Induced orders on sets

The goal of this post is to understand when a map from a poset to an unordered set induces a partial order, and how that applies to the specific case of the set of simplicial complexes. Thanks to Yanlong Hao for spotting some mistakes in my seminar talk on the same topic yesterday.

Definition 1: Let $(A,\leqslant_A)$ be a poset and $f\colon A\to B$ a map of sets. The relation $\leqslant_B$ on $B$, with $a\leqslant_Aa'$ implying $f(a)\leqslant_Bf(a')$, is the relation induced by $f$ on $B$. The map $f$ is monotonic if whenever $b\leqslant_b b'$,

1. if $a\in f^{-1}(b)$, $a'\in f^{-1}(b)$ are comparable, then $a\leqslant_A a'$, and
2. if $a'\in f^{-1}(b')$, then there exists $a\in f^{-1}(b)$ such that $a\leqslant_A a'$.

Since $f$ may not be surjective, there may be $b\in B$ with $f^{-1}(b) = \emptyset$. For such $b$ we only have $b\leqslant_B b$ and $b$ is not comparable to any other element of $B$.

Lemma 2: If $f\colon A\to B$ is monotonic, then the induced relation $\leqslant_B$ is a partial order on $B$.

Proof: For reflexivity, take any $a\in A$, which has $a\leqslant_A a$ by reflexivity of $\leqslant_A$. Then $f(a)\leqslant_Bf(a)$, so every $b\in \im(f)$ satisfies reflexivity. Every $b\not\in\im(f)$ also satisfies reflexivity by the comment above.

For anti-symmetry, suppose that $b\leqslant_Bb'$ and $b'\leqslant_Bb$. Since $b\leqslant_B b'$, there is some $a\in f^{-1}(b)$ and $a'\in f^{-1}(b')$ such that $a\leqslant_A a'$. Similarly, there is some $c'\in f^{-1}(b')$ and $c\in f^{-1}(b)$ such that $c'\leqslant_A c$. Since $c\in f^{-1}(b)$ and $c'\in f^{-1}(b')$ are comparable, and the first assumed relation is $b\leqslant_Bb'$, by property 1 of Definition 1, we must have $c\leqslant_A c'$. By anti-symmetry of $A$, we now have that $c=c'$, so it follows that $b=f(c)=f(c')=b'$.

For transitivity, suppose that $b\leqslant_Bb'$ and $b'\leqslant_Bb''$. Take $a''\in f^{-1}(b'')$, for which property 2 of Definition 1 guarantees that there exists $a'\in f^{-1}(b')$ such that $a'\leqslant_Aa''$. Similarly, the first assumed relation and the same property guarantees there exists $a\in f^{-1}(b)$ such that $a\leqslant_Aa'$. By transitivity of $A$, we have $a\leqslant_A a''$. By the definition of $\leqslant_B$, we have $b=f(a) \leqslant_B f(a'') = b''$. $\square$

Let $M$ be a piecewise linear, compact, connected, embedded manifold in $\R^N$, and $SC$ the category of simplicial complexes. Let $A= \{1<2a>2b<3\}$. The product $A^N$ has the product order. Fix $n\in \Z_{>0}$ and let $T$ be the set of all distinct 2-,3-,...,$n$-tuples in $\{1,\dots,n\}$, or $T := \bigcup_{k=2}^n\left(\{1,\dots,n\}^k\setminus \Delta\right)/_{S_k}$. This set has size $\sum_{k=2}^n \binom nk = 2^n-n-1$. Assume every $v\in T$ is ordered in the canonical way. Then $v$ induces a natural projection $\pi_v\colon M^n \to M^{v}$, as well as another map \[ \begin{array}{r c l}
\pi_v'\colon M^n \times \R_{>0} & \to & A, \\
(P,t) & \mapsto & \begin{cases}
1 & \forall\ i,j, \pi_v(P)_i = \pi_v(P)_j, \\
2a & \exists\ i,j \text{ s.t. }\pi_v(P)_i\neq\pi_v(P)_j \text{ and } \textstyle\bigcap_{i=1}^{|v|} B(\pi_v(P)_i,t) \neq \emptyset, \\
2b & \exists\ i,j \text{ s.t. }\pi_v(P)_i\neq\pi_v(P)_j \text{ and } \textstyle\bigcap_{i=1}^{|v|} B(\pi_v(P)_i,t)=*, \\
3 & \exists\ i,j \text{ s.t. }\pi_v(P)_i\neq\pi_v(P)_j \text{ and } \textstyle\bigcap_{i=1}^{|v|} B(\pi_v(P)_i,t) = \emptyset.
\end{cases}
\end{array} \] Here all the balls $B$ are closed, and $M^n$ has the Hausdorff topology.

