Exit paths and entry paths through $\infty$-categories

Let $X$ be a topological space, $(A,\leqslant)$ a poset, and $f: X\to (A,\leqslant)$ a continuous map.

Definition: An exit path in an $A$-stratified space $X$ is a continuous map $\sigma: |\Delta^n|\to X$ for which there exists a chain $a_0\leqslant \cdots \leqslant a_n$ in $A$ such that $f(\sigma(t_0,\dots,t_i,0,\dots,0))=a_i$ for $t_i\neq 0$. An entry path is a continuous map $\tau: |\Delta^n|\to X$ for which there exists a chain $b_0\leqslant \cdots \leqslant b_n$ in $A$ such that $f(\tau(0,\dots,0,t_i,\dots,t_n))=b_i$ for $t_i\neq 0$.

Up to reordering of vertices of $\Delta^n$ and induced reordering of the realization $|\Delta^n|$, an exit path is the same as an entry path. The next example describes this equivalence.

Example: The standard 2-simplex $|\Delta^2|$ is uniquely an exit path and an entry path with a chain of 3 distinct elements, stratfied in the ways described below.

Recall the following algebraic constructions, through Joyal's quasi-category model:

• A simplicial set is a functor $\Delta^{op}\to \Set$.
• A Kan complex is a simplicial set satisfying the inner horn condition for all $0\leqslant k\leqslant n$. That is, the $k$th $n$-horn lifts (can be filled in) to a map on $\Delta^n$.
• An $\infty$-category is a simplicial set satisfying the inner horn condition for all $0<k<n$.

Moreover, if the lift is unique, then the Kan complex is the nerve of some category. Recall also the category $\Sing(X) = \{$continuous $\sigma: |\Delta^n|\to X\}$, which can be combined with the stratification $f: X\to A$ of $X$

Remark: The subcategory $\Sing^A(X)$ of exit paths and the subcategory $\Sing_A(X)$ of entry paths are full subcategories of $\Sing(X)$, with $(\Sing^A(X))^{op} = \Sing_A(X)$. If the stratification is conical, then these two categories are $\infty$-categories.

Recall the nerve construction of a category. Here we are interested in the nerve of the category $SC$ of simplicial complexes, so $N(SC)_n = \{$sequences of $n$ composable simplicial maps$\}$. Recall the $k$th $n$-horns, which are compatible diagrams of elements of $N(SC)_n$. In general, they are colimits of a diagram in the category $\Delta$. That is, \[ \Lambda^n_k := \colim \left(\bigsqcup_{0\leqslant i<j\leqslant n} \Delta^{n-2} \rightrightarrows \bigsqcup_{0\leqslant i\leqslant n \atop i\neq k} \Delta^{n-1}\right). \] Example: The images of the 3 different types of 2-horns and 4 different types of 3-horns in $SC$ are given below. Note that they are not unique, and depend on the choice of simplices $S_i$ (equivalently, on the choice of functor $\Delta^{op}\to SC$).

For example, the 0th 2-horn $\Lambda^2_0$ can be filled in if there exists a simplicial map $h: S_1\to S_2$ in $SC$ (that is, an element of $N(SC)_1$) such that $h\circ f = g$. Similarly, the 1st 3-horn $\Lambda^3_1$ can be filled in if there exists a functor $F: [0<1<2]\to SC$ for which $F(0<1)=f_{02}$, $F(0<2)=f_{03}$, and $F(1<2)=f_{23}$ (equivalently, a compatible collection of elements of $N(SC)_2$).

Definition: Let $A,B$ be $\infty$-categories. A functor $F: A\to B$ is a morphism of the simplicial sets $A,B$. That is, $F:A\to B$ is a natural transformation for $A,B\in \text{Fun}(\Delta^{op},\Set)$.

A functor of simplicial sets of a particular type can be identified with a functor of 1-categories. Recall the nerve of a 1-category, which turns it into an $\infty$-category. This construction has a left adjoint.

