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    • CommentRowNumber1.
    • CommentAuthorDean
    • CommentTimeSep 9th 2020
    • (edited Sep 9th 2020)

    Consider either the category of sheaves over the big zariski site, or some setting for Lawvere’s synthetic differential geometry.

    What does the category of objects XX with a fixed reduction X 0X_0 look like. The morphisms are morphisms XYX \rightarrow Y such that X redY redX_{red} \rightarrow Y_{red} is an isomorphism.

    These are like the fibers of TypeType red\text{Type} \rightarrow \text{Type}_{red} where Type red\text{Type}_{red} consists of the reduced objects of Type\text{Type}, under modal homotopy type theory.

    In the geometric setting, we should get some kind of linear category, like the category of quasicoherent sheaves. But I can’t find a place where this is discussed.

    It’s a bit like how QuasicoherentSheavesScheme\text{QuasicoherentSheaves} \rightarrow \text{Scheme} is like a 22-fibration, and the fiber of a scheme is its category of quasicoherent sheaves.

    Or the “2-fibration” of a tangent category over a given category, possibly adjoint to a certain 22-context extension.

    One thing to note is that all of these fibers of categories are settings for some linear type theory (possibly without objects being dualizable).

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeSep 9th 2020

    This is discussed in Section 2.3.4 (p. 35) of arXiv:1402.7041.

    • CommentRowNumber3.
    • CommentAuthorDean
    • CommentTimeSep 10th 2020
    • (edited Sep 10th 2020)

    The algebras for the maybe monad canonically have a monoidal structure. I can’t figure out why this extends to a monoidal structure on all of the infinitesimal extensions. Particularly, how to get the monoidal structure of jets, or a monoidal closed structure. Is there a way to get that canonically?

    I’m slowly learning the theory here. Thanks for the excellent reference. I think in a few months I might approach the problem you mentioned earlier, to see if I can make use of myself. Thanks for all of your help, Urs.