6
$\begingroup$

Recall:

Given a category $A$, and two classes of morphisms $S,S'$, we say that $S$ is right-cancellative with respect to $S'$ if for any pair of maps $f\in S, g\in S'$ such that $gf$ is defined, we have the implication $gf\in S \Rightarrow g\in S$.

Recall that the class of inner-anodyne morphisms in the category of simplicial sets is defined to be the class $\operatorname{llp}(\operatorname{rlp}(E))$, where $E$ denotes the set of inner-horn inclusions $\iota^n_k:\Lambda^n_k \hookrightarrow \Delta^n$ for $0<k<n.$

Question:

Is the class of inner-anodynes right-cancellative with respect to the class of monomorphisms? This is certainly the case for the Joyal-trivial cofibrations, but being inner-anodyne is a rather stronger condition.

$\endgroup$
2
  • $\begingroup$ Incidentally, do you know if right plus left anodyne implies inner anodyne? $\endgroup$ – Akhil Mathew Dec 3 '11 at 7:32
  • 2
    $\begingroup$ @Akhil: The map $\Delta[1] \to \Delta[2]$ induced by $\delta_1 : [1] \to [2]$ is both left and right anodyne, but it's not inner anodyne since it's not bijective on $0$-simplices. $\endgroup$ – Karol Szumiło Dec 3 '11 at 8:41
3
$\begingroup$

The fact that in $\operatorname{Set}_{\Delta}$ the class of inner anodyne maps has the right cancellation property (within the class of monomorphisms) was proven by Danny Stevenson in his 2016 paper Stability for inner fibrations revisited, which was also published in TAC.

$\endgroup$
3
  • $\begingroup$ Absolute necromancer! Better late than never! Thanks! $\endgroup$ – Harry Gindi Jul 20 '18 at 11:22
  • 1
    $\begingroup$ The reason why this is particularly nice is that it shows that the smallest weakly-saturated right-cancellative class of morphisms containing the spine inclusions is exactly the inner anodynes. Joyal was only able to show this for trivial cofibrations, but this is a much stronger result! $\endgroup$ – Harry Gindi Jul 20 '18 at 11:30
  • 1
    $\begingroup$ Indeed. Cf. Lemma 3.5 in Joyal and Tierney's Quasi-categories vs. Segal spaces. $\endgroup$ – Daniel Gerigk Jul 20 '18 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.