Lemma 6.125.
For every effective epimorphism \(\bigsqcup_{i=1}^n Y_i \to X\) in \(\Shv(\Lambda)\), the map \(\bigsqcup_{i=1}^n \phi_*(Y_i) \to \phi_*(X)\) is an effective epimorphism in \(\Shv(\Spec(\Lambda))\).
Proof
The category \(\Shv(\Spec(\Lambda))\) is generated under colimits by the image of \(u\): as \(\Spec(\Lambda)\) is a spectral space, it is generated by the quasi-compact opens. Since we have
\[\phi_*(Y_i) \times_{\phi_*(X)} u(\lambda) = \phi_*(Y_i \times_X \lambda),\]
we may assume that \(X = \lambda \in \Lambda\). Now, consider the epi-mono factorization \[Y_i \overset{u_i}{\twoheadrightarrow} \lambda_i \hookrightarrow \lambda.\]
Note that the map \(\lambda_i \hookrightarrow \lambda\) is representable, because any subobject of a representable is representable (since \(\Lambda\) is precisely the subobjects of the terminal sheaf).Since \(\Lambda_{/\lambda_i}\) has dimension at most \(0\), the map \(u_i\) has a section \(s_i\). So, we then have the composite \[\bigsqcup_{i=1}^n u(\lambda_i) \xrightarrow{(s_i)} \bigsqcup_{i=1}^n \phi_*(Y_i) \to u(\lambda),\]
which is an effective epimorphism in \(\Shv(\Spec(\Lambda))\), because \(u\) preserves finite joins. But then it follows that also the second map is an effective epimorphism.