The Stacks project

Definition 21.5.1. Let $\mathcal{C}$ be a site. Let $\mathcal{G}$ be a sheaf of (possibly non-commutative) groups on $\mathcal{C}$. A pseudo torsor, or more precisely a pseudo $\mathcal{G}$-torsor, is a sheaf of sets $\mathcal{F}$ on $\mathcal{C}$ endowed with an action $\mathcal{G} \times \mathcal{F} \to \mathcal{F}$ such that

  1. whenever $\mathcal{F}(U)$ is nonempty the action $\mathcal{G}(U) \times \mathcal{F}(U) \to \mathcal{F}(U)$ is simply transitive.

A morphism of pseudo $\mathcal{G}$-torsors $\mathcal{F} \to \mathcal{F}'$ is simply a morphism of sheaves of sets compatible with the $\mathcal{G}$-actions. A torsor, or more precisely a $\mathcal{G}$-torsor, is a pseudo $\mathcal{G}$-torsor such that in addition

  1. for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ there exists a covering $\{ U_ i \to U\} _{i \in I}$ of $U$ such that $\mathcal{F}(U_ i)$ is nonempty for all $i \in I$.

A morphism of $\mathcal{G}$-torsors is simply a morphism of pseudo $\mathcal{G}$-torsors. The trivial $\mathcal{G}$-torsor is the sheaf $\mathcal{G}$ endowed with the obvious left $\mathcal{G}$-action.


Comments (2)

Comment #90 by Keenan Kidwell on

My comment actually refers to the sentence which comes just after this definition ends (sorry), saying that it is clear that a morphism of torsors is necessarily an isomorphism. It is clear that for such a morphism, if , then the simply transitive condition immediately implies that is bijective, but when I had to write down the equalizer diagrams for a covering such that in order to see that . Is this the argument you had in mind?

Comment #91 by on

Yeah, I guess there is something to it. You can argue as you say, or you can use that sheaves are local things, so a map of sheaves which is "locally" an isomorphism is an isomorphism. See Sites, Section 7.26.

There are also:

  • 2 comment(s) on Section 21.5: First cohomology and torsors

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