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Given a simplicial set $X_\bullet$, define its powerset simplicial set $\mathcal{P}_\bullet(X)$ as the composition $$\Delta^\mathsf{op}\xrightarrow{X_\bullet}\mathsf{Set}\xrightarrow{\mathcal{P}}\mathsf{Set},$$ where $\mathcal{P}$ is the covariant powerset functor.

How homotopically well-behaved is the powerset simplicial set construction?

  1. If $X_\bullet$ is a Kan complex, must $\mathcal{P}_\bullet(X)$ also be one?
  2. Given a Kan complex $X_\bullet$, are the homotopy groups of $X_\bullet$ related to those of $\mathrm{Ex}^\infty_\bullet(\mathcal{P}_\bullet(X))$?
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    $\begingroup$ If $X$ is a set, then $\mathcal P(X)$ is the free infinitary idempotent commutative monoid on $X$. It follows that if $X$ is a simplicial set, then $\mathcal P_\bullet(X)$ is the free infinitary idempotent commutative simplicial monoid on $X$. The unit of $\mathcal P_\bullet(X)$, given by $\emptyset$, is a disjoint basepoint for $\mathcal P_\bullet(X)$. I might guess that $\mathcal P_\bullet(X)$ is weakly homotopy equivalent to $\mathcal P(\pi_0(X))$ (taken discretely). $\endgroup$
    – Tim Campion
    Commented Feb 21, 2023 at 20:36

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The first question has a negative answer, given by the simplicial set $\def\Exi{{\sf Ex}^{\sf\infty}}X=\Exi Y$, where $Y$ is a simplicial set generated by vertices $a,b,b',c,c',d,d'$, 1-simplices $ab,ab',ac,ac',ad,ad'$, 2-simplices $abc,acd,abd,ab'c',ac'd',ab'd',abc'$, and 3-simplices $abcd,ab'c'd'$.

We specify a 3-horn $Λ^3_0→P(X)$ by setting its 1st face to $\{acd,ac'd'\}$, 2nd face to $\{abd,ab'd'\}$, 3rd face to $\{abc,ab'c',abc'\}$.

Any 3-simplex of $P(X)$ that fills in this horn must in particular contain a 3-simplex of $X$ whose 3rd face is $abc'$. This forces its 1st face to be $ac'd'$ and its 2nd face to be $abd$, which means that the 1-simplex connecting 0th and 3rd vertices would have to be simulataneously $ad$ and $ad'$, which is impossible.

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  • $\begingroup$ Thanks, Dmitri! I'll keep the question open for a bit to see if anyone has something to say about the remaining questions (and if not I'll split it in two and mark your answer with the green tick =) $\endgroup$
    – Emily
    Commented Feb 19, 2023 at 20:58

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