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This might be too easy but I cannot proof it easily. Any reference or hint will be great.

Q: Suppose P is a poset in which every chain is finite and $\Delta P$ is the poset complex associated to it. (i.e. faces of $\Delta P$ are the chains of the poset.). Suppose I reverse the order in the poset and suppose it again gives me a poset say $\bar{P}$. Let $\bar{P}$ also has the property that every chain is finite. Does $\Delta P$ and $\Delta \bar{P}$ are homeomorphic.

Simple examples shows me that the second one gives me a subdivision of the first one, but is it true in general. Thank's in advance.

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  • $\begingroup$ There are a lot of "suppose" here... The reverse of an order is always an order, so you don't have to suppose that $\overline{P}$ is a poset, that's always true. Along the same lines: The chains of $P$ are the same as the chains in $\overline{P}$ so again the assumption that $\overline{P}$ has finite chains is not really an assumption, it's always true. And then we arrive at what @BorisNovikov said: Since the chains are the same, the complexes are the same (as simplicial complexes not just homemorphic). What was the question again? $\endgroup$
    – Johannes Hahn
    Commented Oct 24, 2013 at 13:59
  • $\begingroup$ This should be asked at mse. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 24, 2013 at 14:32

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Let $f:P\to \bar{P}$ is the antiisomorphism. It maps a chain $a_1<\ldots< a_n$ into $a_n<\ldots a_1$ so it maps every simplex into itself.

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