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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeApr 29th 2010
   • (edited Apr 29th 2010)
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   Author: Harry Gindi
   Format: MarkdownOn 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?

   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?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeApr 29th 2010
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   Author: Mike Shulman
   Format: MarkdownItexA skeletal category can still have nontrivial *automorphisms*. Otherwise no category with nontrivial automorphisms could have a skeleton.

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

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeApr 29th 2010
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   Author: Harry Gindi
   Format: MarkdownOh 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).

   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).

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeApr 29th 2010
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   Author: Mike Shulman
   Format: MarkdownItexYeah, 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 $n$ are the group $S_n$, and there are no other morphisms."

   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 $n$ are the group $S_n$, and there are no other morphisms.”

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeApr 29th 2010
   • (edited Apr 29th 2010)
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   Author: Todd_Trimble
   Format: MarkdownItexMike 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 $Set^{FB^{op}}$ with appropriate Day convolution satisfies the same 2-universal property as $Set^{\mathbb{P}^{op}}$, so you just carry out the same argument with $FB$ 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 $[n]$ to be the coproduct of $n$ copies of a chosen 1-element set. Regard these as objects of the underlying groupoid $FB$, and take the full subcategory of $FB$ whose objects are these $[n]$. That can be your skeletal subcategory. Now say $D$ is symmetric-monoidally cocomplete. Given an object $d$, define a functor $$FB \to D$$ which takes a finite set $S$ to $hom_{FB}([n], S) \otimes_{S_n} d^{\otimes n}$, where $S_n$ acts on each side according to symmetric monoidal structure. Call this thingy $d^{\otimes S}$. All this applies in the special case where $D$ is $Set^{FB^{op}}$ equipped with its Day convolution. Now, given $F, G: FB^{op} \to Set$, if you work through the relevant abstract nonsense, the substitution product $F \circ G: FB^{op} \to Set$ will be given by the formula $$(F \circ G)(T) = \int^{S \in FB} F(S) \times G^{\otimes S}(T)$$ and an operad will be a monoid with respect to this substitution product.

   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 $Set^{FB^{op}}$ with appropriate Day convolution satisfies the same 2-universal property as $Set^{\mathbb{P}^{op}}$, so you just carry out the same argument with $FB$ 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 $[n]$ to be the coproduct of $n$ copies of a chosen 1-element set. Regard these as objects of the underlying groupoid $FB$, and take the full subcategory of $FB$ whose objects are these $[n]$. That can be your skeletal subcategory.

   Now say $D$ is symmetric-monoidally cocomplete. Given an object $d$, define a functor

   $$FB \to D$$

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

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

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

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