Definition: Let $\mathcal C$ be an $\infty$-category. The homotopy category $h\mathcal C$ of $\mathcal C$ has objects $\mathcal C_0$ and morphisms $\Hom_{h\mathcal C}(X,Y) = \pi_0(\text{Map}_{\mathcal C}(X,Y))$.

By Lurie, $h$ is left-adjoint to $N$. That is, $h : \sSet \rightleftarrows \text{Cat} : N$, or $\text{Map}_{\sSet}(\mathcal C,N(\mathcal D)) \cong \text{Map}_{\text{Cat}}(h\mathcal C, \mathcal D)$, for any $\infty$-category $\mathcal C$ and any 1-category $\mathcal D$. Our next goal is to describe a functor $\Sing_A(X)\to N(SC)$, maybe through this adjunction, where $SC$ is the 1-category of simplicial complexes and simplicial maps.

References: Lurie (Higher topos theory, Sections 1.1.3 and 1.2.3), Lurie (Higher algebra, Appendix A.6), Goerss and Jardine (Simplicial homotopy theory, Section I.3), Joyal (Quasi-categories and Kan complexes)

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)

The point-counting stratification of the Ran space is conical (really though)

This post is meant to correct the issues of a previous post ("The point-counting stratification of the Ran space is conical," 2017-11-06). The setting is the same as before.

A key object here will be the Ran space of a disconnected space. If $M$ has $k$ connected components, then $\Ran^{\leq n}(M)$, for $k\leq n$, has $2^k$ connected components (as we must choose whether or not to have at least one point in each component). When $M$ is disconnected with $k\leq n$ components, let $\Ran^{\leq n}(M)'$ be the largest connected component of $\Ran^{\leq n}(M)$, that is, the one with at least one point in every component of $M$.

Proposition 1: The point-counting stratification $f:X\to A$ is conical.

Proof: Fix $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)$ and set $2\epsilon = \min_{i<j}d(P_i,P_j)$. Set
\[
Z := \prod_{i=1}^k B^{\R^m}_{\epsilon/3}(0),
\hspace{.5cm}
Y := \Bigg\{(Q^1,\dots,Q^k)\in \underbrace{\Ran^{\leq n}\left(\coprod_{i=1}^k B^{\R^m}_{2\epsilon/3}(0)\right)'}_{W}\ :\ \overbrace{\sum_{i=1}^k \mathbf d(\{0\},Q^i)=\frac{2\epsilon}3}^{\text{the cone condition}}\ ,\ \hspace{-5pt}\underbrace{\sum_{j=1}^{|Q^i|}Q_j^i =0}_{\text{the centroid condition}}\hspace{-5pt}\ \forall\ i\Bigg\}.
\]
The cone condition ensures the right topology at the cone point of $C(Y)$. The centroid condition ensures injectivity of $\varphi_P$ when multiplying by $Z$. Reasoning for the constants $\epsilon/3$ and $2\epsilon/3$ is given in Proposition 2. The balls $B_{\epsilon/3}$ in $Z$ are open and the balls $B_{2\epsilon/3}$ in $Y$ are closed. The collection $Q^i$ is all the points in the $i$th ball $B_{2\epsilon/3}$. The product $Z\times C(Y)$ is given the topology for which a set $U$ is open if $\varphi_P(U)$ is open in the subspace topology on $\im(\varphi_P)\subseteq X$.

