This is  [ 1 , Theorem II.7 ] .

We can assume that \(X=\mathbb {P}^n\) (since \(X\) is projective). And, by changing coordinates, we can assume that \(\phi (S)\) is not contained in any hyperplane. Then \(\phi \) is given by a linear system \(P\subseteq |D|\) with no fixed component, where \(D\in P\) is a divisor on \(X\). Let’s prove the theorem by complete induction on \(D^2\). When \(D^2 {\lt} 0\), there’s nothing to prove.

If \(P\) has no base locus, then there is also nothing to prove. So assume \(x\in \operatorname{\mathrm{bs}}P\), and let \(\epsilon :S_1\to S\) be the blowup of \(S\) at \(x\). Then we get a linear system \(\epsilon ^* P\subseteq \epsilon ^*\operatorname{\mathrm{H}}^0(S,\mathcal{O}_S(D))\) on \(S_1\). The fixed locus of \(\epsilon ^* P\) is then equal to the exceptional divisor \(E\) of \(\epsilon \) because it is empty outside of \(E\), but all sectionsin \(\epsilon ^* P\) vanish along \(E\) (as \(x\) is in the base locus of \(P\)). Let \(k\) be the minimum of the multiplicities of \(E\) occuring in the divisors \(F\in \epsilon ^* P\). So, necessarily, \(k\geq 1\), and we get a linear system \(P_1 = \epsilon ^* P - kE \subset |\epsilon ^* D|\) by subtracting \(kE\) from every element of \(\epsilon ^* P\). But then we get a rational map \(\phi _1 = \phi \circ \epsilon : S_1\dashrightarrow \mathbb {P}^n\) corresponding to the linear system \(P_1\). Moreover, \(D_1 = \epsilon ^* D - kE\) is an element of \(P_1\) and we have \(D_1^2 = D^2 - k^2 {\lt} D^2\). So, by induction, there exists a composite \(\eta _1:S'\to S_1\) of finitely many monoidal tranformations and a morphism \(f:S'\to X\) such that \(f = \phi _1\circ \eta _1\). So set \(\eta = \epsilon \circ \eta _1\). We get that \(\phi \circ \eta = \phi \circ \epsilon \circ \eta _1 = \phi _1\circ \eta _1 = f\) as desired.