Timeline for Is the natural isomorphism $|FX_\bullet| \cong F|X_\bullet|$ lax symmetric monoidal?
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22 events
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Jul 1, 2016 at 22:25 | comment | added | Dimitri Chikhladze | Above I meant "the more naturally defined structure on |-| is colax". | |
Jul 1, 2016 at 22:20 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 22:07 | comment | added | Dimitri Chikhladze | In the topological and simplicial cases the inverses to the lax structure of |-| are as described in the last part of my question. However they do not have straightforward monoidal analogues. | |
Jul 1, 2016 at 22:05 | comment | added | Dimitri Chikhladze | That is the point I was making. The more naturally defined structure on $|-|$ is lax. If your $F$ is lax, the compositions will not be either lax or colax. However, you can still form the diagram for $\tau$ by inverting the directions of the monoidal structure maps of $|-|$. I am talking about the commutativity of that diagram. Now, if $|-|$ is invertible, which happens to be the case in the topological and simplicial cases, then |-| is also lax and your original diagrams will commute. | |
Jul 1, 2016 at 18:29 | comment | added | Bruno Stonek | I'm sorry, but again there is something eluding me. At the beginning you define a colax structure on $|-|$. However, you're still assuming (and using) that $F$ is lax, so I don't see what monoidal structure the functors $F|-|$ and $|F-|$ could have (in particular, how to make sense of a monoidal transformation between them?) | |
Jul 1, 2016 at 16:41 | comment | added | Dimitri Chikhladze | Good. But, a product of the topological n and m dimensional simplices is isomorphic to the n+m dimensional simplex. | |
Jul 1, 2016 at 16:31 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 16:10 | comment | added | Bruno Stonek | It's clearer now, thanks! $$$$ You meant "lax structure of F" rather than "colax", and in the end, you meant "depends on the structure of $\Delta^n \times \Delta^m$" (which is not $\Delta^{n+m})$). As for your last comment, I believe what's key is that $|-|:sSet\to Top$ is strong symmetric monoidal: that's where the non-trivial topology comes in. From this and a couple more properties of $|-|$, I think one can deduce formally that $|-|:sTop\to Top$ is strong symmetric monoidal. I agree with you that we cannot expect the result to hold in general without getting some control on the $|-|$'s. | |
Jul 1, 2016 at 15:07 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 13:57 | comment | added | Dimitri Chikhladze | Sorry I have been making some typos. I think I corrected them all now. Hope now it is clearer. | |
Jul 1, 2016 at 13:51 | comment | added | Dimitri Chikhladze | The colimit $\int^{n,m} X_n \otimes Y_m \otimes \Delta^n \otimes \Delta^m$ still contains "summunds" $X_n\otimes Y_n\otimes \Delta^n \otimes \Delta^n$ (that is when $n = m$). So one can go to $X_n\otimes Y_n\otimes \Delta^n \otimes \Delta^n$ and still end up in the two argument coend. | |
Jul 1, 2016 at 13:44 | comment | added | Bruno Stonek | Thanks for the clarification. However, I don't see why you end up in a coend over $\Delta\times \Delta$. Shouldn't the arrow point to $\int^n X_n \otimes Y_n \otimes \Delta^n \otimes \Delta^n$? | |
Jul 1, 2016 at 13:39 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 13:34 | comment | added | Dimitri Chikhladze | The coend is defined using the natural family $1\otimes 1\otimes \delta_{\Delta^n} : X_n\otimes Y_n\otimes\Delta^n \rightarrow X_n\otimes Y_n\otimes\Delta^n\otimes \Delta^n$ | |
Jul 1, 2016 at 13:22 | comment | added | Bruno Stonek | I'm confused... How do you get the morphism in your displayed coends? | |
Jul 1, 2016 at 12:28 | comment | added | Dimitri Chikhladze | Ok, added some details. | |
Jul 1, 2016 at 12:26 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 12:01 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 10:51 | comment | added | Bruno Stonek | Thanks for taking the time! Sorry, but when you say "these diagrams will commute" -- could you elaborate? (also, I think there's an $m$ which should be an $n$ in your displayed coends) | |
Jul 1, 2016 at 9:52 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 9:46 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
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Jul 1, 2016 at 0:27 | history | answered | Dimitri Chikhladze | CC BY-SA 3.0 |