Timeline for Do disjoint unions of stacks commute with finite fibre products?
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| Mar 10, 2020 at 13:01 | comment | added | S. Carnahan♦ | @sdigr To get $y$ from $y_i$, you pull back along the isomorphism $T \to \coprod_i T_i$. As far as stackification is concerned, if you have a finer covering, then we are concerned with gluing objects over $T_i$ in $X_i$, and descent for these objects is handled by the stack property of $X_i$. | |
| Mar 7, 2020 at 22:06 | comment | added | sdigr | I don't understand the description of the $T$-valued points of a disjoint union $\coprod_{i\in I}\mathcal{X}_i$ of stacks as the choice of a decomposition $T=\coprod_{i\in I}T_i$ and $(x_i)_{i\in I}, x_i\in {\mathcal{X}_i}_{T_i}$, because if I take the construction of stackification as in Lemma 02ZN literally, there does not seem to be an obvious reason why one can choose the covering indexed by the same index set $I$ and it is also not clear to me, why the descent data is not mentioned in this description. I think this causes extra confusion to me. | |
| Mar 7, 2020 at 21:50 | comment | added | sdigr | Thanks for the explanation. I understand, that one can pullback $y$ along $\coprod_i T_i\to T$ because of the fibre product property (after a choice of pullbacks was made in the very beginning). But given a family $(y_i)$ of $y_i\in \mathcal{Y}_{T_i}$ as on the left. How can one obtain $y$? I guess that more is needed than only the property that pullbacks exist uniquely. I guess that we have to keep track of the descent data coming from the stackification process aswell, do we not need this? | |
| Mar 7, 2020 at 19:53 | history | answered | S. Carnahan♦ | CC BY-SA 4.0 |