Lemma 3: The map $\pi_v$ is continuous on $M^v\times \R_{>0}$.

Proof: Every $(Q,s)\in (\pi_v')^{-1}(3)$ has an open ball of radius $\max_{i,j}\{d(\pi_v(Q)_i,\pi_v(Q)_j)\}/2-s$ around it that is still contained within $(\pi_v')^{-1}(3)$. Similarly, every $(Q,s)\in (\pi_v')^{-1}(2a)$ has an open ball of radius \[ \min\left\{\frac12\text{diam}\left(\bigcap_{i=1}^{|v|}B(\pi_v(Q)_i,s)\right),\max_{i,j}\{d(\pi_v(Q)_i,\pi_v(Q)_j)\}\right\} \hspace{2cm} (1) \] around it that is still contained within $(\pi_v')^{-1}(2a)$. The first expression in the $\min$ makes sure the intersection is non-empty, and the second expression makes sure all elements of $Q$ are not the same.

The set $(\pi_v')^{-1}(1<2a)$ is open by the same argument as for $2a\in A$, enlarging the open ball by removing the second expression in the $\min$ of expression (1). Finally, the set $(\pi_v')^{-1}(2a>2b<3)$ is open by the same argument, now enlarging the ball used for $2a\in A$ by removing the first expression in the $\min$ of expression (1). $\square$

Let $q\colon M^n\to \Ran^{\leqslant n}(M)$ be the natural quotient map, and $\check C\colon \Ran^{\leqslant n}(M)\times \R_{>0}\to SC$ be the Čech simplical complex map. For the next propositions, we will use two maps $f$ and $g$ defined as \[ \begin{array}{r c l}
f\colon M^n\times \R_{>0} & \to & A^{2^n-n-1}, \\
(P,t) & \mapsto & \prod_{v\in T} \pi_v'(P,t),
\end{array}
\hspace{2cm}
\begin{array}{r c l}
g\colon \im(f) & \to & SC, \\
f(P,t) & \mapsto & \check C(q(P),t).\end{array} \] The map $g$ is well-defined because $a\in A^{2^n-n-1}$ with non-empty preimage in $M^n\times \R_{>0}$ specifies whether or not every $k$-tuple of points has a simplex spanning it, for all $k=2,\dots,n$. This defines a unique simplicial complex, so choosing any $(P,t)\in f^{-1}(a)$ will give the same Čech complex, up to renaming of vertices.

Proposition: The map $f\colon M^n\times \R_{>0} \to A^{2^n-n-1}$ is continuous.

Proof: Let $a\in A^{2^n-n-1}$ and suppose that $f^{-1}(a)\neq \emptyset$. Let $a_i\in A$ be in the $i$th factor of $a$, and $r_i$ the radius of the open ball decreed by Lemma 3 to still be within $(\pi_v')^{-1}(a_i)$, where $v$ is the $i$th tuple in the chosen order on $T$. Then every $(P,t)\in f^{-1}(a)$ has an open ball of radius $\min_i\{r_i\}$ around it that is still contained within $f^{-1}(a)$, so $f$ is continuous. $\square$

Proposition: The map $g$ is monotonic.

