**Proof.**
Since $X$ is integral, we see that $U$ is integral. Hence $Y$ is integral, and $(1, f)(U) \subset Y$ is an open dense subscheme. Also, note that the closed subscheme $Y \subset X \times _ S \mathbf{P}^1_ S$ does not depend on the choice of the open $U$, since after all it is the closure of the one point set $\{ \eta '\} = \{ (1, f)(\eta )\} $ where $\eta \in X$ is the generic point. Having said this let us prove the assertions of the lemma.

For (1) note that $p$ is the composition of the closed immersion $Y \to X \times _ S \mathbf{P}^1_ S = \mathbf{P}^1_ X$ with the proper morphism $\mathbf{P}^1_ X \to X$. As a composition of proper morphisms is proper (Morphisms, Lemma 29.41.4) we conclude.

It is clear that $Y \cap U \times _ S \mathbf{P}^1_ S = (1, f)(U)$. Thus (2) follows. It also follows that $\dim _\delta (Y) = n$.

Note that $q(\eta ') = f(\eta )$ is not contained in $D_0$ or $D_\infty $ since $f \in R(X)^*$. Hence (3) by Divisors, Lemma 31.13.13. We obtain $\dim _\delta (Y_0) = n - 1$ and $\dim _\delta (Y_\infty ) = n - 1$ from Lemma 42.16.1.

Consider the effective Cartier divisor $Y_0$. At every point $\xi \in Y_0$ we have $f \in \mathcal{O}_{Y, \xi }$ and the local equation for $Y_0$ is given by $f$. In particular, if $\delta (\xi ) = n - 1$ so $\xi $ is the generic point of a integral closed subscheme $Z$ of $\delta $-dimension $n - 1$, then we see that the coefficient of $[Z]$ in $\text{div}_ Y(f)$ is

\[ \text{ord}_ Z(f) = \text{length}_{\mathcal{O}_{Y, \xi }} (\mathcal{O}_{Y, \xi }/f\mathcal{O}_{Y, \xi }) = \text{length}_{\mathcal{O}_{Y, \xi }} (\mathcal{O}_{Y_0, \xi }) \]

which is the coefficient of $[Z]$ in $[Y_0]_{n - 1}$. A similar argument using the rational function $1/f$ shows that $-[Y_\infty ]$ agrees with the terms with negative coefficients in the expression for $\text{div}_ Y(f)$. Hence (4) follows.

Note that $D_0 \to S$ is an isomorphism. Hence we see that $X \times _ S D_0 \to X$ is an isomorphism as well. Clearly we have $Y_0 = Y \cap X \times _ S D_0$ (scheme theoretic intersection) inside $X \times _ S \mathbf{P}^1_ S$. Hence it is really the case that $Y_0 \to X$ is a closed immersion. It follows that

\[ p_*\mathcal{O}_{Y_0} = \mathcal{O}_{Y'_0} \]

where $Y'_0 \subset X$ is the image of $Y_0 \to X$. By Lemma 42.12.4 we have $p_*[Y_0]_{n - 1} = [Y'_0]_{n - 1}$. The same is true for $D_\infty $ and $Y_\infty $. Hence (6) is a consequence of (5). Finally, (5) follows immediately from Lemma 42.18.1.
$\square$

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