Definition 26.3.1. Let f : X \to Y be a morphism of locally ringed spaces. We say that f is an open immersion if f is a homeomorphism of X onto an open subset of Y, and the map f^{-1}\mathcal{O}_ Y \to \mathcal{O}_ X is an isomorphism.
26.3 Open immersions of locally ringed spaces
The following construction is parallel to Sheaves, Definition 6.31.2 (3).
Example 26.3.2. Let X be a locally ringed space. Let U \subset X be an open subset. Let \mathcal{O}_ U = \mathcal{O}_ X|_ U be the restriction of \mathcal{O}_ X to U. For u \in U the stalk \mathcal{O}_{U, u} is equal to the stalk \mathcal{O}_{X, u}, and hence is a local ring. Thus (U, \mathcal{O}_ U) is a locally ringed space and the morphism j : (U, \mathcal{O}_ U) \to (X, \mathcal{O}_ X) is an open immersion.
Definition 26.3.3. Let X be a locally ringed space. Let U \subset X be an open subset. The locally ringed space (U, \mathcal{O}_ U) of Example 26.3.2 above is the open subspace of X associated to U.
Lemma 26.3.4. Let f : X \to Y be an open immersion of locally ringed spaces. Let j : V = f(X) \to Y be the open subspace of Y associated to the image of f. There is a unique isomorphism f' : X \cong V of locally ringed spaces such that f = j \circ f'.
Proof. Let f' be the homeomorphism between X and V induced by f. Then f = j \circ f' as maps of topological spaces. Since there is an isomorphism of sheaves f^\sharp : f^{-1}(\mathcal{O}_ Y) \to \mathcal{O}_ X, there is an isomorphism of rings f^\sharp : \Gamma (U, f^{-1}(\mathcal{O}_ Y)) \to \Gamma (U, \mathcal{O}_ X) for each open subset U \subset X. Since \mathcal{O}_ V = j^{-1}\mathcal{O}_ Y and f^{-1} = f'^{-1} j^{-1} (Sheaves, Lemma 6.21.6) we see that f^{-1}\mathcal{O}_ Y = f'^{-1}\mathcal{O}_ V, hence \Gamma (U, f'^{-1}(\mathcal{O}_ V)) \to \Gamma (U, f^{-1}(\mathcal{O}_ Y)) is an isomorphism for every U \subset X open. By composing these we get an isomorphism of rings
for each open subset U \subset X, and therefore an isomorphism of sheaves f^{-1}(\mathcal{O}_ V) \to \mathcal{O}_ X. In other words, we have an isomorphism f'^{\sharp } : f'^{-1}(\mathcal{O}_ V) \to \mathcal{O}_ X and therefore an isomorphism of locally ringed spaces (f', f'^{\sharp }) : (X, \mathcal{O}_ X) \to (V, \mathcal{O}_ V) (use Lemma 26.2.2). Note that f = j \circ f' as morphisms of locally ringed spaces by construction.
Suppose we have another morphism f'' : (X, \mathcal{O}_ X) \to (V, \mathcal{O}_ V) such that f = j \circ f''. At any point x \in X, we have j(f'(x)) = j(f''(x)) from which it follows that f'(x) = f''(x) since j is the inclusion map; therefore f' and f'' are the same as morphisms of topological spaces. On structure sheaves, for each open subset U \subset X we have a commutative diagram
from which we see that f'^\sharp and f''^\sharp define the same morphism of sheaves. \square
From now on we do not distinguish between open subsets and their associated subspaces.
Lemma 26.3.5. Let f : X \to Y be a morphism of locally ringed spaces. Let U \subset X, and V \subset Y be open subsets. Suppose that f(U) \subset V. There exists a unique morphism of locally ringed spaces f|_ U : U \to V such that the following diagram is a commutative square of locally ringed spaces
Proof. Omitted. \square
In the following we will use without further mention the following fact which follows from the lemma above. Given any morphism f : Y \to X of locally ringed spaces, and any open subset U \subset X such that f(Y) \subset U, then there exists a unique morphism of locally ringed spaces Y \to U such that the composition Y \to U \to X is equal to f. In fact, we will even by abuse of notation write f : Y \to U since this rarely gives rise to confusion.
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