The Stacks project

Proof. There is an immediate reduction to the case where $X$ and hence $Y$ are affine. Say $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(A)$. Then $X' = \mathop{\mathrm{Spec}}(A')$, where $A'$ is the integral closure of $R$ in $A$, see Definitions 29.53.2 and 29.53.3. By Algebra, Theorem 10.123.12 for every $y \in Y$ at which $f$ is quasi-finite, there exists an open $U'_ y \subset X'$ such that $(f')^{-1}(U'_ y) \to U'_ y$ is an isomorphism. Set $U' = \bigcup U'_ y$ where $y \in Y$ ranges over all points where $f$ is quasi-finite. It remains to show that $f$ is quasi-finite at all points of $(f')^{-1}(U')$. If $y \in (f')^{-1}(U')$ with image $x \in X$, then we see that $Y_ x \to X'_ x$ is an isomorphism in a neighbourhood of $y$. Hence there is no point of $Y_ x$ which specializes to $y$, since this is true for $f'(y)$ in $X'_ x$, see Lemma 29.44.8. By Lemma 29.20.6 part (3) this implies $f$ is quasi-finite at $y$. $\square$

Proof. Suppose $f$ is quasi-finite at $x$. Let $x \in U = \mathop{\mathrm{Spec}}(A) \subset X$, $V = \mathop{\mathrm{Spec}}(R) \subset S$ be affine opens as in Definition 29.20.1. By either Theorem 29.56.1 above or Algebra, Lemma 10.123.13, the set of primes $\mathfrak q$ at which $R \to A$ is quasi-finite is open in $\mathop{\mathrm{Spec}}(A)$. Since these all correspond to points of $X$ where $f$ is quasi-finite we get the first statement. The second statement is obvious. $\square$

Proof. This is Algebra, Lemma 10.123.14 reformulated in the language of schemes. $\square$

Proof. As in the proof of Lemma 29.48.7 this reduces immediately to the case where the base $X$ is affine. In this case $Y = \bigcup _{i = 1, \ldots , n} Y_ i$ is a finite union of affine opens (as $f$ is quasi-compact). Since each $T \cap Y_ i$ is nowhere dense, and since a finite union of nowhere dense sets is nowhere dense (see Topology, Lemma 5.21.2), it suffices to prove that the image $f(T \cap Y_ i)$ is nowhere dense in $X$. This reduces us to the case where both $X$ and $Y$ are affine. At this point we apply Lemma 29.56.3 above to get a diagram

with $Z$ affine, $\pi$ finite and $j$ an open immersion. Set $\overline{T} = \overline{j(T)} \subset Z$. By Topology, Lemma 5.21.3 we see $\overline{T}$ is nowhere dense in $Z$. Since $f(T) \subset \pi (\overline{T})$ the lemma follows from the corresponding result in the finite case, see Lemma 29.48.7. $\square$