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Feb 21, 2023 at 20:36 comment added Tim Campion 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).
Feb 21, 2023 at 19:12 vote accept Emily
Feb 21, 2023 at 19:11 history edited Emily CC BY-SA 4.0
Split the question in two
Feb 19, 2023 at 19:20 answer added Dmitri Pavlov timeline score: 2
Feb 19, 2023 at 17:27 history edited Emily CC BY-SA 4.0
edited title
Feb 19, 2023 at 17:09 history asked Emily CC BY-SA 4.0