 Note that any relation $S\leqslant_{SC}S'$ may be split up as a chain of relations $S=T_1\leqslant_{SC} \cdots \leqslant_{SC} T_\ell=S'$, where the only differences between $T_i$ and $T_{i+1}$ are either (i) $T_i$ has a $k$-simplex $\sigma$ that $T_{i+1}$ does not have, or (ii) where $T_i$ has a single 0-simplex where a $k$-simplex $\sigma$ and all its faces used to be in $T_{i+1}$. Hence it suffices to show that properties 1 and 2 of Definition 1 are satisfied in cases (i) and (ii).

Proof: Case (i): Suppose that $S\leqslant_{SC}S'$, and take $a\in g^{-1}(S)$, $a'\in g^{-1}(S')$ with $a\leqslant_A a'$. If there is $b\in g^{-1}(S)$ and $b'\in g^{-1}(S')$ such that $b'\leqslant_A b$, then $g(b)$ has the $k$-simplex $\sigma$ that $g(b')$ does not have, but since $b'$ is ordered lower than $b$, it must be that this $k$-simplex has collapsed to a point. Then we would be in case (ii), a contradiction, so property 1 holds in this case.

Now let $i_1,\dots,i_{\sigma}$ be the indices of $a'$ and $a$ representing the $(k+1)$-fold intersection that describes $\sigma$, so $a'_j = 3$ and $a_j = 2b$ for all $j=i_1,\dots,i_\sigma$. Take any $b'\in g^{-1}(S')$, which also has some indices $\ell_1,\dots,\ell_\sigma$ representing this same $(k+1)$-fold intersection, so $b'_j=3$ at all $j=\ell_1,\dots,\ell_\sigma$. Let $b\in A^{2^n-n-1}$ be the element with all the same factors as $b'$, except at indices $\ell_1,\dots,\ell_\sigma$, which have been changed to $2b$. This element $b$ is still in $\im(f)$ as removing only this $k$-simplex still leaves the well-defined simplex $S'$ we assumed at the beginning. Hence $g(b)=S'$ and property 2 holds. \\

Case (ii): Suppose that $S\leqslant_{SC}S'$, and take $a\in g^{-1}(S)$, $a'\in g^{-1}(S')$ with $a\leqslant_A a'$. If there is $b\in g^{-1}(S)$ and $b'\in g^{-1}(S')$ such that $b'\leqslant_A b$, then $g(b')$ has the $k$-simplex $\sigma$ and all its faces that $g(b)$ does not have, but since $b'$ is ordered lower than $b$, it must be that we have introduced $\sigma$ and all its faces. Then we would be in case (i), or a chain of case (i) situations, a contradiction, so property \ref{1mon} holds in this case.

Now let $i_1,\dots,i_{\sigma}$ be the indices of $a'$ and $a$ representing the $(k+1)$-fold intersection that describes $\sigma$, and all the implied $(f+1)$-fold intersections that describe the $f$-faces of $\sigma$, $f>0$. That is, $a'_j = 2a$ and $a_j = 1$ for all $j=i_1,\dots,i_\sigma$. Take any $b'\in g^{-1}(S')$, which also has some indices $\ell_1,\dots,\ell_\sigma$ representing this same $(k+1)$-fold (and lower) intersection, so $b'_j=3$ at all $j=\ell_1,\dots,\ell_\sigma$. Let $b\in A^{2^n-n-1}$ be the element with all the same factors as $b'$, except at indices $\ell_1,\dots,\ell_\sigma$, which have been changed to $1$. This element $b$ is still in $\im(f)$ as collapsing this $k$-simplex and all its faces to a single 0-simplex still leaves the well-defined simplex $S'$ we assumed at the beginning. Hence $g(b)=S'$ and property 2 holds. $\square$

Since $g$ is monotonic, by Lemma 2 the relation $\leqslant_{SC}$ is a partial order on $SC$.