If $k=n$, then $Y=\emptyset$ because of the cone and centroid conditions. Then $C(Y)=*$ and $Z\times * \cong Z$ is by construction an open set in $X$, so $X$ is conically stratified at $P$. If $k<n$, then $Y$ has a natural $A_{>k}$ stratification\[(Q^1,\dots,Q^k) \mapsto \sum_{i=1}^k |Q^i| \in \{k+1,\dots,n\},\]
with the image landing in $A_{>k}=\{k+1,\dots,n\}$ because there are at least $k$ points in every element of $W$, and to satisfy the cone condition at least one $Q^i$ must have at least two points. We now claim that
\[\begin{array}{r c l}
\varphi_P\ :\ Z \times C(Y) & \to & X, \\
(R,(Q^1,\dots,Q^k),t\neq 0) & \mapsto & \{P_1+R_1+tQ^1,\dots,P_k+R_k+tQ^k\},\\
(R,*,0) & \mapsto & \{P_1+R_1,\dots,P_k+R_k\}
\end{array}\]
is an open embedding. Here $t\in [0,1)$ is the cone component and $P_i+R_i+Q^i$ is the collection of all $P_i+R_i+Q^i_j,$ for $j=1,\dots,|Q^i|$. Injectivity follows from the cone condition and centroid condition on $Y$: it is clear that for fixed $R\in Z$ the map $\varphi_P$ is injective, as we are taking points at different distances from $P_i+R_i$ (the cone condition). Because of the centroid condition, moving around to different $R$ means the centroid will also move, so we will not get a collection of points we previously had.

Further, $Z\times C(Y)$ gets mapped homeomorphically into $X$ because of the topology forced upon it - continuity and openness of the map follow immediately. Finally, since $\im(\varphi_P)$ is open in $X$, $\varphi_P$ is an open embedding, so $X$ is conically stratified at $P$. $\square$

The following proposition gives a better idea of where $\im(\varphi_P)$ actually lands in $X$.

Proposition 2: For $\varphi_P$ as above, we have $B^X_{2\epsilon/3k}(P)\subseteq \im(\varphi_P)\subseteq B^X_{\epsilon}(P)$.

Proof: For the first inclusion, take $S\in B^X_{2\epsilon/3k}(P)$ and for every $1\leq i\leq k$ set
\begin{align*}S^i & := \{s\in S\ :\ d(s,P_i)<d(s,P_j)\ \forall\ j\neq i\}, & (\text{points closest to $P_i$}) \\
T_i & := \frac1{|S^i|}\sum_{j=1}^{|S^i|} S^i_j, & (\text{centroid of the $S^i_j$}) \\
c_i & := \mathbf d(\{T_i\},S^i). & (\text{distance from the $S^i_j$ to their centroid})\end{align*}
If $|S|=k$, then $c_i=0$ for all $i$, and $t=0$, so we are at the cone point, and $S = \varphi_P(T-P,*,0)$. If $|S|>k$, then $0<\sum_i c_i <2\epsilon/3$, as $c_i < 2\epsilon/{3k}$ for all $i$, so there is some $t'\in (0,1)$ such that $\sum_i c_i = t'2\epsilon/3$. Then
\begin{align*}\varphi_P\left(T-P, \left( {\textstyle \frac1{t'}} (S^1-T_1),\dots, {\textstyle \frac1{t'}}(S^k-T_k) \right), t'\right)
& = \left\{ P_1+(T-P)_1 + t'{\textstyle \frac1{t'}}(S^1-T_1), \dots, P_k+(T-P)_k + t'{\textstyle \frac1{t'}}(S^k-T_k)\right\} \\
& = \left\{ P_1+T_1-P_1 + S^1-T_1, \dots, P_k+T_k-P_k + S^k-T_k\right\} \\
& = \left\{S^1, \dots, S^k\right\} \\
& = S.\end{align*}
Note that since $T_i$ is the centroid of the $S^i_j$, and $S^i \subset B^M_{2\epsilon/3k}(P_i)$, and the centroid of a collection of points is in their convex hull, we also have $T_i\in B^M_{2\epsilon/3k}(P_i)$. Since $\frac{2\epsilon}{3k}<\frac{\epsilon}3$ when $k>2$, we have that $P_i-T_i\in B^{\R^m}_{\epsilon/3}(0)$. If $k\leq 2$, then use $k+2$ instead of $k$. Finally, since $\sum_i c_i/t' = 2\epsilon/3$ and $c_i,t\geq 0$, we have that $c_i/t'\leq 2\epsilon/3$, meaning that $\frac1{t'}(S^i-T_i)\subset B^{\R^m}_{2\epsilon/3}(0)$. Hence the argument of $\varphi_P$ given above is in the domain of $\varphi_P$.

