On the barycentric subdivision of a poset - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:33:32Z http://mathoverflow.net/feeds/question/57684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57684/on-the-barycentric-subdivision-of-a-poset On the barycentric subdivision of a poset Giacomo d'Antonio 2011-03-07T15:43:48Z 2011-03-07T16:47:25Z <p>Hi everybody,</p> <p>I'm interested in the barycentric subdivision of a poset $P$, defined as the face poset of the corresponding order complex $\mathcal F(\Delta(P))$ (or alternatively, the poset of chains in $P$ ordered by inclusion). Actually, I'm interested in the subdivision of small categories, but I'm also happy understanding the poset case.</p> <p>I was wondering if the functor $sd:\mbox{Poset} \to \mbox{Poset}$ has a right adjoint. I am aware of the Ex functor, which is right adjoint to the corresponding functor of simplicial sets. Can one define a similar object for posets?</p> http://mathoverflow.net/questions/57684/on-the-barycentric-subdivision-of-a-poset/57694#57694 Answer by Karol Szumiło for On the barycentric subdivision of a poset Karol Szumiło 2011-03-07T16:47:25Z 2011-03-07T16:47:25Z <p>This functor has no right adjoint, since it doesn't preserve colimits. Let $[m]$ denote the linearly ordered set <code>$\{0,\ldots,m\}$</code> and consider the pushout of $[0]\overset{0}{\to}[1]$ and $[0]\overset{1}{\to}[1]$, which can be identified with $[2]$. Now $\mathrm{sd}[2]$ is the poset of chains in $[2]$, but the pushout of $\mathrm{sd}[1]\leftarrow\mathrm{sd}[0]\to\mathrm{sd}[1]$ has only five elements, in fact it can be identified with the subset of $\mathrm{sd}[2]$ consisting of chains <code>$\{0\},\{1\},\{2\},\{0,1\},\{1,2\}$</code>.</p>