Let $P$ denote the face poset of a simplicial complex, $\Delta$ the order complex of a poset, and $\sim$ homotopy equivalence. It's known that for any finite simplicial complex $\mathcal{K}$ that $\Delta(P(\mathcal{K}))$ is homeomorphic to $\mathcal{K}$ (it's the barycentric subdivision). See for example Bjorner's Topological Methods. But instead suppose you have a finite simplicial complex $\mathcal{K}$ and a subcomplex $A\subseteq\mathcal{K}$. Can it be shown that $\Delta(P(\mathcal{K}\setminus A))\sim\mathcal{K}\setminus A$? The face poset of $\mathcal{K}\setminus A$ is well defined. In fact $P(\mathcal{K}\setminus A)=P(\mathcal{K})\setminus P(A)$, but $\mathcal{K}\setminus A$ isn't actually a simplicial complex. Also if this is true could the results be extended to regular CW-complexes?

I asked this on math exchange the other day and it didn't get any response. I'd be happy just for a reference if it exists.

  • $\begingroup$ I computed a handful of examples, and did not find a counterexample. In some cases the homotopy equivalence was not so obvious, e.g. when K is the boundary of a tetrahedron and A is its vertex set. So there appears to be something interesting here. $\endgroup$
    – Dan Ramras
    Oct 14, 2014 at 3:23
  • $\begingroup$ Kyle, what examples have you computed already? $\endgroup$
    – Dan Ramras
    Oct 14, 2014 at 3:24
  • $\begingroup$ @Dan, in the examples I've worked out K is a cellular decomposition of S^2 where the faces are an nxn grid of squares along with one "outer" face and A is a spanning tree on the 1-skeleton of K. This is what I need for what I'm working on, but I figured it was more likely that this had been proven for simplicial complexes than for regular CW-complexes. $\endgroup$ Oct 14, 2014 at 3:58
  • $\begingroup$ Does $P$ have excision property? $\endgroup$
    – user43326
    Oct 14, 2014 at 7:13
  • $\begingroup$ @user43326 I'm sorry. I don't know what you mean. Could you clarify your question? $\endgroup$ Oct 14, 2014 at 19:43

1 Answer 1


So I believe I have a solution to my question. It seems to be true for regular CW-complexes. I've changed my notation slightly as I've been working on the problem. Let me know if my answer could be improved in any way.

Theorem: Let $X$ be a regular CW-complex and $L\in X$ a collection of open faces, then $X\setminus L \sim P(X\setminus L)$.

This will follow from a sequence of definitions and lemmas.

Definition: Let $\Delta^n$ be an n-simplex and $\Delta^{n-1}$ a facet. Then $A_n = \Delta^n\setminus \Delta^{n-1}$ is a pavilion of $\Delta^n$ with $dim(A_n)=n$. Continuing inductively we can write $\Delta^n = A_n\cup A_{n-1}\cup\ldots\cup A_0$ with $A_i$ a pavilion (of $\Delta^n$) and $dim(A_i)=i$. Call this a pavilion decomposition of $\Delta^n$.

Note: For $A_k$ a pavilion, $A_k$ contains exactly one vertex. So if $A_n\cup A_{n-1}\cup\ldots\cup A_0$ is a pavilion decomposition of $\Delta^n$, then each $A_i$ contains exactly one vertex of $\Delta^n$. We can then specify a pavilion decomposition with a bijection $f:vert(\Delta^n)\to [n]\cup\lbrace 0\rbrace$ where for $a\in A_i$, $f(a) = dim(A_i)$.

Lemma: Let $A_n\cup A_{n-1}\cup\ldots\cup A_0$ be a pavilion decomposition of $\Delta^n$ and let $I\subseteq [n]\cup\lbrace 0\rbrace$, then $\bigcup_{i\in I}A_i$ is convex.

Proof: We induct on $|I|$. Note that a pavilion is contractible. For the inductive step, note that the line between a point in a pavilion $A$ and any point in $cl(A)\setminus A$ lies entirely inside $A$ except for the one endpoint.$\square$

Definition: Let $X$ be a simplicial complex and $\chi = sd(X)$ be its barycentric subdivision. Let $\alpha\in\chi$ be a closed face. Define the principal pavilion of $\alpha$ to be the pavilion $\Gamma$ of $\alpha$ such that $dim(\Gamma)=dim(\alpha)$ and $\lbrace max(\alpha)\rbrace\in\Gamma$ (remember $\alpha$ is a chain of faces of $X$). Applying this inductively produces the principal pavilion decomposition of $\alpha$ and also of $\chi$.

Note: For $\alpha$ a closed face of $\chi$, there exists a chain $c$ for which $\alpha$ is the collection of all subchains of $c$ (in particular $c$ is the single element of $int(\alpha)$) and the principal pavilion of $\alpha$ is $\lbrace d\,|\, d \text{ is a subchain of $c$ and } max(c)\in d\rbrace = \lbrace d\,|\,d\text{ is a subchain of $c$ and } max(d)=max(c)\rbrace$.

