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
X=\bigcup _{i\in I} U_ i,
\varphi _ i(U_{ij}) = U_ i \cap U_ j, and
\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
Comments (1)
Comment #2160 by Daniel Smolkin on
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