Timeline for Do disjoint unions of stacks commute with finite fibre products?
Current License: CC BY-SA 4.0
18 events
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| Mar 11, 2020 at 9:17 | comment | added | David Roberts♦ | Another way to think about it is how the extensive topology interacts with stackification for superextensive topologies. Daniel Schäppi showed something you might want to check out, see discussion at nforum.ncatlab.org/discussion/3907/… | |
| Mar 10, 2020 at 12:55 | comment | added | S. Carnahan♦ | @sdigr My mistake - your guess is correct. | |
| Mar 8, 2020 at 23:27 | answer | added | Harry Gindi | timeline score: 2 | |
| Mar 8, 2020 at 1:49 | comment | added | Harry Gindi | @RizaHawkeye The nontrivial part of this statement is not the universality of colimits in the 2,1-topos of groupoid stacks on Sch_fppf. It's that the inclusion of Artin stacks into this category preserves coproducts. This is an important property of the topology, and it is clearly false if one chooses the trivial topology (more or less Scott's comment). | |
| Mar 8, 2020 at 1:46 | comment | added | Harry Gindi | @S.Carnahan The definition of the coproduct of categories fibred in groupoids that sdigr is indeed correct. The point is that the (2,1)-Yoneda embedding doesn't preserve colimits (nor does the (1,1)-Yoneda embedding!). The stackified (2,1)-Yoneda embedding, however, does commute with coproducts. | |
| Mar 8, 2020 at 0:10 | history | edited | sdigr | CC BY-SA 4.0 |
countable index set is not necessary as M.Brandenburg pointed out
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| Mar 7, 2020 at 23:16 | comment | added | sdigr | @Riza Hawkeye Thanks, this is nice to know, but I did not learn about higher topos yet and thats why I try to understand this first in a more elementary way. | |
| Mar 7, 2020 at 23:14 | comment | added | sdigr | @S. Carnahan So the disjoint union of categories fibred in groupoids is defined not this way? I found this definition in the book [Champs algébriques, G.Laumon/ L.Moret-Bailly] in chapter two under (2.2.1). This is the only reasonable definition I can think of. I use this clumsy notation, because I want to understand how things work in very detail (I am an average level master student, not a working mathematician, and I deal with this the first time). | |
| Mar 7, 2020 at 22:46 | comment | added | user147129 | This is just universality of colimits in a higher topos. | |
| Mar 7, 2020 at 19:53 | answer | added | S. Carnahan♦ | timeline score: 1 | |
| Mar 7, 2020 at 19:53 | comment | added | S. Carnahan♦ | Your first guess is wrong even for schemes: Let each $X_i$ be a point, and let $T$ be 2 points. The description in the next comment is correct (modulo clumsy notation). | |
| Mar 7, 2020 at 14:30 | comment | added | sdigr | Let $\coprod_i -$ be the stack in groupoids associated to the coproduct in the $2$-category of categories fibred in groupoids. If I did understand the process of stackification correctly, then the map $\mathcal{X}_i(T)\to (\coprod_i \mathcal{X}_i)(T)$ is given by choosing $T=T_i$ and $T_j=\emptyset$ for $j\not= i$ and then mapping $x_i\in \mathcal{X}_i(T)$ to the $T$-valued point $(T=\coprod_i T_i=T_i, (x_i\mid_{T_i})_{i\in I}=\{x_i\mid_{T_i}\})$!? | |
| Mar 7, 2020 at 14:14 | comment | added | sdigr | However I do not see, how the tuple on the left is mapped to the one on the right. This is likely related to the fact, that I do not understand how disjoint unions work in the $2$-category of categories fibred in groupoids. For example, how does one define $\mathcal{X}_i\to \coprod_i \mathcal{X}_i$ ? My first guess would be to define $(\coprod_i \mathcal{X}_i)(T)\colon\!\!=\coprod_i( \mathcal{X}_i(T))$ for categories fibred in groupoids? | |
| Mar 7, 2020 at 14:11 | comment | added | sdigr | @S.Carnahan I have tried to write down the morphism on $T$-valued points. Given $T$, on the left hand side we have $(T=\coprod_i T_i, (\xi_i)_{i\in I})$ where $\xi_i\in (\mathcal{X}_i\times_{\mathcal{Z}}\mathcal{Y})(T_i)$ i.e. $\xi_i=(x_i,y_i,\varphi_i)$ where $\varphi_i$ is an isomorphism $\rho_{\mathcal{X}_i}(x_i)\overset{\sim}{\to}\rho_{\mathcal{Y}}(y_i)$ in $\mathcal{Z}_{T_i}$ if $\rho_{\mathcal{X}_i},\rho_{\mathcal{Y}}$ denote the given morphisms $\mathcal{X}_i\to \mathcal{Z}$ and $\mathcal{Y}\to \mathcal{Z}$ respectively. On the right we have $((T=\coprod_i T'_i, x'_i),y,\varphi)$. | |
| Mar 1, 2020 at 21:31 | comment | added | Martin Brandenburg | Of course $\mathbb{Z}$ can be replaced by any index set. | |
| Mar 1, 2020 at 14:08 | comment | added | S. Carnahan♦ | Have you tried writing down suitable functors between the two sides? | |
| Mar 1, 2020 at 13:47 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added top-level tag
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| Mar 1, 2020 at 13:03 | history | asked | sdigr | CC BY-SA 4.0 |