Lemma: For $X$ a simplicial complex, $\chi=sd(X)$, $a\in X$ an open face, and $\gamma\in\chi$ a principal pavilion of some face of $\chi$, $\gamma\cap sd(a) = \left\{ \begin{matrix} \gamma & \quad \text{if }\lbrace a\rbrace\in\gamma\\ \emptyset & \quad \text{if }\lbrace a\rbrace\notin\gamma \end{matrix} \right.$

Proof: $sd(a)$ is the collection of all chains with max element $a$. Let $\alpha$ a closed face of $\chi$ be such that $\gamma$ is the principal pavilion of $\alpha$. Let $c$ be a chain such that $\alpha$ is all subchains of $c$, then $\gamma$ is all subchains of c with maximum element $max(c)$. So if $a = max(c)$, then $\gamma\subseteq sd(a)$, otherwise $\gamma\cap sd(a) = \emptyset$.$\square$

Lemma: For $X$ a simplicial complex, $\chi = sd(X)$, $L\subseteq X$ a collection of open faces, $\Lambda=sd(L)\subseteq\chi$, and $\alpha\in\chi$ a closed face, if $int(\alpha)\cap\Lambda =\emptyset$ then $\alpha\cap\Lambda$ is contractible and contained in the boundary of $\alpha$.

Proof: Clearly if $int(\alpha)\cap\Lambda =\emptyset$ then $\alpha\cap\Lambda\subseteq bd(\alpha)$. Furthermore, for each $l\in L$ and $\Gamma_i$ the $i$-dimensional pavilion in the principal pavilion decomposition of $\alpha$, $\Gamma_i\cap sd(l)$ is either $\emptyset$ or $\Gamma_i$. So $\Gamma_i\cap\Lambda$ is either $\emptyset$ or $\Gamma_i$, and $\alpha\cap\Lambda = \cup_{i\in I}\Gamma_i$ where $I\subseteq [dim(\alpha)]\cup\{0\}$. So $\alpha\cap\Lambda$ is contractible.$\square$

Proof of Theorem: Let $\chi=sd(X)$ and $\Lambda=sd(L)\subseteq\chi$. We know that $sd:X\to\chi$ is a homeomorphism. Restricting to $X\setminus L$, we see $X\setminus L\simeq\chi\setminus\Lambda$. Now $\chi\setminus\Lambda$ is a simplicial complex missing some number of open faces. Let $\alpha\in\chi\setminus\Lambda$ be a maximal open face with $cl(\alpha)\cap\Lambda\neq\emptyset$. By the previous lemma $cl(\alpha)\cap\Lambda$ is a contractible portion of the boundary of $\alpha$. So $cl(\alpha)\setminus\Lambda$ is homeomorphic to a closed disk minus a contractible portion of its boundary, and thus deformation retracts on to the remainder of its boundary. That is, $cl(\alpha)\setminus\Lambda\sim cl(\alpha)\setminus(\Lambda\cup\alpha)$. Since $\alpha$ is maximal, the deformation retraction on $cl(\alpha)\setminus\Lambda$ extends trivially to $\chi\setminus\Lambda$. Thus we have $\chi\setminus\Lambda\sim \chi\setminus(\Lambda\cup\alpha)= del_{\chi\setminus\Lambda}(\alpha)$. Let $\Phi= \{\alpha\in\chi\setminus\Lambda \,|\, \alpha \text{ is open and maximal and } cl(\alpha)\cap\Lambda\neq\emptyset\}$. Applying the previous process to all the elements of $\Phi$ at once we have $\chi\setminus\Lambda\sim del_{\chi\setminus\Lambda}(\Phi)$. Now this process has decreased the maximum dimension of maximal faces whose closure intersects $\Lambda$. Iterating this process we find $\chi\setminus\Lambda\sim del_{\chi\setminus\Lambda}(\{\alpha\in\chi\setminus\Lambda \,|\, \alpha\text{ is an open face and }cl(\alpha)\cap\Lambda\neq\emptyset\})= del_{\chi}(\Lambda)$. (Note when you delete an open face from a complex, you delete all of its cofaces but not any of its faces.) Now $\alpha\in del_{\chi}(\Lambda) \iff cl(\alpha)\cap\Lambda=\emptyset \iff \forall a\in L\text{, } cl(\alpha)\cap sd(a)=\emptyset \iff \forall a\in L\text{, } a\notin\alpha \text{(viewing $\alpha$ as a chain of faces of $X$)} \iff \alpha\cap L=\emptyset \iff \alpha\in P(X\setminus L)$. That is $del_{\chi}(\Lambda)$ and $P(X\setminus L)$ are both chains of faces of $X$ that do not include elements of $L$. So we have $X\setminus L\sim del_{\chi}(\Lambda)= P(X\setminus L)$.$\square$


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