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Nov 26 at 7:46 vote accept naahiv
Nov 26 at 6:40 answer added naahiv timeline score: 0
Nov 25 at 21:03 comment added Tim Campion $Ch(CMon) = Psh_{CMon}(Ch)$ is a $CMon$-enriched presheaf category, where $Ch$ is the $CMon$-enriched category whose object set is $\mathbb N$, and $Hom(n,m) = \begin{cases} \mathbb N & m \in \{n, n +1\} \\ 0 & \text{otherwise} \end{cases}$. Composition is defined in the unique possible way.
Nov 25 at 8:38 comment added naahiv @TimCampion Thanks, sorry can you specify the result you're using for "compute the left adjoint on representables pretty easily to see it agrees with the dold kan functor"? I tried to use the fact that every colimit-preserving functor out of a presheaf category has a right adjoint (determined by value on representables), but DK is the left adjoint. Or are you using that Ch(CMon) is embedded in a presheaf category...?
Nov 25 at 6:21 comment added Tim Campion Chris’s functor obviously preserves limits (since it’s defined using kernels) (and the kernels are split so it also commutes with colimits) so it has a left adjoint and a right adjoint. Similarly for the dold kan functor. You should be able to compute the left adjoint on representables pretty easily to see it agrees with the dold kan functor. Then the isomorphism of composite functors you have is presumably the unit so it’s fully faithful (though iirc it’s not actually necessary to check this last bit)
Nov 25 at 3:37 history edited naahiv CC BY-SA 4.0
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Nov 25 at 1:37 comment added naahiv @ChrisSchommer-Pries I agree that your construction gives a left inverse to $\mathrm{DK}$. To prove $\mathrm{DK}$ is full, I seem to need that your construction is a faithful functor. Is this obvious/true?
Nov 24 at 21:54 history edited naahiv CC BY-SA 4.0
removed an incorrect remark of mine.
Nov 24 at 21:54 comment added naahiv Ah, of course, thank you both for the clarification! In that case I believe the original logic stands, and Chris's suggestion answers my question allowing me to generalize full faithfulness to any semiadditive category.
Nov 24 at 15:28 comment added Zhen Lin Commutative monoids in a symmetric monoidal category have coproduct given by tensor product in the base category. In this case the base category is the category of sets with the cartesian product.
Nov 24 at 14:49 comment added Maxime Ramzi I don't understand your Edit 2 : CMon is certainly semiadditive, and the definition/verification of that fact does not involve the tensor product
Nov 24 at 14:25 history edited naahiv CC BY-SA 4.0
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Nov 22 at 22:18 comment added naahiv @ChrisSchommer-Pries Update: I believe you are correct. Some confusion stemmed from different notation between Lurie and Goerss-Jardine. Regardless, your functor is a proper left inverse to what I've called $\mathrm{DK}$ and I will be happy if you post as an answer.
Nov 22 at 22:15 history edited naahiv CC BY-SA 4.0
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Nov 22 at 21:33 comment added naahiv That is indeed a chain complex and functorial. It occurs to me that the original Dold-Kan defines this complex with differential $(-1)^{n}d_n$ only so that the natural map into the Moore complex is a chain map, which is needed for the other direction of equivalence. In any case, I think your functor $\widetilde{N}$ is indeed a left inverse to $\mathrm{DK}$ via the usual proof. In this case I hope you will post this as an answer.
Nov 22 at 20:12 comment added Chris Schommer-Pries If I have a simplicial commutative monoid, $A_\bullet$, let $B_n$ be the intersection in $A_n$ of the kernels of all the $d_i$ except $d_n$, and let the differential be the last map $d_n$. Doesn't this define a chain complex $B_*$?
Nov 22 at 12:50 answer added Tim Porter timeline score: 0
Nov 22 at 12:34 history edited naahiv CC BY-SA 4.0
fixed definition
Nov 22 at 10:51 history edited naahiv
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Nov 22 at 10:03 history asked naahiv CC BY-SA 4.0