Timeline for Can a functorial factorization be modified so that it fixes the initial object?
Current License: CC BY-SA 4.0
12 events
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| Nov 20, 2018 at 19:05 | comment | added | Dylan Wilson | I think he’s using empty to mean there’s no maps to the initial object other than the identity- so that your counterexample doesn’t apply. The lil arrow is ‘left cone’ ie that category with one new object that maps to everything else uniquely and has no maps back. | |
| Nov 20, 2018 at 19:02 | comment | added | Bruno Stonek | @AaronMazel-Gee If $*$ is a zero object (is this what you meant by “empty”?), then a similar naïve definition of a $Q’$ as above also won’t work. Again, consider $X\to *\to Y$; on one hand it should go to $QX\to Q*\to QY$, and on the other hand it should go to $QX\to *\to QY$. This leads to exactly the same problem as before. What does the little triangle mean in your displayed arrow? | |
| Nov 20, 2018 at 18:21 | comment | added | Aaron Mazel-Gee | @BrunoStonek thanks for spelling that out -- of course, this becomes simpler when the initial object is empty. I think this is equivalent to the canonical functor $(C \backslash \{ \emptyset \})^\triangleleft \to C$ being an equivalence, which is where Harry's "extension by zero" would apply. | |
| Nov 20, 2018 at 14:32 | history | edited | Bruno Stonek | CC BY-SA 4.0 |
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| Nov 20, 2018 at 13:17 | comment | added | Harry Gindi | @BrunoStonek That's why I didn't give you an answer =). I still think that the functor can be rectified, but not sure. I think something could possibly work if you have a zero object, but not sure about the case of 'rings'. | |
| Nov 20, 2018 at 12:50 | comment | added | Bruno Stonek | @HarryGindi (cont.) Indeed, consider a composition $X\stackrel{f}{\to} *\stackrel{g}{\to} Z$ where $X$ and $Z$ are not $*$. Then $Q’$ maps the composition to $QX\to Q*\to QZ$. On the other hand, the composition of the $Q’$’s of these arrows is $QX\to Q*\to *\to QZ$. So now the question is whether $Qg:Q*\to QZ$ is equal to $Q*\to *\to QZ$, which is a case of assertion (A) above, which doesn’t hold for $Q$ (if it did, there would be no need to go through all this). | |
| Nov 20, 2018 at 12:50 | comment | added | Bruno Stonek | @AaronMazel-Gee I don't see how this would work. For simplicity, suppose instead of modifying the whole functorial factorization we just modify the functor $Q$, so we ask ourselves whether we can construct $Q'$, an endofunctor of $\mathcal C$ such that $Q’X$ is $QX$ if $X\not= *$ and is $*$ if $X=*$. OK, then on an arrow $X\to Y$ it should be $QX\to QY$ if $X,Y\not= *$, it should be $*\to QY$ if $X=*$, and it should be $QX\to Q*\to *$ if $X\not=*$ and $Y=*$. But this is not always a functor. | |
| Nov 19, 2018 at 17:55 | comment | added | Harry Gindi | @AaronMazel-Gee That's what I said in chat last week, but I wasn't sure enough to give it as an answer. It's like a cheeky extension by zero of the restriction to $C-\{*\}$ | |
| Nov 19, 2018 at 17:40 | comment | added | Aaron Mazel-Gee | Ah, in this case is it not possible to just re-define $Q\ast$ to be $\ast$ itself, and leave the rest of the values of $Q$ unchanged? I think all of the relevant naturality squares will automatically commute, since $\ast$ is initial. | |
| Nov 19, 2018 at 16:54 | history | edited | Bruno Stonek | CC BY-SA 4.0 |
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| Nov 19, 2018 at 16:53 | comment | added | Bruno Stonek | @HarryGindi That's indeed the case! I'll edit it in, thank you. | |
| Nov 19, 2018 at 15:49 | history | asked | Bruno Stonek | CC BY-SA 4.0 |