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1. • CommentRowNumber1.
   • CommentAuthorZhen Lin
   • CommentTimeFeb 22nd 2013
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   Author: Zhen Lin
   Format: MarkdownItexIt seems that there is a general sense in which $\pi_0$ preserves finite products, whether $\pi_0$ is acting on topological spaces or simplicial sets or categories or groupoids. I believe taking the skeleton of a category or a preorder also preserves finite products. Is there a common reason for all this? Tangentially related, does $Ho$ preserve finite products, considered as a functor from relative categories (in the sense of Barwick and Kan [2012]) to categories? If it did then we would have a quintuple adjoint string $Ho \dashv min \dashv und \dashv max \dashv weq : \mathbf{RelCat} \to \mathbf{Cat}$ with the leftmost adjoint preserving finite products -- and so we would have something that looks very much like a cohesion structure.

   It seems that there is a general sense in which preserves finite products, whether is acting on topological spaces or simplicial sets or categories or groupoids. I believe taking the skeleton of a category or a preorder also preserves finite products. Is there a common reason for all this?

   Tangentially related, does preserve finite products, considered as a functor from relative categories (in the sense of Barwick and Kan [2012]) to categories? If it did then we would have a quintuple adjoint string with the leftmost adjoint preserving finite products – and so we would have something that looks very much like a cohesion structure.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeFeb 22nd 2013
   • (edited Feb 22nd 2013)
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   Author: Todd_Trimble
   Format: MarkdownItexLet's see: $\pi_0$ for spaces is a reflexive coequalizer, starting with an obvious coreflexive pair involving the interval $I$: $$Top(I, X) \stackrel{\to}{\to} Top(1, X) \to \pi_0(X)$$ and of course reflexive coequalizers commute with finite limits. Suppose we replace the interval category $\mathbf{2}$ with the codiscrete groupoid $K(2)$ on two objects? It seems to me the corresponding reflexive coequalizer of $$C^{K(2)} \stackrel{\to}{\to} C^1$$ should be the skeleton of $C$. (I'd never thought of it this way, though.) Edit: possibly not. I just woke up.

   Let’s see: for spaces is a reflexive coequalizer, starting with an obvious coreflexive pair involving the interval :

   and of course reflexive coequalizers commute with finite limits. Suppose we replace the interval category with the codiscrete groupoid on two objects? It seems to me the corresponding reflexive coequalizer of

   should be the skeleton of . (I’d never thought of it this way, though.)

   Edit: possibly not. I just woke up.

3. • CommentRowNumber3.
   • CommentAuthorZhen Lin
   • CommentTimeFeb 22nd 2013
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   Author: Zhen Lin
   Format: MarkdownItexTaking the coequaliser of $C^{K(2)} \rightrightarrows C^1$ kills all the automorphisms, though. But this certainly explains why taking the _set_ of isomorphism classes in a category preserves products. Thanks!

   Taking the coequaliser of kills all the automorphisms, though. But this certainly explains why taking the set of isomorphism classes in a category preserves products. Thanks!

4. • CommentRowNumber4.
   • CommentAuthorKarol Szumiło
   • CommentTimeFeb 22nd 2013
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   Author: Karol Szumiło
   Format: MarkdownItexThe functor $Ho : \mathbf{RelCat} \to \mathbf{Cat}$ does preserve finite products. This is an exercise in adjointness using the fact that both $\mathbf{RelCat}$ and $\mathbf{Cat}$ are cartesian closed.

   The functor does preserve finite products. This is an exercise in adjointness using the fact that both and are cartesian closed.

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeFeb 22nd 2013
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   Author: Todd_Trimble
   Format: MarkdownItexYeah, that was what I was worried about. But it works for preorders.

   Yeah, that was what I was worried about. But it works for preorders.

6. • CommentRowNumber6.
   • CommentAuthorZhen Lin
   • CommentTimeFeb 22nd 2013
   • (edited Feb 22nd 2013)
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   Author: Zhen Lin
   Format: MarkdownItex@Karol: I'm afraid I don't follow. I can see that $min$ is a cartesian closed functor, which shows that $Ho$ satisfies various Frobenius reciprocity conditions, e.g. $$und ([Y, min Z]_{rel}) \cong [Ho Y, Z]$$ $$X \times Ho Y \cong Ho (min X \times Y)$$ and also that $\operatorname{Ho} 1 \cong 1$, but don't we need to know a little bit more before we can conclude that $Ho$ preserves binary products? **Postscript.** Ah, I see now. We need to use the fact that $[Y, min Z]_{rel} \cong min([Ho Y, Z])$ as relative categories. What's the abstract nonsense way of saying this?

   @Karol: I’m afraid I don’t follow. I can see that is a cartesian closed functor, which shows that satisfies various Frobenius reciprocity conditions, e.g.

   and also that , but don’t we need to know a little bit more before we can conclude that preserves binary products?

