Lemma 29.56.4. Let f : Y \to X be a quasi-finite morphism of schemes. Let T \subset Y be a closed nowhere dense subset of Y. Then f(T) \subset X is a nowhere dense subset of X.
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
\xymatrix{ Y \ar[rd]_ f \ar[rr]_ j & & Z \ar[ld]^\pi \\ & X & }
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
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