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    • CommentRowNumber1.
    • CommentAuthorzskoda
    • CommentTimeApr 25th 2011

    New entry adjoint monad with redirect adjoint comonad. Last night I had to struggle with proving the lemma at the bottom so I claimed the credit in brackets.

    • CommentRowNumber2.
    • CommentAuthorYaron
    • CommentTimeApr 25th 2011

    Added a query box in adjoint monad stating that the the construction in item 2 is what Mac Lane calls conjugate natural transformations (it’s in a query box because I’m not entirely sure if this is useful information for this entry - feel free to delete this).

    Also, I was a little confused by “there is a bijection of sets between natural transformations ϕ\phi and natural transformations ψ\psi … such that …” (in item 2), because the condition involves both ϕ\phi and ψ\psi. Wouldn’t it be better to say somethig like “given ϕ\phi there is a unique ψ\psi such that…” and vice versa?

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeApr 25th 2011
    • (edited Apr 25th 2011)

    Eilenberg and Moore write symbol \dashv “adjoint”, i.e. ϕψ\phi\dashv \psi: this conjugation, or duality depends on two adjunctions to start with. Thanks for confirming other terminology.

    In my opinion both formulations are equivalent. One can start with any ϕ\phi and find ψ\psi and other way around. Now what is such refering to. If it is refering to ψ\psi then it is precisely what you said. If it is misunderstood that such refers to the bijection, does not matter, as all ϕ\phi-s and all ψ\psi-s are involved and the shape of bijection is determined by the condition. We can make different formulation, though.

    • CommentRowNumber4.
    • CommentAuthorzskoda
    • CommentTimeApr 25th 2011

    OK, I incorporated your query into text as

    Eilenberg and Moore would write ϕψ\phi\dashv\psi and talk about “adjointness for morphisms” (of functors), which is of course relative to the given adjunctions among functors. MacLane calls the correspondence conjugation (Categories for Working Mathematician, 99-102).

    • CommentRowNumber5.
    • CommentAuthorYaron
    • CommentTimeApr 25th 2011


    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeApr 25th 2011

    By the way, to make the various square diagrams display with correct alignement, one can use the




    command to make arow labels stick out to the left or the right of the arrow alignement, respectively. For instance

          A (S,-) &\to& B(-,T)
         {}^{\mathllap{A(\phi,-)}}\downarrow &&\downarrow {}^{\mathrlap{B(-,\psi)}}
         A(S',-)&\to & B(-,T')


    A(S,) B(,T) A(ϕ,) B(,ψ) A(S,) B(,T) \array{ A (S,-) &\to& B(-,T) \\ {}^{\mathllap{A(\phi,-)}}\downarrow &&\downarrow {}^{\mathrlap{B(-,\psi)}} \\ A(S',-)&\to & B(-,T') }
    • CommentRowNumber7.
    • CommentAuthorzskoda
    • CommentTimeApr 25th 2011

    Yes, great, you told me once, but I could not recall the command and had no time to search through old images.

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeApr 26th 2011

    As I said at the other thread, I think this really is exactly a special case of what the Australians call mates; why do you say it isn’t? And isn’t the lemma at the end just a special case of the fact that mates are natural/functorial? There’s nothing special about the fact that the three adjunctions involved are all composites of the same given adjunction.

    the category End(A) of endofunctors… is monoidal with respect to the composition… Hence we can consider operads and algebras over operads

    I don’t know how to define an operad in a monoidal category that isn’t at least braided.