The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

25.3 Open immersions of locally ringed spaces

Definition 25.3.1. 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 25.3.2. 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 25.3.3. 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 25.3.2 above is the open subspace of $X$ associated to $U$.

Lemma 25.3.4. 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 25.2.2). Note that $f = j \circ f'$ as morphisms of locally ringed spaces by construction.

Suppose we have another morphism $f'' : (X, \mathcal{O}_ X) \to (V, \mathcal{O}_ Y)$ 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 25.3.5. 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 (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01HD. Beware of the difference between the letter 'O' and the digit '0'.