3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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 $X_i \cap X_j = U_{i,j} = U_{j,i}$. $\endgroup$ – Zhen Lin Mar 23 '16 at 14:26
  • $\begingroup$ @ZhenLin: That is a good answer. $\endgroup$ – Manos Mar 23 '16 at 14:30
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.