For the second inclusion, first fix $i$. For an element in the image of $\varphi_P$, note that $d(P_i,P_i+R_i)\leqslant \epsilon/3$ and $\mathbf d(\{R_i\},tQ^i)<\mathbf d(\{R_i\},Q^i)\leq 2\epsilon/3$. Since $d(P_i,R_i) = \mathbf d(\{P_i\},\{R_i\})$, we have that
\begin{align*}\mathbf d(\{P_i\},P_i+R_i+tQ^i) & \leqslant \mathbf d(\{P_i\},\{P_i+R_i\})+\mathbf d(\{P_i+R_i\},P_i+R_i+tQ^i) & (\text{triangle inequality}) \\
& = \mathbf d(\{0\},\{R_i\}) + \mathbf d(\{0\},tQ^i) & (\text{linearity of $\mathbf d$}) \\
& = \mathbf d(\{0\},\{R_i\}) + t\mathbf d(\{0\},Q^i) & (\text{linearity of $\mathbf d$}) \\
& < \frac{\epsilon}3 +\frac{2\epsilon}3 & (\text{assumption}) \\
& = \epsilon.\end{align*} $\square$

The following diagram describes the last calculation in the proof.

The point-counting stratification of the Ran space is conical

Note: There are problems with the proof here, in particular with making the map $\varphi$ an embedding. The mistakes are corrected in a later post ("The point-counting stratification of the Ran space is conical (really though)," 2017-11-15).

---

This post completes the effort of several previous posts to show that $f:\Ran^{\leqslant n}(M)\to A=\{1,\dots,n\}$ is a conically stratified space, where $f$ is the point-counting map, for $M$ a compact smooth $m$-manifold embedded in $\R^N$.

Remark: Since $M$ is a manifold, we will work on $M$ or through charts in $\R^m$, as necessary, without explicitly mentioning the charts or domains. Balls $B^M_\lambda, B^{\R^m}_\lambda$ of radius $\lambda$ will be closed and $\mathcal B^{\R^m}_\lambda,\mathcal B^X_\lambda$ will be open. We write $d$ for distance between points of $M$ (or $\R^m$) and $\mathbf d$ for distance between finite subsets of $\R^m$. This is essentially the definition given by Remark 5.5.1.5 of Lurie: \[ \mathbf d(P,Q) = \frac 12 \left(\sup_{p\in P}\inf_{q\in Q} d(p,q) + \sup_{q\in Q}\inf_{p\in P}d(p,q)\right). \] We add the $\frac 12$ so that $\mathbf d(\{p\},\{q\}) = d(p,q)$. Note also $\sup,\inf$ may be replaced by $\max,\min$ in the finite case.

Remark: In our context, given $P\in X$, $\mathbf d$ may be thought of as how far away have new points split off from the $P_i$. That is, if $Q\in X$ is close to $P$ representing the $P_i$ splitting up, then $\mathbf d(P,Q)$ is (half) the sum of the distance to the farthest point splitting off from the $P_i$ and to the farthest point among every $P_i$'s closest point. The diagram below gives the idea.

