Face poset of a subcomplex complement 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.  
 A: 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 $X\setminus L\sim del_{\chi}(\Lambda)= P(X\setminus L)$.$\square$
