Lemma 29.55.4. 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 29.55.3 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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