Conical stratifications via semialgebraic sets

The goal of this post is to describe a conical stratification of $\Ran_{\leqslant n}(M)\times \R_{\geqslant 0}$ that refines the stratification previously seen (in "Exit paths, part 2," 2017-09-28, and "Refining stratifiations," 2018-03-11). Thanks to Shmuel Weinberger for the key observation that the strata under consideration are nothing more than semialgebraic sets, which are triangulable, and so admit a conical stratification via this triangulation.

Remark: Fix $n\in \Z_{>0}$, let $M$ be a smooth, compact, connected, embedded submanifold in $\R^N$, and let $M^n$ have the Hausdorff topology. We will be interested in $M^n\times \R_{>0}$, though this will be viewed as the compact set $M^n\times [0,K]\subseteq \R^{nN+1}$ for some $K$ large enough (for instance, larger than the diameter of $M$) when necessary. The point 0 is added for compactness.

Stratification of the Ran space by semialgebraic sets

We begin by stratifying $M^n\times \R_{>0}$ by a poset $A$, creating strata based on the pairwise distance between points in each $M$ component. Then we take that to a stratification of the quotient $\Ran^{\leqslant n}(M)\times \R_{>0}$ via the action of the symmetric group $S_n$ and overcounting of points.

Definition: Define a partial order $\leqslant$ on the set $A = \big\{$partitions of ($\{1,\dots,n\}^2\setminus \Delta)/S_2$ into 4 parts$\big\}$ of ordered 4-tuples of sets by \[ (Q,R,S,T) \leqslant \left(Q\setminus Q',\ R\cup Q' \cup S',\ S\setminus (S'\cup S''),\ T\cup S''\right), \] for all $Q'\subseteq Q$ and $S',S''\subseteq S$, with $S'\cap S'' = \emptyset$.

The diagram to keep in mind is the one below, with arrows pointing from lower-ordered elements to higher-ordered elements. Once we pass to valuing the 4-tuple in simplicial complexes, moving between $Q$ and $R$ will not change the simplicial complex type (this comes from the definition of the Vietoris--Rips complex).

Lemma 1: The map $f: M^n\times \R_{>0}\to (A,\leqslant)$ defined by \begin{align*} (\{P_1,\dots,P_n\},t)\mapsto \bigg( \{(i,j>i)\ & :\ P_i=P_j\},\ \{(i,j>i)\ :\ d_M(P_i,P_j)<t\},\\
& \{(i,j>i)\ :\ d_M(P_i,P_j)=t\},\ \{(i,j>i)\ :\ d_M(P_i,P_j)>t\}\bigg) \end{align*} is continuous in the upset topology on $(A,\leqslant)$.

Proof: Choose $(Q,R,S,T)\in A$ and consider the open set $U = U_{(Q,R,S,T)}$ based at $(Q,R,S,T)$. Take $(P,t)\in f^{-1}(U)$, which we claim has a small neighborhood still contained within $f^{-1}(U)$. If we move a point $P_i$ slightly that was exactly distance $t$ away from $P_j$, then the pair $(i,j)$ was in $S$, but is now in either $R$ or $T$, and both $(Q,R\cup\{(i,j)\},S\setminus \{(i,j)\},T)$ and $(Q,R,S\setminus \{(i,j)\},T\cup \{(i,j)\})$ are ordered higher than $(Q,R,S,T)$, so the perturbed point is still in $f^{-1}(U)$. If $P_i=P_j$ in $P$ and we move them apart slightly, since $t\in \R_{>0}$, the pair $(i,j)$ will move from $Q$ to $R$, and $(Q,R,S,T) \leqslant (Q\setminus \{(i,j)\},R\cup \{(i,j)\},S,T)$, so the perturbed point is still in $f^{-1}(U)$. For all pairs $(i,j)$ in $R$ or $T$, the distances can be changed slightly so that the pair still stays in $R$ or $T$, respectively. Hence $f$ is continuous. $\square$

This shows that $M^n\times \R_{>0}$ is stratified by $(A,\leqslant)$, using Lurie's definition of a (poset) stratification, which just needs a continuous map to a poset. Our goal is to work with the Ran space of $M$, instead of the $n$-fold product of $M$, which are related by the natural projection map $\pi : M^n\to \Ran^{\leqslant n}(M)$, taking $P=\{P_1,\dots,P_n\}$ to the unordered set of distinct elements in $P$. We also would like to stratify $\Ran^{\leqslant n}(M)\times \R_{>0}$ by simplicial complex type, so we need the following map.