   Postscript. Ah, I see now. We need to use the fact that as relative categories. What’s the abstract nonsense way of saying this?

7. • CommentRowNumber7.
   • CommentAuthorKarol Szumiło
   • CommentTimeFeb 22nd 2013
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   Author: Karol Szumiło
   Format: MarkdownItexI'm not sure I follow your arguments and I'm also not sure I understand your notation. Is $min$ supposed to stand for "minimal relative structure"? If so then $Ho$ is not a left adjoint of $min$. Minimal relative structure has just identities as weak equivalences and $Ho$ is a left adjoint of a functor $iso$ that sends a category to a relative category with isomorphisms as weak equivalences. Now, the exercise in adjointness I had in mind is the following. Fix a relative category $A$ and consider a square $$ \array{ \mathbf{RelCat} & \stackrel{- \times A}{\to} & \mathbf{RelCat} \\ \downarrow^{Ho} & & \downarrow^{Ho} \\ \mathbf{Cat} & \stackrel{- \times Ho A}{\to} & \mathbf{Cat} \\ } $$ which can be filled with a natural transformation coming from the universal property of products. We want to show that it is an isomorphism. Since all the functors here are left adjoints it will suffice to verify that the corresponding transformation filling the square of their right adjoints is an isomorphism and this follows by a direct inspection.

   I’m not sure I follow your arguments and I’m also not sure I understand your notation. Is supposed to stand for “minimal relative structure”? If so then is not a left adjoint of . Minimal relative structure has just identities as weak equivalences and is a left adjoint of a functor that sends a category to a relative category with isomorphisms as weak equivalences.

   Now, the exercise in adjointness I had in mind is the following. Fix a relative category and consider a square

   which can be filled with a natural transformation coming from the universal property of products. We want to show that it is an isomorphism. Since all the functors here are left adjoints it will suffice to verify that the corresponding transformation filling the square of their right adjoints is an isomorphism and this follows by a direct inspection.

8. • CommentRowNumber8.
   • CommentAuthorZhen Lin
   • CommentTimeFeb 22nd 2013
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   Author: Zhen Lin
   Format: MarkdownItex> I’m not sure I follow your arguments and I’m also not sure I understand your notation. Is min supposed to stand for “minimal relative structure”? If so then Ho is not a left adjoint of min. Minimal relative structure has just identities as weak equivalences and Ho is a left adjoint of a functor iso that sends a category to a relative category with isomorphisms as weak equivalences. Ah, right, yes. I was assuming that every isomorphism is automatically a weak equivalence. Perhaps I was secretly thinking of homotopical categories rather than relative categories.

   I’m not sure I follow your arguments and I’m also not sure I understand your notation. Is min supposed to stand for “minimal relative structure”? If so then Ho is not a left adjoint of min. Minimal relative structure has just identities as weak equivalences and Ho is a left adjoint of a functor iso that sends a category to a relative category with isomorphisms as weak equivalences.

   Ah, right, yes. I was assuming that every isomorphism is automatically a weak equivalence. Perhaps I was secretly thinking of homotopical categories rather than relative categories.

9. • CommentRowNumber9.
   • CommentAuthorZhen Lin
   • CommentTimeMar 20th 2013
   • (edited Mar 20th 2013)
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   Author: Zhen Lin
   Format: MarkdownItexAfter some further thought, there is actually just one reason why the leftmost adjoints in all of these strings of adjoints preserves finite products: $$\pi_0 \dashv disc \dashv ob \dashv codisc : \mathbf{Set} \to \mathbf{Grpd}$$ $$\mathbf{I} \dashv und \dashv iso : \mathbf{Grpd} \to \mathbf{Cat}$$ $$\tau_1 \dashv N : \mathbf{Cat} \to \mathbf{SSet}$$ $$\pi_1 \dashv N : \mathbf{Grpd} \to \mathbf{SSet}$$ $$\pi_0 \dashv disc \dashv (-)_0 \dashv cosk_0 : \mathbf{Set} \to \mathbf{SSet}$$ $$Ho \dashv min^+ : \mathbf{Cat} \to \mathbf{RelCat}$$ Namely, because each one is the reflector of an [[exponential ideal]]. Admittedly, it seems to be easier to prove directly that $\tau_1 : \mathbf{SSet} \to \mathbf{Cat}$ preserves finite products, but everything else follows by nonsense. This also explains why $skel = \pi_0 iso : \mathbf{Cat} \to \mathbf{Set}$ preserves finite products.

   After some further thought, there is actually just one reason why the leftmost adjoints in all of these strings of adjoints preserves finite products:

   Namely, because each one is the reflector of an exponential ideal. Admittedly, it seems to be easier to prove directly that preserves finite products, but everything else follows by nonsense. This also explains why preserves finite products.

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeMar 21st 2013
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    Author: Mike Shulman
    Format: MarkdownItexNice!

    Nice!