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Dec 20, 2021 at 22:04 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Nov 20, 2021 at 21:34 answer added Doron Grossman-Naples timeline score: 1
Nov 16, 2021 at 21:23 comment added Raphaël Kalfon 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 ?
Nov 16, 2021 at 20:03 comment added Gregory Arone 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.
Nov 16, 2021 at 19:53 comment added Gregory Arone 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
Nov 16, 2021 at 19:49 comment added Andreas Blass 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."
S Nov 16, 2021 at 19:36 review First questions
Nov 16, 2021 at 19:59
S Nov 16, 2021 at 19:36 history asked Raphaël Kalfon CC BY-SA 4.0