] of \(P\), we have \(g\circ\phi=f\) on \(U'\).

Because \(\phi\) is a homeomorphism, \(W=\phi(U')\) is an open neighborhood of \(Q\) contained in \(V\). On \(W\) we therefore have \[ f\circ\psi=g, \] so \(f\circ\psi\) is regular near \(Q\). Since \(Q\) and \(f\) were arbitrary, \(\psi\) is a morphism. Thus \(\phi\) is an isomorphism, completing (b).

Let \(U\subseteq X\) be nonempty and open. Since a variety is irreducible, \(U\) is dense in \(X\). Let \(C=\overline{\phi(U)}\subseteq Y\). Then \(\phi^{-1}(C)\) is a closed subset of \(X\) containing \(U\), so density of \(U\) gives \(\phi^{-1}(C)=X\). Hence \(\phi(X)\subseteq C\). By hypothesis \(\phi(X)\) is dense in \(Y\), and therefore \(C=Y\).

Fix \(P\in X\), put \(Q=\phi(P)\), and suppose \([f]_Q\in\mco_{Q,Y}\) satisfies \[ \phi_P^*([f]_Q)=0. \] Choose a representative \(f\) regular on an open neighborhood \(V\) of \(Q\). The vanishing of the pullback germ gives a nonempty open neighborhood \[ U\subseteq\phi^{-1}(V) \] of \(P\) on which \(f\circ\phi=0\). By step 4, \(\phi(U)\) is dense in \(Y\), hence dense in the open subspace \(V\). The zero set of the regular function \(f\) is closed in \(V\) and contains \(\phi(U)\), so it is all of \(V\). Thus \(f=0\) on \(V\), whence \([f]_Q=0\). Therefore \(\phi_P^*\) is injective.

Step 1 proves (a), steps <1>2–<1>3 prove (b), and steps <1>4–<1>5 prove (c).