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The inertia of a stack in setoids is trivial
Statement
Let be a stack in groupoids over (Descent data, prestacks and stacks in groupoids over the fppf site) all of whose fibre categories are setoids, i.e. all of whose automorphism groups are trivial. Then the projection from the inertia stack (Algebraic stacks and their inertia stacks) is an equivalence of stacks in groupoids (Categories fibred in groupoids over a site); conversely, if this projection is an equivalence, then every fibre category of is a setoid. In particular, for an algebraic space over (Algebraic spaces over a scheme, defined as fppf sheaves), the fibre category of over is the discrete groupoid on , so .
Facts & Assumptions
Given: A stack in groupoids over the fppf site, its inertia stack with projection , and, in the last clause, the stack in setoids of an algebraic space .
has objects with and , for , a morphism from over to over is a base arrow over with ; in a fixed fibre this is exactly an isomorphism intertwining the two automorphisms; the projection forgets (Algebraic stacks and their inertia stacks).
An equivalence of categories fibred in groupoids induces fully faithful, essentially surjective functors on every fibre; an explicit inverse over the base up to natural isomorphisms establishes equivalence without making choices; a stack in setoids has only identity automorphisms, and the stack in setoids of an algebraic space has fibre category the discrete groupoid on the set of morphisms (Categories fibred in groupoids over a site, Descent data, prestacks and stacks in groupoids over the fppf site, Algebraic spaces over a scheme, defined as fppf sheaves).
Proof
Full faithfulness in the setoid case. Suppose every fibre category of is a setoid. Then the only objects of are . For any two such objects, every isomorphism in satisfies , so it lifts uniquely to a morphism . Thus the projection is fully faithful on each fibre.
Essential surjectivity in the setoid case. For every , the object of maps to , so the projection is essentially surjective on every fibre. The functor , sending an arrow to the same arrow , is an explicit inverse over the base: the inertia condition holds for identity automorphisms, and both composites are identities because every automorphism is the identity. Thus is an equivalence of stacks in groupoids, without using a choice-based converse to fibrewise essential surjectivity. Conversely, suppose is an equivalence. For any and , full faithfulness applied to and lifts the identity to a morphism between them. The inertia-morphism condition in [F1] then gives , so every fibre category is a setoid.
The stack in setoids of an algebraic space. If then is the discrete groupoid on by [F2], so its only automorphisms are identities and step 1.2 shows that is an equivalence; the fibre category of over is therefore the discrete groupoid on , which is exactly the fibre category of .
Depends on
Used by
- A quotient stack need not be a scheme Counterexample
Dependency tree · two levels
21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- The Stacks Project, Chapter 8 (Stacks), Section 8.7 and Chapter 4 (Categories), Section 4.34 (standard reference, not scraped)