Definition: Let $g: (A,\leqslant)\to SC$ be the map into simplicial complexes that takes $(Q,R,S,T)$ to the clique complex of the simple graph $C$ on $n-k$ vertices, for $|Q|=k(k+1)/2$, defined as follows: 

• $V(C) = \{[i]\ :\ i=1,\dots,n,\ [j]= [i] \text{\ iff\ } (i,j)\in Q\}$,
• $E(C) = \{([i],[j])\ :\ (i,j)\in R\cup S\}$.

We require $C$ to be simple, so if $(i,j)\in Q$ and $(i,\ell),(j,\ell)\in R\cup S$, we only add one edge $([i],[\ell])=([j],[\ell])$ to $C$.

The map $g$ induces a partial order $\leqslant$ on $SC$ from the partial order on $A$, with $C\leqslant C'$ in $SC$ whenever there is $(Q,R,S,T)\in g^{-1}(C)$ and $(Q',R',S',T')\in g^{-1}(C')$ such that $(Q,R,S,T) \leqslant (Q',R',S',T')$ in $A$. Note that if $C\in SC$ is not in the image of $g$, then it is not related to any other element of $SC$. By the universal property of the quotient and continuity of $f$ and $g$ (as $A$ and $SC$ are discrete), there is a continuous map $h:\Ran^{\leqslant n}(M)\times \R_{>0}\to (SC,\leqslant)$ such that the diagram

commutes. Hence $\Ran^{\leqslant n}(M)\times \R_{>0}$ is stratified by $(SC,\leqslant)$.

Remark: The map $\pi$ can be thought of as a quotient by the action of the symmetric group $S_n$, followed by the quotient of the equivalence relation \[ \{P^1_1,\dots,P^{\ell_1}_1,P^1_2,\dots,P^{\ell_2}_2,P^1_3,\dots,P^{\ell_k}_k\} \ \ \sim\ \
\{P^1_1,\dots,P^{\ell_1-1}_1,P^1_2,\dots,P^{\ell_2+1}_2,P^1_3,\dots,P^{\ell_k}_k\} \] on $M^n$, for all possible combinations $\ell_1+\cdots + \ell_k =n$ and $1\leqslant k\leqslant n-1$, where $P_m^i=P_m^j$ for all $1\leqslant i<j\leqslant \ell_m$.

Semialgebraic geometry

Next we move into the world of semialgebraic sets and triangulations, following Shiota. Here we come across a more restrictive notion of stratification of a manifold $X$, which requires a partition of $X$ into submanifolds $\{X_i\}$. If Lurie's stratification $f:X\to A$ gives back submanifolds $\{f^{-1}(a)\}_{a\in A}$, then we have Shiota's stratification. Conversely, the poset $(\{X_i\},\leqslant)$, for $X_i \leqslant X_j$ iff $X_i \subseteq \closure(X_j)$ is always a stratification in the sense of Lurie.

Definition 2: A semialgebraic set in $\R^N$ is a set of the form \[ \bigcup_{\text{finite}} \{x\in \R^N\ :\ f_1(x)=0,f_2(x)>0,\dots,f_m(x)>0\},\] for polynomial functions $f_1,\dots,f_m$ on $\R^N$. A semialgebraic stratification of a space $X\subseteq \R^N$ is a partition $\{X_i\}$ of $X$ into submanifolds that are semialgebraic sets.

