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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJun 25th 2017
    • (edited Jun 25th 2017)

    In some application I am running into a situation as follows, and it feels like this must be an example of a more general setup that might go by some name already. Please let me know if the following reminds of you of anything that you have seen before or thought about before.

    So in a context of homotopy theory I am faced with a pair of adjoint functors LR and a morphism of the form f:L(X)L(Y). Then it turns out that I want to be looking for a top horizontal morphism making the following diagram commute:

    XYηXηYRL(X)R(f)RL(Y)

    Here η denotes the adjunction unit. Hence if f were itself the image of some g under L, then this would naturally be filled by g, but the f I am looking at is not in the image of L.

    Another way to state the problem faced is: Lift the adjunct of f through the adjunction unit.

    Or rather, what I am really interested in is the following:

    In the situation at hand, an actual lift does not exist. But a “lift to first Goodwillie linear order” does exist. Namely it happens that I find a commuting diagram of the form

    YXϕ0BηXRL(X)R(f)RL(Y)

    and then I find a canonical lift

    ΩBΣBYϕ1Xϕ0BηXRL(X)R(f)RL(Y)

    where ΣBΩB denotes stabilization in the slice over B (so that ΣBY is a parameterized spectrum over B).

    So it seems evident that what I am looking at is a solution to “Solve the lifting problem to first Goodwillie linear order around a 0-order solution ϕ0”.

    In the concrete application that I am looking at this simply turns out to be the right thing to do for specific reasons of the setup. What I would like to understand is if this is also the “right thing to do on more general grounds”: That kind of lifting problem of the adjunct of some f as above through the adjunction unit, is this something that occurs elsewhere?

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeJun 25th 2017

    Well, in the simpler version of the problem, if such a g exists, then necessarily f=L(g) by the universal property of η.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJun 25th 2017
    • (edited Jun 25th 2017)

    Right, so I should just say right away that f is not in the image of L.

    Also I should state the form of the “first order solution” that I am looking at in a better way:

    The base space B really appears as a projection of RL(Y)

    RL(X)RL(Y)B

    which induces the map YB and the “first order factorization” is of the form

    XΩBΣB(Y)ηXΩBΣBηYRL(X)ΩBΣBRL(Y)

    I think. That should begin to make more sense.

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeJun 25th 2017

    Doesn’t ring any bells for me.