The Stacks project

Lemma 26.24.3. A composition of immersions of schemes is an immersion, a composition of closed immersions of schemes is a closed immersion, and a composition of open immersions of schemes is an open immersion.

Proof. This is clear for the case of open immersions since an open subspace of an open subspace is also an open subspace.

Suppose $a : Z \to Y$ and $b : Y \to X$ are closed immersions of schemes. We will verify that $c = b \circ a$ is also a closed immersion. The assumption implies that $a$ and $b$ are homeomorphisms onto closed subsets, and hence also $c = b \circ a$ is a homeomorphism onto a closed subset. Moreover, the map $\mathcal{O}_ X \to c_*\mathcal{O}_ Z$ is surjective since it factors as the composition of the surjective maps $\mathcal{O}_ X \to b_*\mathcal{O}_ Y$ and $b_*\mathcal{O}_ Y \to b_*a_*\mathcal{O}_ Z$ (surjective as $b_*$ is exact, see Modules, Lemma 17.6.1). Hence by Lemma 26.24.2 above $c$ is a closed immersion.

Finally, we come to the case of immersions. Suppose $a : Z \to Y$ and $b : Y \to X$ are immersions of schemes. This means there exist open subschemes $V \subset Y$ and $U \subset X$ such that $a(Z) \subset V$, $b(Y) \subset U$ and $a : Z \to V$ and $b : Y \to U$ are closed immersions. Since the topology on $Y$ is induced from the topology on $U$ we can find an open $U' \subset U$ such that $V = b^{-1}(U')$. Then we see that $Z \to V = b^{-1}(U') \to U'$ is a composition of closed immersions and hence a closed immersion. This proves that $Z \to X$ is an immersion and we win. $\square$


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