Next we observe that the strata of $M^n\times \R_{>0}$ are semialgebraic sets, with the preimage theorem and I.2.9.1 of Shiota, which says that the intersection of semialgebraic sets is semialgebraic. Take $(Q,R,S,T)\in A$  and note that \[ f^{-1}(Q,R,S,T) = \left\{(\{P_1,\dots,P_n\},t)\in M^n\times \R_{>0}\ :\ \begin{array}{r l}
d(P_i,P_j) = 0 & \forall (i,j)\in Q,\\
t-d(P_i,P_j) = 0 & \forall (i,j)\in S, \\
t-d(P_i,P_j) > 0 & \forall (i,j)\in R, \\
d(P_i,P_j) - t > 0 & \forall (i,j)\in T.
\end{array}\right\} \] Here $d$ means distance on the manifold, and we assume the metric to be analytic. Alternatively, $d$ could be Euclidean distance between points on the embedding of $M^n\times \R_{>0}$, induced by the assumed embedding of $M$.

For his main Theorem II.4.2, Shiota uses cells, but we opt for simplices instead, and for cell complexes we use simplicial complexes. Every cell and cell complex admits a decomposition into simplicial complexes, even without introducing new 0-cells (by Lemma I.3.12), so we do not lose any generality.

Definition 3: Let $X,Y$ be semialgebraic sets.

• A map $f: X\to Y$ is semialgebraic if the graph of $f$ is semialgebraic.
• A semialgebraic cell triangulation of a semialgebraic set $X$ is a pair $(C,\pi)$, where $C$ is a simplicial complex and $\pi: |C|\to X$ is a semialgebraic homeomorphism for which $\pi|_{\interior(\sigma)}$ is a diffeomorphism onto its image.
• A semialgebraic cell triangulation $(C,\pi)$ is compatible with a family $\{X_i\}$ of semialgebraic sets if $\pi(\interior(\sigma))\subseteq X_i$ or $\pi(\interior(\sigma))\cap X_i = \emptyset$ for all $\sigma\in C$ and all $X_i$.

A semialgebraic cell triangulation $(C,\pi)$ of $X$ induces a stratification $X\to (C_0 \cup \{\pi(\interior(\sigma))\},\leqslant)$, where the order is the one mentioned just before Definition 2. We use the induced stratification and the cell triangulation interchangeably, specifically in Proposition 4.

A compatible conical stratification

Finally we put everything together to get a conical stratification of $\Ran^{\leqslant n}(M)\times \R_{>0}$. Unfortunately we have to restrict ourselves to piecewise linear manifolds, or PL manifolds, which are homeomorphic images of geometric realizations of simplicial complexes, as otherwise we cannot claim $M$ is a semialgebraic set. We can also just let $M=\R^k$, as the point samples we are given could be coming from an unknown space.

Proposition 4: Let $M$ be a PL manifold embedded in $\R^N$. There is a conical stratification $\widetilde h:\Ran^{\leqslant n}(M)\times \R_{>0}\to (B,\leqslant)$ compatible with the stratification $h: \Ran^{\leqslant n}(M)\times \R_{>0}\to (SC,\leqslant)$.

Proof: (Sketch) The main lifting is done by Theorem II.4.2 of Shiota. Since $M$ is PL, it is semialgebraic, and so $M^n\times \R_{>0}\subseteq \R^{nN+1}$ is semialgebraic, by I.2.9.1 of Shiota. Since the quotient $\pi$ of diagram (1) is semialgebraic, the space $\Ran^{\leqslant n}(M)\times \R_{>0}$ is semialgebraic, by Scheiderer. Similarly, $\{f^{-1}(a)\}_{a\in A}$ is a family of semialgebraic sets, where $f$ is the map from Lemma 1.  Theorem II.4.2 gives that $\Ran^{\leqslant n}(M)\times \R_{>0}$ admits a cell triangulation $(K,\tau)$ compatible with $\{h^{-1}(S)\}_{S\in SC}$. By the comment after Definition \ref{semialgdef}, this means we have a stratification $\Ran^{\leqslant n}(M)\times \R_{>0}\to (K_0\cup \{\tau(\interior(\sigma))\}_{\sigma\in K},\leqslant)$. Further, by Proposition A.6.8 of Lurie, we have a conical stratification $|K|\to (B,\leqslant)$. This is all described by the solid arrow diagram below.

