Timeline for Internal logic of the topos of simplicial sets
Current License: CC BY-SA 3.0
38 events
| when toggle format | what | by | license | comment | |
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| Oct 29 at 2:46 | comment | added | HDB | @Gro-Tsen Ok, now I am confused about what I was confused about. Nothing you said is unknown to me but weirdly I lived under the unquestioned impression of a totally incorrect definition of an abstract simplicial complex, which implied in particular that it would essentially correspond to a simplicial set which is $k$-coskeletal for all $k$, i.e. looking like a complete graph in all dimensions on its vertices. I am confused why this could have happened but at least my issue is resolved now. Thanks! | |
| Oct 27 at 9:31 | comment | added | Gro-Tsen | @HDB: Regarding degenerate simplices, their presence or absence in a simplicial subset $S\subseteq\Delta^n$ is completely determined by the presence or absence in $S$ of the unique nondegenerate simplex of $\Delta^n$ of which they are a degeneration (sanity check: the term “simplicial subset” refers to a subset of the simplices which is chosen in a manner compatible with all face and degeneracy maps), so a simplicial subset $S$ of $\Delta^n$ is determined by the choice of nondegenerate simplices in $S$ (which is exactly a simplicial complex). | |
| Oct 26 at 20:42 | comment | added | HDB | @Gro-Tsen It is true that the non-degenerate subsimplices are exactly the abstract simplicial complexes but not all elements of $\mathsf{sSet}(\Delta^n, \Omega)$ are non-degenerate. In general, there are multiple maps $\Delta^n \to \Omega$, having the same image and this is my confusion here. | |
| Oct 25 at 13:20 | comment | added | Gro-Tsen | @HDB: Regarding your comment of 2023-04-03 13:13:59Z, it seems to me that an abstract simplicial complex on $[n] := \{0,\ldots,n\}$ is precisely the same thing as a simplicial subset of $\Delta^n$ (considering the nondegenerate simplices, in both cases the condition is that if you take a $k$-simplex you must take all its $\ell$-simplices for $\ell\leq k$). So your formula for $\Omega(n)$ is correct, and proves what is stated in this answer. | |
| Apr 22, 2023 at 1:02 | comment | added | François G. Dorais | @Bixxli I don't do this stuff regularly enough to troubleshoot your situation. I don't have much free time right now and I don't think I can help you at this time. | |
| Apr 17, 2023 at 18:00 | comment | added | HDB | @FrançoisG.Dorais Then which isomorphism is wrong? The first one is the Yoneda lemma, and the second is the definition of a subobject classifier. Now a subobject of an object $X$ in any topos $\mathcal{E}$ is an isomorphism class in the full subcategory of the slice category $\mathcal{E}/X$ on the monomorphisms, which in this case is a simplicial subset, or is it? | |
| Apr 4, 2023 at 2:12 | comment | added | François G. Dorais | @Bixxli No, your calculation is incorrect. The base topos is Set not sSet. See Mike's answer for clarification. | |
| Apr 3, 2023 at 13:13 | comment | added | HDB | I am confused, don’t we have $\Omega(n) = \mathsf{sSet}(\Delta^n, \Omega) = \mathrm{Sub}(\Delta^n)$? That is, the $n$-simplices of the subobject classifier are all simplicial subsets of $\Delta^n$, not only the simplicial complexes. | |
| Apr 25, 2018 at 14:53 | comment | added | David Spivak | Just wanted to add the tag "Dedekind number" here. en.wikipedia.org/wiki/Dedekind_number | |
| Mar 12, 2014 at 21:43 | vote | accept | Mike Shulman | ||
| Mar 12, 2014 at 14:35 | vote | accept | Mike Shulman | ||
| Mar 12, 2014 at 14:54 | |||||
| Mar 12, 2014 at 0:26 | history | edited | François G. Dorais | CC BY-SA 3.0 |
cleanup
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| Mar 11, 2014 at 23:43 | comment | added | Zhen Lin | But the construction of $\Omega$ tells you precisely that $\Omega (n)$ is (isomorphic to) the lattice of sieves on $[n]$...? Unless you mean to draw attention to the fact that the representable presheaves form a separating set. | |
| Mar 11, 2014 at 23:26 | comment | added | François G. Dorais | @ZhenLin: Out of curiosity, I wonder if there was a better way to explain all of this. The intuition from the beginning was that everything is computed pointwise in a presheaf topos, so all you have to know is what the lattice of sieves at $[n]$ looks like and what identities hold for all of these. Was there a way to explain this without going through the construction of $\Omega$? | |
| Mar 11, 2014 at 23:08 | comment | added | François G. Dorais | @ZhenLin: You're right! There is a free top but no free bottom. So it's the free distributive lattice with a free top or the free bounded distributive lattice without the free bottom. So the count is off by one whichever way you think about it... | |
| Mar 11, 2014 at 23:06 | history | edited | François G. Dorais | CC BY-SA 3.0 |
final correction, I hope!
