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We have the categories, $S$, of simplicial sets and $SS$, of symmetric simplicial sets (whose simplices are unordered). There are functors:

$H:S\to SS$ forgetting the ordering on simplices and

$L:SS\to S$ mapping an unordered simplex to all possible ordered simplices that could correspond to it (so an unordered $n$-simplex maps to $n!$ ordered simplices).

A monograph of Denis-Charles Cisinski defined a model-structure on $SS$ and showed that these functors define a Quillen equivalence with the usual model structure on $S$.

For any simplicial set $X$, there is a canonical cofibration

$\eta_X:X\to LHX$ --- every simplex of $X$ is augmented by all possible other orderings of its vertices.

Question: Is there a class of simplicial sets (like fibrant ones, for instance) for which $\eta_X$ is a weak equivalence?

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Denis-Charles Cisinski proves that, if $X$ is a symmetric simplicial set that is fibrant, the canonical fibration: $HLX\to X$ is a weak equivalence. (it "multiplies" every $n$-simplex by $n!$ –  Justin Smith Jun 28 '13 at 15:11
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In Remark 3.5 of "Left-determined model categories and universal homotopy theories" (AMS Transactions vol 355) J. Rosicky and W. Tholen assert that $\eta_X$ is a trivial cofibration for all $X$. This is easily refuted (other statements in this paper have been shown to be false). –  Justin Smith Jun 29 '13 at 21:18
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1 Answer 1

I think the answer to my question is yes. If X is a fibrant simplicial set, I think there is a trivial fibration

$LHX\to X$.

Here's my reasoning: The conclusion is certainly true if $X=LY$ for some fibrant $Y$ in $SS$ because there exists a trivial fibration $HLY\to Y$. Now apply the $L$ functor to get $LHLY\to LY$ or $LHX\to X$. I think that any fibrant $X$ satisfies the condition above. It is well-known that fibrant simplicial sets have every possible orientation of every simplex in them (i.e., if $\Delta$ is an $n$-simplex in a fibrant simplicial set, a prism argument shows that there are $n!-1$ other simplices that are merely different orientations of $\Delta$.

It follows that $X=LY$ where $Y$ is the result of replacing each of these sets of oriented simplices by un-oriented simplices.

It is also not too hard to show that $Y$ is fibrant in $SS$.

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Strictly speaking, this doesn't answer my question, but is just as good as a yes-answer (for the application I have in mind). –  Justin Smith Jun 30 '13 at 13:30
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