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I'm reading a book about Kan Complexes in Simplicial Homotopy Theory (Curtis), and came across this theorem at the beginning : if $f : \Delta[n] \times \Lambda^k[m] \to K$ is a simplicial map and $K$ is a Kan complex, then $f$ has an extension from $\Delta[n] \times \Delta[m] \to K$.

The proof is fairly confusing to me : "$\Delta[n] \times \Lambda^k[m]$ can be obtained from $\Delta[n] \times \Delta[m]$ by successively adjoining a simplex and one of its faces, all other faces already lying in a subcomplex. Iterated application of the extension condition gives the lemma."

The extension condition is that if $f : \Lambda^k[m] \to K$ is a simplicial map and $K$ is a Kan complex, then $f$ has an extension from $\Delta[m] \to K$.

Could you help me with this ? I do not understand what he means by "successively adjoining a simplex", what "subcomplex" he's talking about and how he applies the extension condition. I'm a very beginner in the domain and any help would be very appreciated !

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    $\begingroup$ As far as I can see, the statement "$\Delta[n] \times \Lambda^k[m]$ can be obtained from $\Delta[n] \times \Delta[m]$ by successively adjoining a simplex and one of its faces" has the two complexes interchanged; it should instead say that "$\Delta[n] \times \Delta^k[m]$ can be obtained from $\Delta[n] \times \Lambda[m]$ by successively adjoining a simplex and one of its faces." $\endgroup$
    – Andreas Blass
    Commented Nov 16, 2021 at 19:49
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    $\begingroup$ Furthermore, the point is that since $\Lambda^k[m]\hookrightarrow \Delta[m]$ is an anodyne extension, so is $\Delta[n]\times \Lambda^k[m]\hookrightarrow \Delta[n]\times \Delta[m]$. In fact, you could replace $\Delta[n]$ with any simplicial set. See, for example, this nlab page ncatlab.org/nlab/show/anodyne+morphism, in particular Proposition 3.1 $\endgroup$
    – Gregory Arone
    Commented Nov 16, 2021 at 19:53
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    $\begingroup$ I looked up Curtis' paper, and he does seem to have it backwards (lemma 1.14 in the version published in Advances in Mathematics). Must be just a typo. $A$ and $B$ should be switched. $\endgroup$
    – Gregory Arone
    Commented Nov 16, 2021 at 20:03
  • $\begingroup$ Yes that's what I thought, thank you. It's a bit less confusing now, however I cannot figure out a rigorous proof for this statement. I could for example extend the map restricted to elements of the form $(*,x)$ where $*$ are degenerates of $\Delta[n]$ obtained from a vertex and $x$ are points in $\Lambda^k[m]$. But I have no way to know whether or not this extension extends to all of $\Delta[n] \times \Delta[m]$. I would be very grateful for a detailed proof. More particularly, to what precise maps does he apply the extension conditions ? $\endgroup$
    – Raphaël Kalfon
    Commented Nov 16, 2021 at 21:23

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$\require{mathtools}$

The reason this works is that the inclusion map $i:\Delta[n]\times\Lambda^k[m]\hookrightarrow{}\Delta[n]\times\Delta[m]$ can be viewed as a composition. The maps in this composition are all monomorphisms of the form $i_{\ell}:X\hookrightarrow{}X\cup_{S}\Delta[m+n]$; that is, they adjoin one of the missing simplices, which all happen to be of dimension $m+n$. In particular, this is a pushout of a (generalized) horn inclusion, so it has the left lifting property with respect to Kan complexes. A detailed account of this decomposition can be found in, e.g., the appendix to Rezk's notes on quasicategories, but it's a good exercise to work it out yourself.

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