The vertical induced map comes as the poset $B$ has the exact same structure as the abstract suimplicial complex $K$. The diagonal induced map comes as the map $|K|\to \Ran^{\leqslant n}(M)\times \R_{>0}$ is a homeomorphism, and so has a continuous inverse. Composing the inverse with the conical sratification of Lurie, we get a conical stratification of $\Ran^{\leqslant n}(M)\times \R_{>0}$. Composing the vertical induced arrow and the maps to $(SC,\leqslant)$ show that there is a conical stratification of $\Ran^{\leqslant n}\times \R_{>0}$ compatible with its simplicial complex stratification from diagram (1). $\square$

Shiota actually requires that the space that admits a triangulation be closed semialgebraic, and having $\R_{>0}$ violates that condition. Replacing this piece with $\R_{\geqslant 0}$, then applying Shiota, and afterwards removing the $t=0$ piece we get the same result.

Remark: Every (sufficiently nice) manifold admits a triangulation, so it may be possible to extend this result to a larger class of manifolds, but it seems more sophisticated technology is needed.

References: Shiota (Geometry of subanalytic and semialgebraic sets, Chapters I.2, I.3, II.4), Scheiderer (Quotients of semi-algebraic spaces), Lurie (Higher algebra, Appendix A.6)

Refining stratifications

The goal of this post is to describe a natural stratification associated to any stratification, with hopes of it being conical. Let $X$ be a topological space, $(A,\leqslant_A)$ a finite partially ordered set, and $f:X\to A$ a stratifying map. For every $x\in X$, write $A_{>f(x)} = \{a\in A\ :\ a>f(x)\}\subseteq A$, and analogously for $A_{\geqslant f(x)}$. For every $a\in A$, write $X_a = \{x\in X\ :\ f(x)=a\}$.

Definition: For any other stratified space $g\colon Y\to B$, a stratified map $\varphi\colon (X\to A) \to (Y\to B)$ is a pair of maps $\varphi_{XY}\in \Hom_{\Top}(X,Y)$  and $\varphi_{AB}\in \Hom_{\Set}(A,B)$ such that the diagram

commutes. A stratified map $\varphi$ is an open embedding if both $\varphi_{XY}$ and $\varphi_{XY}|_{X_a}\colon X_a\to Y_{\varphi_{AB}(a)}$ are open embeddings.

Recall the cone $C(Y)$ of a space $Y$ is defined as $Y\times [0,1) / Y\times \{0\}$.

Definition: A stratification $f\colon X\to A$ is conical at $x\in X$ if there exist

• a stratified space $f_x \colon Y\to A_{>f(x)}$,
• a topological space $Z$, and
• an open embedding $Z\times C(Y)\hookrightarrow X$ of stratified spaces whose image contains $x$.

The cone $C(Y)$ has a natural stratification $f_x' \colon C(Y)\to A_{\geqslant f(x)}$, as does the product $Z\times C(Y)$. The space $X$ itself is \emph{conically stratified} if it is conically stratfied at every $x\in X$.

The image to have in mind is that $Z$ is a neighborhood of $x$ in its stratum $X_{f(x)}$, and $C(Y)$ is an upwards-directed neighborhood of $f(x)$ in $A$. Now we describe how to refine the stratification of an arbitrary stratified space to make it conical.

Definition: Let $\leqslant_{\mathbf P(A)}$ be the partial order on $\mathbf P(A)$ defined in the following way:

• For every $x,y\in A$, set $x\leqslant_{\mathbf P(A)} y$ whenever $x\leqslant_A y$, and
• for every $C\in \mathbf P(A)$, set $C\leqslant_{\mathbf P(A)} C'$ for all $C'\in \mathbf P(C)$.

