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Let $\mathsf C$ be a category and $G$ an internal group. Suppose $\mathsf C$ is finitely complete, so that $\pi_2:G\times B\to B$ is an internal group in $\mathsf C_{/B}$ for every $B$. A $G$-bundle over $B$ is a $\pi_2$-action in $\mathsf C_{/B}$.

Say a $G$-bundle is a trivial $G$-torsor over $B$ if it's isomorphic to the multiplicative action of $\pi_2$ on itself. A $G$-bundle $\alpha:A\to B$ over $B$ is a trivial $G$-torsor over $B$ iff it has a global section and the canonical map $G\times A\to A\times _B A$ is an isomorphism of the bundles $\pi_2:G\times A\to A$ and $\alpha^\ast\alpha$.

Proposition 4.43 of Vistoli's notes on descent gives an analogous characterization for locally triviail $G$-torsors where "locally" is expressed in terms of a subcanonical site structure on the underlying category, using lemma 4.44.

I want to define "locally" in terms of descent: say a $G$-bundle $\alpha:A\to B$ is a $G$-torsor if it admits an effective descent morphism $p$ pulling it back to a trivial $G$-torsor. I then expect the following conditions to be equivalent for a $G$-bundle to be a $G$-torsor.

There's an effective descent morphism $p$ such that:

  • $p^\ast \alpha $ admits a global section;
  • the canonical map $G\times A\to A\times _B A$ is an isomorphism of the bundles $\pi_2:G\times A\to A$ and $\alpha^\ast\alpha$.

That is, a locally trivial torsor still trivializes itself. Are the conditions indeed equivalent? How to prove the equivalence?

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  • $\begingroup$ One direction is clear - take $p=\alpha$. In the other, I think it is enough to require of $p$ that $p^*$ reflects isomorphisms: if $p^*\alpha$ has a global section then it can be used to provide an inverse for $p^*(G\times A\to A\times_BA)$, so the latter is an isomorphism. $\endgroup$ Commented Jan 19, 2017 at 14:52
  • $\begingroup$ @მამუკაჯიბლაძე Doesn't pullback along effective descent morphisms always reflect isomorphisms? $\endgroup$
    – Arrow
    Commented Jan 19, 2017 at 15:55
  • $\begingroup$ Yes I believe it does $\endgroup$ Commented Jan 19, 2017 at 16:05
  • $\begingroup$ So is this question answered? $\endgroup$
    – HeinrichD
    Commented Mar 10, 2017 at 11:11

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