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    • CommentRowNumber1.
    • CommentAuthorziggurism
    • CommentTimeMar 6th 2020

    A torsor or principal bundle may be characterized as a group action ρ:P×GP\rho: P\times G\to P such that the associated map 1×ρ:P×GP×P1\times\rho:P\times G \to P\times P is an isomorphism (in whatever category, often a comma category C/XC/X especially when using bundle language)

    A groupoid may be defined as two objects G 1G_1 (arrows) and G 0G_0 (points) and two maps s,t:G 1G 0s,t\colon G_1\to G_0, the source and target morphisms (along with some other data and axioms). These two morphisms may be assembled into a product (s,t):G 1G 0×G 0.(s,t)\colon G_1\to G_0\times G_0. If the groupoid is discrete (only arrows are identity arrows) then ss and tt are both the identity morphism and s×ts\times t is the diagonal map. If the groupoid is codiscrete (aka pair groupoid, every pair of points has a unique arrow), then ss and tt are projections and s×ts\times t is equality.

    If you consider the action groupoid of the torsor ρ:P×GP\rho: P\times G\to P, and assemble its source and target morphisms into a product as above, you recover the map 1×ρ:P×GP×P1\times\rho:P\times G \to P\times P which is stipulated to be an isomorphism. So a codiscrete groupoid has a s×ts\times t map which is equality, whereas a torsor has a s×ts\times t map which is an isomorphism.

    So would it be correct to characterize a torsor as an “essentially codiscrete” groupoid? As in, it’s codiscrete only up to isomorphism, instead of equality?

    Is there a bigger context to this P×GP×PP\times G \to P\times P map associated to a group action? Is it just a coincidence that it plays an important role both in the definition of the torsor, and the action groupoid?