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 $X\setminus L\sim del_{\chi}(\Lambda)= P(X\setminus L)$.$\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.