This is  [ 4 , Theorem III.11.4 ] .

It’s a consequence of the Theorem on formal functions, which can be used to show that \(f_*\mathcal{O}_X = \mathcal{O}_Y\) implies the connectedness of the fibers of \(f\).

This is  [ 4 , Lemma V.5.1 ] and we’ve already used it for surfaces, where it says that the set of fundamental points is finite. It a consequence of the valuative criterion.

This is  [ 4 , Theorem V.5.2 ] .

Write \(\Gamma \) for the graph of \(T_{|\operatorname{\mathrm{dom}}U}\) as above. Then \(p_X:\Gamma \to X\) satisfies the conditions of Theorem 4.25. So \(p_X^{-1}(P)\) is connected. It \(\dim p_X^{-1}(P) = 0\), then there is a Zariski open neighborhood \(V\) of \(P\), where the induced morphism \(p_X^{-1} V\to V\) is finite and (necessarily) birtional. But then, since \(X\) is normal, it must be an isomorphism. But then \(T\) is defined at \(P\), contradicting our assumption that \(P\) was a fundamental point.

So then \(p_Y(p_X^{-1}(P)\) is connected, and the map \(p_Y: p_X^{-1}(P)\to Y\) is always an isomorphism onto its image. (Just think set-theoretically about that.) So we’re done.

We can let \(S_1\) denote the closure of the graph of \(\phi \) with \(q\) and \(q'\) the projections on \(S\) and \(S'\) respectively. So \(\phi \circ q = q'\) as rational maps. Since \(\phi ^{-1}\) is undefined at \(p\), \((q')^{-1}\) is also undefined at \(p\). So there is an irreducible curve \(C_1\subset S_1\) with \(q'(C_1) = \{ p\} \) by Lemma 4.34. But then \(q(C_1)\) is a curve \(C\) in \(S\) (isomrophic to \(C_1\) via the map \(q\)).

This is  [ 4 , Theorem V.5.3 ] .

Write \(T = \epsilon ^{-1}\circ f: X'\dashrightarrow \widetilde{X}\). The goal is to show that \(T\) is a morphism.

If not, there is a fundamental point \(P'\) of \(T\). And, by Lemma 4.27, there is a curve \(C\) in \(\widetilde{X}\) such that \(T^{-1}[C] = P'\). We must have \(f(P') = P\), since \(\epsilon \) is an isomorphism outside of the inverse image of \(P\). So we get that \(\epsilon (C) = P\), and, thus, \(C = E\), where \(E\) is the exceptional divisor of \(\epsilon \).

Now, \(T^{-1}: Y' \dashrightarrow X\) is defined at all but finitely many points of \(Y'\). So we can find a closed point \(Q\in E\), where \(T^{-1}\) is defined. Thus \(T^{-1}(Q) = P'\).

Choosing local coordinates \(x,y\in \mathcal{O}_{X,P}\), we can find an open neighborhood \(V\) of \(P\) above which \(\epsilon \) is defined in \(\mathbb {P}^1_V\) by the equation \(sx = ty\) (in terms of homogenous coordinates \(s\) and \(t\) on \(\mathbb {P}^1\). Without loss of generality, we can assume that \(Q\) is given by \([s,t] = [1,0]\). Then \(t\) and \(y\) are local coordinates in \(\mathcal{O}_{\widetilde{X}, Q}\). Moreover, the map \(\epsilon :\epsilon ^{-1}V \to V\) is given by \((t,y)\mapsto (ty, y)\).

Since \(P\) is a fundamental point of \(f^{-1}\), \(f^{-1}(P)\) is connected and one-dimensional. So we can find a curve \(C'\) in \(X'\) containing \(P'\). Let \(z=0\) be the local equation for \(C'\) at \(P'\).

Then \(f^{-1}(P)\) is defined by the equation \(x=y=0\) in \(f^{-1}(V)\). Therefore, \(f^(x)\) and \(f^*(y)\) are both in the ideal generated by \(z\). So, working in \(K = K(X)\), we can find \(a,b\in \mathcal{O}_{X',P'}\) such that \(x = az\) and \(y = bz\).

On the other hand, since \(t, y\) are local coordinates at \(Q\), \(y\not\in \mathfrak {m}_{Q}^2\). Therefore, \(y\not\in \mathfrak {m}_{P'}^2\) either as the morphism \(T^{-1}\) is defined at \(Q\) and, therefore, \(\mathcal{O}_{Q}\) dominates \(\mathcal{O}_{P'}\). But then, since \(y = bz\in \mathcal{O}_{X',P'}\) and \(z\in \mathfrak {m}_{P'}\), it follows that \(b\in \mathcal{O}_{P'}^2\). So \(t = x/y = a/b\in \mathcal{O}_{P'}\). And, since \(t\in \mathfrak {m}_Q\), we must have \(t\in \mathfrak {m}_{P'}\).

But then, since \(T(P') = E\), we must have \(w\in (y)\) for any \(w\in \mathfrak {m}_{P'}\) as \(E = V(y)\). So, taking \(w = t\), we find that \(t\in (y)\). But that contradicts the assumption that \(t\) and \(y\) were local coordinates at \(Q\).

This is  [ 4 , Corollary V.5.4 ] .

For the finiteness, note that, if \(f(C')\) is a point \(P\) in \(X\), then \(P\) is a fundamental point of \(f^{-1}\). So, since the number of fundamental points of \(f^{-1}\) is finite, so is \(n(f)\).

Now induct on \(n(f)\). If \(n(f) = 0\), then \(f:X'\to X\) is a birational morphism contracting no curves. So, by Zariski’s main theorem, it is an isomorphism. On the other hand, suppose \(f(C') = P\). Then, by Theorem 4.29, we have \(f = f_1\circ \epsilon \), where \(\epsilon : \widehat{X}\to X\) is the blow up of \(P\) and \(f_1:X'\to \widehat{X}\) is a morphism. We claim that \(n(f_1) {\lt} n(f)\). (In fact, \(n(f_1) = n(f) - 1\).)

To see this, set \(N(f) = \{ C'\in X' : f(C')\) is a point \(\} \), and similarly, for \(N(f_1)\). Then, if \(C'\in N(f_1)\), we have that \(f(C') = \epsilon \circ f_1 (C')\) is also a point. So \(N(f_1)\subseteq N(f)\). On the other hand, since \(f_1^{-1}\) is defined outside of finitely many points, there is a unique curve \(E'\) which is the proper transform of the exceptional divisor \(E\subseteq \widehat X\) under \(f_1\). We have \(E\in N(f)\setminus N(f_1)\). So we’re done.

This is is  [ 4 , Theorem V.5.5 ] .

Let’s skip it for now.

This is  [ 4 , Theorem V.5.8 ] .

Suppose, to get a contradiction that the result doesn’t hold. Then we can find a counterexample \(X\) with smallest possible Picard number \(\rho _X\).

Since \(X\) is a counterexample, \(X\) cannot itself be a relative minimal model. So we can find a birational morphism \(f:X\to X'\) to another nonsingular projective surface \(X'\). But then we can factor \(f\) as \(f = f_1\circ \epsilon \), where \(\epsilon :X\to X_1\) is the blowup of a point \(P_1\in X_1\) and \(f_1:X_1\to X'\) is a morphism. But then \(\rho _{X_1} = \rho _{X} - 1\). So \(X_1\) admits a relative minimal model \(X_1\to X'\), and we get a contradiction.