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

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