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| Mar 11, 2014 at 23:01 | comment | added | Zhen Lin | @FrançoisG.Dorais I think I agree with everything now, except for the claim that $\Omega (n)$ is a free distributive lattice. By my count, $\Omega (0)$ has 2 elements, $\Omega (1)$ has 5 elements, and $\Omega (2)$ has 19 elements. This is one less than what OEIS says free distributive lattices should have. | |
| Mar 11, 2014 at 22:31 | comment | added | François G. Dorais | @MattF. That was quick! Looks like both Wikipedia and I got it wrong. The pictures on Wikipedia represent the free bounded distributive lattices. Though I originally had the word bounded in there, $D_n$ is actually the free distributive lattice on $n$ generators (which just happens to be bounded but isn't freely bounded). | |
| Mar 11, 2014 at 22:27 | history | edited | François G. Dorais | CC BY-SA 3.0 |
deleted 8 characters in body
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| Mar 11, 2014 at 22:24 | comment | added | François G. Dorais | @MattF. The weak excluded middle is not true in the lattices here. There is a well known subtlety with free distributive lattices that impacts what $0$ and $1$ end up being; I'll check later tonight to make sure I got the distinction right... | |
| Mar 11, 2014 at 22:03 | comment | added | user44143 | Does the Wikipedia diagram of free distributive lattices show what you intend for $D_0, D_1, D_2, D_3$? In those lattices it looks like $\neg P \vee \neg \neg P$ holds, which would be simpler. The diagrams are at en.wikipedia.org/wiki/… | |
| Mar 11, 2014 at 20:59 | history | edited | François G. Dorais | CC BY-SA 3.0 |
Removed extra word
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| Mar 11, 2014 at 20:25 | history | edited | François G. Dorais | CC BY-SA 3.0 |
minor edits
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| Mar 11, 2014 at 20:11 | comment | added | François G. Dorais | @ZhenLin: I forced myself to not skip any steps and I think it's much better now... | |
| Mar 11, 2014 at 20:10 | history | undeleted | François G. Dorais | ||
| Mar 11, 2014 at 20:10 | history | edited | François G. Dorais | CC BY-SA 3.0 |
Major rewrite!
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| Mar 11, 2014 at 19:19 | history | deleted | François G. Dorais | via Vote | |
| Mar 11, 2014 at 19:19 | comment | added | François G. Dorais | @ZhenLin: Yes, of course! Let me retry this... | |
| Mar 11, 2014 at 19:11 | comment | added | Zhen Lin | Ah. But then as I said, there are only two morphisms $1 \to \Omega$. | |
| Mar 11, 2014 at 18:50 | comment | added | François G. Dorais | @ZhenLin: Hm, I didn't mean to imply anything like that. I'm just trying to get at truth values $1 \to \Omega$ (without having to describe all of $\Omega$ or subobjects in general). | |
| Mar 11, 2014 at 18:45 | comment | added | Zhen Lin | Hmmm. Actually, I'm not so sure about the main assertion – that the lattice of subobjects of $\Delta^n$ is a free distributive lattice. There are 5 subobjects of $\Delta^1$, for instance. | |
| Mar 11, 2014 at 18:45 | comment | added | François G. Dorais | @ZhenLin: Right, I wasn't giving a full characterization of $\Omega$ since I also left out the functoriality conditions. Sorry for the confusion. Feel free to suggest better wording. | |
| Mar 11, 2014 at 18:43 | history | edited | François G. Dorais | CC BY-SA 3.0 |
fixed again
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| Mar 11, 2014 at 18:36 | comment | added | Zhen Lin | No, what I meant was that an $n$-simplex of $\Omega$ (= generalised element of $\Omega$ with domain of variation $\Delta^n$) is a sieve on $[n]$, hence can be identified with etc.; in particular global elements of $\Omega$ are just sieves on $[0]$. I'm not sure what you mean by "element of $\Omega$" without qualification. | |
| Mar 11, 2014 at 18:26 | history | edited | François G. Dorais | CC BY-SA 3.0 |
fixed after Zhen Lin's comment
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| Mar 11, 2014 at 18:25 | comment | added | Zhen Lin | I'm slightly confused by your statement about sieves. Surely what you mean is that a sieve on $[n]$ can be identified with an abstract simplicial complex with vertices drawn from $\{ 0, \ldots, n \}$? | |
| Mar 11, 2014 at 18:04 | history | edited | François G. Dorais | CC BY-SA 3.0 |
added 36 characters in body
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| Mar 11, 2014 at 17:55 | history | answered | François G. Dorais | CC BY-SA 3.0 |