The Stacks project

26.3 Open immersions of locally ringed spaces

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.

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

\[ \Gamma (U, f'^{-1}(\mathcal{O}_ V)) \to \Gamma (U, \mathcal{O}_ X) \]

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

\[ \xymatrix@R=5em{ \Gamma (U, f^{-1}(\mathcal{O}_ Y)) \ar[d]_\cong \ar[r]^\cong & \Gamma (U, \mathcal{O}_ X) \\ \Gamma (U, f'^{-1}(\mathcal{O}_ V)) \ar@/^/[ru]^{f'^\sharp } \ar@/_/[ru]_{f''^\sharp } & } \]

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

\[ \xymatrix{ U \ar[d]_{f|_ U} \ar[r] & X \ar[d]^ f \\ V \ar[r] & Y } \]

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.

Comments (2)

Comment #5894 by Zhaolin on

In the proof of the uniqueness of , I guess there is a small typo of rather than .

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