Lemma 29.48.7. Let $f : Y \to X$ be a finite morphism of schemes. Let $T \subset Y$ be a closed nowhere dense subset of $Y$. Then $f(T) \subset X$ is a closed nowhere dense subset of $X$.
Proof. By Lemma 29.44.11 we know that $f(T) \subset X$ is closed. Let $X = \bigcup X_ i$ be an affine covering. Since $T$ is nowhere dense in $Y$, we see that also $T \cap f^{-1}(X_ i)$ is nowhere dense in $f^{-1}(X_ i)$. Hence if we can prove the theorem in the affine case, then we see that $f(T) \cap X_ i$ is nowhere dense. This then implies that $T$ is nowhere dense in $X$ by Topology, Lemma 5.21.4.
Assume $X$ is affine. Choose a diagram
\[ \xymatrix{ Z' \ar[rd] & Y' \ar[l]^ i \ar[d]^{f'} \ar[r]_ a & Y \ar[d]^ f \\ & X' \ar[r]^ b & X } \]
as in Lemma 29.48.6. The morphisms $a$, $b$ are open since they are finite locally free (Lemmas 29.48.2 and 29.25.10). Hence $T' = a^{-1}(T)$ is nowhere dense, see Topology, Lemma 5.21.6. The morphism $b$ is surjective and open. Hence, if we can prove $f'(T') = b^{-1}(f(T))$ is nowhere dense, then $f(T)$ is nowhere dense, see Topology, Lemma 5.21.6. As $i$ is a closed immersion, by Topology, Lemma 5.21.5 we see that $i(T') \subset Z'$ is closed and nowhere dense. Thus we have reduced the problem to the case discussed in the following paragraph.
Assume that $Y = \bigcup _{i = 1, \ldots , n} Y_ i$ is a finite union of closed subsets, each mapping isomorphically to $X$. Consider $T_ i = Y_ i \cap T$. If each of the $T_ i$ is nowhere dense in $Y_ i$, then each $f(T_ i)$ is nowhere dense in $X$ as $Y_ i \to X$ is an isomorphism. Hence $f(T) = f(T_ i)$ is a finite union of nowhere dense closed subsets of $X$ and we win, see Topology, Lemma 5.21.2. Suppose not, say $T_1$ contains a nonempty open $V \subset Y_1$. We are going to show this leads to a contradiction. Consider $Y_2 \cap V \subset V$. This is either a proper closed subset, or equal to $V$. In the first case we replace $V$ by $V \setminus V \cap Y_2$, so $V \subset T_1$ is open in $Y_1$ and does not meet $Y_2$. In the second case we have $V \subset Y_1 \cap Y_2$ is open in both $Y_1$ and $Y_2$. Repeat sequentially with $i = 3, \ldots , n$. The result is a disjoint union decomposition
\[ \{ 1, \ldots , n\} = I_1 \amalg I_2, \quad 1 \in I_1 \]
and an open $V$ of $Y_1$ contained in $T_1$ such that $V \subset Y_ i$ for $i \in I_1$ and $V \cap Y_ i = \emptyset $ for $i \in I_2$. Set $U = f(V)$. This is an open of $X$ since $f|_{Y_1} : Y_1 \to X$ is an isomorphism. Then
\[ f^{-1}(U) = V\ \amalg \ \bigcup \nolimits _{i \in I_2} (Y_ i \cap f^{-1}(U)) \]
As $\bigcup _{i \in I_2} Y_ i$ is closed, this implies that $V \subset f^{-1}(U)$ is open, hence $V \subset Y$ is open. This contradicts the assumption that $T$ is nowhere dense in $Y$, as desired. $\square$
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