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    • CommentRowNumber1.
    • CommentAuthorHarry Gindi
    • CommentTimeApr 29th 2010
    • (edited Apr 29th 2010)

    On the page operad, in the section on the slick definition, we look at the "groupoid" of finite cardinals with bijections between them. However, pretty much by definition, the category of finite cardinals is skeletal (because a cardinal is an object representing the equipotence class of sets with that cardinality), and therefore its underlying groupoid will be discrete. Should we not use the groupoid of finite sets with bijections between them instead?

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeApr 29th 2010

    A skeletal category can still have nontrivial automorphisms. Otherwise no category with nontrivial automorphisms could have a skeleton.

    • CommentRowNumber3.
    • CommentAuthorHarry Gindi
    • CommentTimeApr 29th 2010

    Oh yes, that's true. Either way, could we still use the groupoid of finite sets instead of groupoid of finite cardinals (how you define these guys depends on your choice of foundations, so I'm wary of defining things in terms of them).

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeApr 29th 2010

    Yeah, when you talk about functors on some category it doesn’t matter if you use an equivalent category. How you define finite cardinals is really irrelevant; another definition of the skeletal category in question is “its objects are natural numbers, the automorphisms of nn are the group S nS_n, and there are no other morphisms.”

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 29th 2010
    • (edited Apr 29th 2010)

    Mike already answered, and I don’t think I can say it more simply than he just did, but just to answer your question more explicitly: you could do it using finite sets alone, if you wanted. The point is that Set FB opSet^{FB^{op}} with appropriate Day convolution satisfies the same 2-universal property as Set opSet^{\mathbb{P}^{op}}, so you just carry out the same argument with FBFB in place of \mathbb{P} mutatis mutandis.

    If you actually cared to write out formulas explicitly in those terms, it looks a little more complicated than it really ought. But just for kicks, let’s do it anyway. If you have a chosen coproduct on the category of finite sets, then define [n][n] to be the coproduct of nn copies of a chosen 1-element set. Regard these as objects of the underlying groupoid FBFB, and take the full subcategory of FBFB whose objects are these [n][n]. That can be your skeletal subcategory.

    Now say DD is symmetric-monoidally cocomplete. Given an object dd, define a functor

    FBDFB \to D

    which takes a finite set SS to hom FB([n],S) S nd nhom_{FB}([n], S) \otimes_{S_n} d^{\otimes n}, where S nS_n acts on each side according to symmetric monoidal structure. Call this thingy d Sd^{\otimes S}. All this applies in the special case where DD is Set FB opSet^{FB^{op}} equipped with its Day convolution.

    Now, given F,G:FB opSetF, G: FB^{op} \to Set, if you work through the relevant abstract nonsense, the substitution product FG:FB opSetF \circ G: FB^{op} \to Set will be given by the formula

    (FG)(T)= SFBF(S)×G S(T)(F \circ G)(T) = \int^{S \in FB} F(S) \times G^{\otimes S}(T)

    and an operad will be a monoid with respect to this substitution product.

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