Note that $(A,\leqslant_A)$ is open in $(\mathbf P(A),\leqslant_{\mathbf P(A)})$ in the upset topology. Hence for $i:A\hookrightarrow \mathbf P(A)$ the inclusion map, $i\circ f:X\to A\hookrightarrow \mathbf P(A)$ is also a stratifying map for $X$. We now define another $\mathbf P(A)$-stratification for $X$.

Definition: Let $f_{\mathbf P} \colon X\to \mathbf P(A)$ be defined by $f_{\mathbf P}(x)= \displaystyle\min_{\left(\mathbf P(A),\leqslant_{\mathbf P(A)}\right)} \left\{C\ :\ x\in \closure(f^{-1}(C'))\ \forall\ C'\in C\right\}$.

This map is well defined because for each $x\in X$ there are finitely many strata $f^{-1}(a)$ which contain $x$ in their closure. The element $C\in \mathbf P(A)$ containing all such $a$ is the $C$ to which $x$ gets mapped. We now claim this is a stratifying map for $X$.

Proposition: The map $f_{\mathbf P}\colon X\to \mathbf P(A)$ is continuous.

Proof: Let $C\in \mathbf P(A)$. We will show that the preimage via $f_{\mathbf P}$ of the open set $U_C = \mathbf P(C)\subseteq \mathbf P(A)$ is open in $X$ (and such sets $U_C$ are a basis of topology for $\mathbf P(A)$). By definition of the map $f_{\mathbf P}$, we have \[ f_{\mathbf P}^{-1}(U_C) = f^{-1}(U_{\min\{C'\in C\}}) \setminus \left(\bigcup_{(D,E)\in A\times (A\setminus C)} \closure( f^{-1}(D))\cap \closure (f^{-1}(E))\right). \] By continuity of $f$, the set $f^{-1}(U_{\min\{C'\in C\}})$ is open in $X$, and the sets we are subtracting from this open set are all closed. Hence $f_{\mathbf P}^{-1}(U_C)$ is open in $X$. $\square$

Unfortunately, this stratification is difficult to work with. Recall the space $\Ran_{\leqslant n}(M)\times \R_+$ for a very nice (smooth, compact, connected, embedded) manifold $M$, along with the map \[ \begin{array}{r c l}
f\colon \Ran_{\leqslant n}(M)\times \R_{\geqslant 0} & \to & SC, \\
(P,t) & \mapsto & VR(P,t),
\end{array} \] for $VR$ the Vietoris-Rips complex on $P$ with radius $t$. To put a partial order on $SC$, we first say that $S\leqslant T$ in $SC$ whenever there is a path $\gamma:I\to X$ satisfying

• $\widetilde f(\gamma(0))=S$ and $\widetilde f(\gamma(1))=T$,
• $\widetilde f(\gamma(t))=\widetilde f(\gamma(1))$ for all $t>1$.

Let $(SC,\leqslant_p)$ denote the partial order on $SC$ generated by all relations of this type. We would like to prove some results about $f_{\mathbf P}$ induced by this $f$, and by any stratifying $f$ in general, but the results seem difficult to prove. We give a list, in order of (percieved) increasing difficulty.

• The stratification $f_{\mathbf P}\colon \Ran_{\leqslant n}(M)\times \R_+ \to \mathbf P(SC)$ is conical.
• The stratification $f_{\mathbf P}\colon X\to \mathbf P(A)$ is conical for any stratified space $f\colon X\to A$.
• If $f\colon X\to A$ is already conical, the map $j\colon A\to \mathbf P(A)$ given by $j(a)= \{b\in A\ :\ f^{-1}(a)\subseteq \closure(f^{-1}(b))\}$ is an isomorphism onto its image, and $f_{\mathbf P} = j\circ f$.

References: Ayala, Francis, Tanaka (Local structure on stratified spaces)