# on the necessity of gluing conditions

Suppose we are given a family of schemes $\left\{X_i\right\}_i$, with $U_{ij}$ open in $X_i$ such that there exists isomoprhism $\phi_{ij}: U_{ij} \rightarrow U_{ji}$. Why do we need the condition $\phi_{ij}(U_{ij} \cap U_{ik}) = U_{ji} \cap U_{jk}$ to glue the schemes? Surely, it is a natural condition, but is it truly necessary?

• Well, it depends on what you expect to get after you glue. Suppose the result is $X$; by abuse of notation, consider each $X_i$ as a subobject of $X$; then one would expect $X_i \cap X_j = U_{i,j} = U_{j,i}$. – Zhen Lin Mar 23 '16 at 14:26
• @ZhenLin: That is a good answer. – Manos Mar 23 '16 at 14:30

If we get \begin{equation} X_0=\coprod_{i\in I}X_i \end{equation} then we can define an equivalence relation $\sim$ on $X_0$ as following: \begin{equation} x,y\in X_0,x\in U_{ij},y\in U_{ji},\,x\sim y\iff\varphi_{ij}(x)=y \end{equation} and we can define \begin{equation} X=X_{0\displaystyle/\sim}. \end{equation} I remember to us that: \begin{gather} \varphi_{ii}=Id_{U_{ii}},\\ \varphi_{ik}=\varphi_{jk}\circ\varphi_{ij},\\ \varphi_{ij}=\varphi_{ji}^{-1}. \end{gather}