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Hi guys, I have a question. Prove or disproof the statement:


Any inner horn $\Lambda[n,k],0< k< n $ admits a filtration $\mathbf{n}<\cdots<\Lambda[n,k]$, such that each step is filling an inner horn.
where $\mathbf{n}$ is the simplicial set $0\to 1 \cdots \to n$ (this notation is bad), its 1-cells are $(i,i+1)$, do not include $(i,j), i+1< j$. For those who does not familiar with simplicial set, one could just think of $\Lambda[n,k]$ as an indexed geometrical horn. So basically this is a combinatary question. For $n=2,3$, this is apparent, for $n=4$ one can still do it by hand. I guest it is true for higher $n$.

Thanks for your reading.

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4 Answers 4

up vote 2 down vote accepted

Joyal and Tierney also prove (extending the result of Cordier and Porter) a large class of statements of this form (including, in addition to this fact, that the inclusion of the spine into a simplex (that is, the inclusion of $n$ into $\Delta^n$ in your notation) is inner anodyne). This is in the paper by Joyal and Tierney comparing quasicategories and complete Segal spaces. I think the title is something like "quasicategories vs segal spaces". They do this by using a formalism of "generalized horns".

Similar results were derived by Rezk in his paper on $\Theta_n$-spaces using a "cover" formalism, which are, for all intents and purposes, generalized spines. This approach has the advantage of admitting a homotopy-theoretic rather than a combinatorial proof. However, while covers do generate the class of inner anodynes, they do not, unfortunately, include the inner horns.

A Rezk cover of a simplex $\Delta[n]$ is a subpresheaf containing the spine $\operatorname{Sp}[n]$ having the right lifting property with respect to all "cospine" inclusions $\Delta[1]=\Delta[\{0,m\}]\hookrightarrow \Delta[m]$. In particular, all covers are stable under pullbacks along "sequential" maps (maps that induce maps on spines).

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I would make the argument that spine inclusions, rather than inner horn inclusions, are the primary inclusions of interest, and are philosophically a more natural type of map in the context of simplicial models for categories. Consider the following statement: A simplicial set $S$ is a quasicategory if and only if for every spine inclusion $\operatorname{Sp}[n]\hookrightarrow \Delta[n]$, the induced map $S^{\Delta[n]}\to S^{\operatorname{Sp}[n]}$ is a trivial fibration of simplicial sets (that is, it has the RLP with respect to all monomorphisms). –  Harry Gindi Sep 20 '11 at 9:44
    
This says that every composable family of $n$ morphisms has a contractible space of composites. This can be sharpened all the way to asking that the above condition hold only in the case that $n=2$. –  Harry Gindi Sep 20 '11 at 10:31
    
@Harry I printed their paper, they have a similar statement Lemma 1.21: $\mathbf{n}\hookrightarrow \Delta^n$ (not far from mine) is inner anodyne, referring to the unpublic work "The theory of quasi-category, I", in preparation. –  Ma Ming Sep 20 '11 at 15:39
    
@Ma Ming: Uhhh.. Then it's in Joyal's notes on quasicategories, maybe? I'll have to get back to you later on an exact reference, but it's in one of those papers by Joyal. –  Harry Gindi Sep 20 '11 at 21:45
    
@Harry Gindi I find that result in Joyal 'The Theory of Quasi-Categories and its Applications'pp85 Proposition 2.13, his proof is basically the same as mine. –  Ma Ming Sep 22 '11 at 16:30
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Essentially this is part of one of the proofs of the paper by Cordier and myself,(J.-M. Cordier and Tim Porter, Vogt’s Theorem on Categories of Homotopy Coherent Diagrams, Math. Proc. Camb. Phil. Soc. 100 (1986) pp. 65-90.) In any case the proofs there contain combinatorial ideas that should be helpful. (As I am not `at home' I have not access to the paper so I cannot check if or where it is discussed therein.)

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Thanks for your answer. I printed your paper yesterday. It seems you are playing the same game, filling/retracting horns. But I am not able to find the statement or extract a proof to my question from your paper. Thanks any way. This is more likely right and known by expert. –  Ma Ming Sep 20 '11 at 8:42
    
I got a induction proof last light. –  Ma Ming Sep 20 '11 at 9:36
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I think that what you are talking about has a cubical version for "partial boxes", and it might be interesting to relate the two versions. The following is Lemma 3.3 of

R. Brown and P.J. Higgins, ``Colimit theorems for relative homotopy groups'', J. Pure Appl. Algebra 22 (1981) 11-41.

and is also given an exposition in our EMS Tract vol 15 "Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids" (2011) Section 11.3.i Fibrant cubical sets.

[Chains of partial boxes] Let $B$, $B'$ be partial boxes in an $r$-cell $C$ of $\mathbb I^n$ such that $B' \subseteq B.$ Then there is a chain $$B = B_s \searrow B_{s-1} \searrow \cdots \searrow B_1 = B'$$ such that

(i) each $B_i$ is a partial box in $C$;

(ii) $B_{i+1} = B_i \cup a_i$ where $a_i$ is an $(r - 1)$-cell of $C$ not in $B_i$;

(iii) $a_i \cap B_i$ is a partial box in $a_i.$

The definition of a partial box is as follows:

Let $C$ be an $r$-cell in the $n$-cube $\mathbb I^n.$ Two $(r - 1)$-faces of $C$ are called opposite if they do not meet (except possibly in degenerate elements). A partial $(r-1)$--box in $C$ is a subcomplex $B$ of $C$ generated by one $(r - 1)$-face $b$ of $C$ (called a base of $B$) and a number, possibly zero, of other $(r - 1)$-faces of $C$ none of which is opposite to $b$. The partial box is a box if its $(r - 1)$-cells consist of all but one of the $(r - 1)$-faces of $C$.

I find the cubical arguments more geometric and so easier for me to find and to follow. The above Lemma is a crucial part of the whole theory. It may be related to some results in Kan's first paper.

There are other reasons for working with cubical sets, mainly the idea of "algebraic inverses to subdivision", using cubical sets with compositions, and also the rule $I^m \times I^n =I^{m+n}$.

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I got a proof, slightly modified after talking to a friend. Thanks @Tim and @Harry.

We prove a stronger statement: any inner horn $\Lambda^n_i$ can be retracted to $\Delta^{n-1}\cup_{\{n-1\}} \Delta^1\{ n-1, n \}$, then by induction the statement follows. To this end, we claim that $\Lambda^n_i$ admits an inner horn retraction to $\Delta^{n-1} \cup_{ \Delta^{n-2}\{1,\cdots,n-1\} } \Delta^{n-1}\{1,\cdots n \} $, then the stronger statement follows also by induction.

To prove the claim, noticing that $\Delta^{n-1} \cup_{ \Delta^{n-2}\{1,\cdots,n-1\} } \Delta^{n-1}\{1,\cdots n \}\hookrightarrow \Lambda^n_i$ is a trivial cofibration, therefore $\Lambda^n_i$ admits a retraction to $\Delta^{n-1} \cup_{ \Delta^{n-2}\{1,\cdots,n-1\} } \Delta^{n-1}\{1,\cdots n \} $. But each step is retracting a horn containing $ \{0 ,n \}$, thus an inner horn retraction.

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@Harry, thanks for pointing out this fault. –  Ma Ming Sep 20 '11 at 13:28
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