Let $\mathscr{P}$ be a polyhedral decomposition of a real vector space $V$. By that I mean that $\mathscr{P}$ is a finite set of polyhedra in $V$ satisfying the following three properties:


1. The union of the polyhedra in $\mathscr{P}$ is equal to $V$.
2. If $P\in \mathscr{P}$ and $Q$ is a face of $P$, then $Q\in\mathcal{P}$.
3. If $P,Q\in\mathscr{P}$, then $P\cap Q$ is a (possibly empty) face of both $P$ and $Q$.


Question: Is the union of the bounded elements of $\mathscr{P}$ contractible?


If $\mathscr{P}$ consists of the set of faces of an essential hyperplane arrangement, then a positive answer is stated in Exercise 1.7.a of Stanley's lecture notes [https://www.cis.upenn.edu/~cis6100/sp06stanley.pdf](https://www.cis.upenn.edu/%7Ecis6100/sp06stanley.pdf), though I don't know the reference for a proof. Note that, when $\mathscr{P}$ does not come from a hyperplane arrangement, then the bounded complex need not be pure.



The bounded complex of a polyhedral decomposition
Below I construct a deformation retract of $V$ onto the bounded sub-complex of $\mathscr P$, proving that the latter is contractible.


For a polyhedral complex $\mathscr P$ resp. a polyhedron $P\in\mathscr P$ I write $\mathscr P^b\subseteq\mathscr P$ resp. $P^b\subseteq P$ for its bounded sub-complex.
Also, by *a retract* of a polyhedron $P$ (or polyhedral complex $\mathscr P$) I mean a [strong deformation retract](https://en.wikipedia.org/wiki/Retraction_(topology)) $\phi\_P: P\times[0,1]\to P^b$ from $P$ onto $P^b\subseteq P$.
Such clearly exist if $P$ is bounded (just use the identity) or if $P$ is a half-infinite edge. But below I explain how to define such for all polyhedra, in such a way that they are compatible on their boundaries and so can be glued together to form a retract for all of $\mathscr P$. The latter is therefore homotopy equivalent to $\bigcup\_{P\in\mathscr P} P=V$, hence contractible.


*Proof.*


* Each $P\in\mathscr P$ can be written as $P=P^b+C\_P$, where $C\_P$ is the [recession cone](https://en.wikipedia.org/wiki/Recession_cone) of $P$. Fix a choice of *generalized barycentric coordinates* (e.g. mean value coordinates, Wachspress coordinates, ...), i.e. for each point $x\in P$ we have a canonical decomposition $x=x\_b+x\_C$ with $x\_b\in P^b$ and $x\_C\in C\_P$ that only depends on the shape of $P$ and the location of $x$ relative to it (see also below). Then set $\phi\_P(x,t):=x\_b+(1-t)x\_C$.
* For $P,Q\in\mathscr P$ the above definition ensures that $\phi\_P(x,t)=\phi\_Q(x,t)$ for all $x\in P\cap Q$, so that we can extend the retract to all of $\mathscr P$.


$\square$




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**Some notes on "the canonical choice"**


As aluded to in the comments, it is sufficient to construct a section $s\_P:P\to P^b\times C\_P$ for the projection


$$\pi\_P:P^b\times C\_P\to P,(x\_b,x\_C)\mapsto x\_b+x\_C,$$


(i.e., $\pi\_P\circ s\_P=\mathrm{id}\_P$) that matches with prescribed sections $s\_\sigma:\sigma\to\sigma^b\times C\_\sigma$ on the faces $\sigma\subseteq P$. Here is one idea for how to do this.


The recession cones of polyhedra in $\mathscr P$ form a full fan $\mathscr F:=\{C\_P\mid P\in\mathscr P\}$ of $V$. For each cone $C\_P\in\mathscr F$ choose a ray $r\_P\subseteq C\_P$.
Now, for a point $x\in P$ let $x\_b'$ be the intersection of $\partial P$ with the translated and inverted ray $-r\_P+x$ (which exists since the linearity space of $C\_P$ is empty). In particular, $x\_b'$ is in some face $\sigma\subseteq \partial P$. Then set $x\_b:=s\_\sigma(x\_b')$ and $x\_C:= x-x\_b$.


