If you have two locally ringed spaces, and a subspace of the first one is isomorphic to a subspace of the other, then you can glue them together into one big locally ringed space.

Lemma 26.14.1. Given any glueing data of locally ringed spaces there exists a locally ringed space $X$ and open subspaces $U_ i \subset X$ together with isomorphisms $\varphi _ i : X_ i \to U_ i$ of locally ringed spaces such that

1. $X=\bigcup _{i\in I} U_ i$,

2. $\varphi _ i(U_{ij}) = U_ i \cap U_ j$, and

3. $\varphi _{ij} = \varphi _ j^{-1}|_{U_ i \cap U_ j} \circ \varphi _ i|_{U_{ij}}$.

The locally ringed space $X$ is characterized by the following mapping properties: Given a locally ringed space $Y$ we have

\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits (X, Y) & = & \{ (f_ i)_{i\in I} \mid f_ i : X_ i \to Y, \ f_ j \circ \varphi _{ij} = f_ i|_{U_{ij}}\} \\ f & \mapsto & (f|_{U_ i} \circ \varphi _ i)_{i \in I} \\ \mathop{\mathrm{Mor}}\nolimits (Y, X) & = & \left\{ \begin{matrix} \text{open covering }Y = \bigcup \nolimits _{i \in I} V_ i\text{ and } (g_ i : V_ i \to X_ i)_{i \in I} \text{ such that} \\ g_ i^{-1}(U_{ij}) = V_ i \cap V_ j \text{ and } g_ j|_{V_ i \cap V_ j} = \varphi _{ij} \circ g_ i|_{V_ i \cap V_ j} \end{matrix} \right\} \\ g & \mapsto & V_ i = g^{-1}(U_ i), \ g_ i = \varphi _ i^{-1} \circ g|_{V_ i} \end{eqnarray*}

Proof. We construct $X$ in stages. As a set we take

$X = (\coprod X_ i) / \sim .$

Here given $x \in X_ i$ and $x' \in X_ j$ we say $x \sim x'$ if and only if $x \in U_{ij}$, $x' \in U_{ji}$ and $\varphi _{ij}(x) = x'$. This is an equivalence relation since if $x \in X_ i$, $x' \in X_ j$, $x'' \in X_ k$, and $x \sim x'$ and $x' \sim x''$, then $x' \in U_{ji} \cap U_{jk}$, hence by condition (1) of a glueing data also $x \in U_{ij} \cap U_{ik}$ and $x'' \in U_{ki} \cap U_{kj}$ and by condition (2) we see that $\varphi _{ik}(x) = x''$. (Reflexivity and symmetry follows from our assumptions that $U_{ii} = X_ i$ and $\varphi _{ii} = \text{id}_{X_ i}$.) Denote $\varphi _ i : X_ i \to X$ the natural maps. Denote $U_ i = \varphi _ i(X_ i) \subset X$. Note that $\varphi _ i : X_ i \to U_ i$ is a bijection.

The topology on $X$ is defined by the rule that $U \subset X$ is open if and only if $\varphi _ i^{-1}(U)$ is open for all $i$. We leave it to the reader to verify that this does indeed define a topology. Note that in particular $U_ i$ is open since $\varphi _ j^{-1}(U_ i) = U_{ji}$ which is open in $X_ j$ for all $j$. Moreover, for any open set $W \subset X_ i$ the image $\varphi _ i(W) \subset U_ i$ is open because $\varphi _ j^{-1}(\varphi _ i(W)) = \varphi _{ji}^{-1}(W \cap U_{ij})$. Therefore $\varphi _ i : X_ i \to U_ i$ is a homeomorphism.

To obtain a locally ringed space we have to construct the sheaf of rings $\mathcal{O}_ X$. We do this by glueing the sheaves of rings $\mathcal{O}_{U_ i} := \varphi _{i, *} \mathcal{O}_ i$. Namely, in the commutative diagram

$\xymatrix{ U_{ij} \ar[rr]_{\varphi _{ij}} \ar[rd]_{\varphi _ i|_{U_{ij}}} & & U_{ji} \ar[ld]^{\varphi _ j|_{U_{ji}}} \\ & U_ i \cap U_ j & }$

the arrow on top is an isomorphism of ringed spaces, and hence we get unique isomorphisms of sheaves of rings

$\mathcal{O}_{U_ i}|_{U_ i \cap U_ j} \longrightarrow \mathcal{O}_{U_ j}|_{U_ i \cap U_ j}.$

These satisfy a cocycle condition as in Sheaves, Section 6.33. By the results of that section we obtain a sheaf of rings $\mathcal{O}_ X$ on $X$ such that $\mathcal{O}_ X|_{U_ i}$ is isomorphic to $\mathcal{O}_{U_ i}$ compatibly with the glueing maps displayed above. In particular $(X, \mathcal{O}_ X)$ is a locally ringed space since the stalks of $\mathcal{O}_ X$ are equal to the stalks of $\mathcal{O}_ i$ at corresponding points.

The proof of the mapping properties is omitted. $\square$

Comment #2160 by Daniel Smolkin on

Suggested slogan: If you have two locally ringed spaces, and a subspace of the first one is isomorphic to a subspace of the other, then you can glue them together into one big locally ringed space.

There are also:

• 9 comment(s) on Section 26.14: Glueing schemes

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.

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 01JB. Beware of the difference between the letter 'O' and the digit '0'.