Then the distance between $P$ and $Q$ is given by
\begin{align*} \mathbf d(P,Q) & = \frac12 \left(\sup_{P_i}\left\{\inf_{Q_j}\left\{d(P_i,Q_j)\right\}\right\} + \sup_{Q_j}\left\{\inf_{P_i}\left\{d(P_i,Q_j)\right\}\right\}\right) \\
& = \frac12\left(\sup \left\{\inf\left\{a,b,c\right\}, \inf\left\{d,e,f,g\right\}\right\} + \sup \left\{ a,b,c,d,e,f,g \right\}\right) \\
& = \frac12\left( \sup\left\{ a,g \right\} + c \right) \\
& = \frac12(a+c). \end{align*}

Now we move on to the main result.

Proposition: The point-counting stratification $f:X\to A$ is conical.

Proof: Fix $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)$ and set $2\epsilon = \min_{i<j}d(P_i,P_j)$. Set \[ Z = \prod_{i=1}^k \mathcal B^{\R^m}_\epsilon(0),
\hspace{2cm}
Y = \coprod_{\sum \ell_i=n \atop \sum t_i = \epsilon} \prod_{i=1}^k\ \left\{Q\in \Ran^{\ell_i}(B^{\R^m}_{t_i}(0))\ :\ \textbf d(0,Q) = t_i,\ \textstyle \sum Q_j = 0 \right\}, \] both of which are topological spaces. The first condition on elements of $Y$ is the cone condition, which ensures the right topology at the cone point in $C(Y)$. The second condition on $Y$ is the centroid condition, which ensures that the point to which 0 maps to (under $\varphi$) is the centroid of points splitting off it, so that we don't overcount when multiplying by $Z$. For $C(Y) = (Y\times [0,1))/(Y\times \{0\})$ the cone of $Y$, define a map \[ \begin{array}{r c l}
\varphi\ :\ C(Y)\times Z & \to & X, \\
\left(\Ran^{\ell_i}(B_{t_i}^{\R^m}(0)),t,R\right) & \mapsto & \Ran^{\ell_i}(B_{tt_i}^M(R_i)),
\end{array} \] where $t\in [0,1)$ is the cone component and $R=\{R_1,\dots,R_k\}\in Z$ is an element of $\Ran^k(M)$ near $P$. It is sufficient to describe where the $\Ran^{\ell_i}$ map to, as all the $Q$ in a fixed $\Ran^{\ell_i}$ map in the same way into $X$.

The map $\varphi$ is continuous by construction, injective by the centroid condition, and a homeomorphism onto its image by the cone condition. Hence $\varphi$ is an embedding, and since the image is open, it is an open embedding. Note that we are taking "open embedding" to mean an embedding whose image is open. Hence every $P\in X$ satisfies Definition A.5.5 of Lurie, so $f:X\to A$ is conically stratified.  $\square$

Remark: Observe that $\mathcal B^X_{\epsilon/k}(P)\subseteq \im(\varphi) \subseteq \mathcal B^X_\epsilon(P)$, both inclusions coming from the $\sum t_i=\epsilon$ condition.

Combined with the proposition of a previous post ("Splitting points in two," 2017-11-02) and Theorem A.9.3 of Lurie, it follows that $A$-constructible sheaves on $X$ are equivalent to functors of $A$-exit paths on $X$ to the category $\mathcal S$ of spaces. A previously given construction (in "Exit paths, part 2," 2017-09-28) gives such a functor, indicating that there exists an $A$-constructible sheaf on $X$.

Next steps may involve applying this approach to the space $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, which was the motivator for all this, or continuing with Lurie's work to see how far this can be taken.

References: Lurie (Higher Algebra, Appendix A), nLab (article "Embedding of topological spaces")

Splitting points in two

The goal of this post is to expand upon some final ideas in a previous post ("Atempts at proving conical stratification," 2017-10-27). Let $M$ be a compact smooth $m$-manifold embedded in $\R^N$, and fix $n\in \Z_{>0}$. Let $X = \Ran^{\leqslant n}(M)$ and $f:X\to A=\{1,\dots,n\}$ the usual point-counting stratification. Let \begin{align*} B^X_\epsilon(P) & = \left\{Q\in X\ :\ 2d_M(P,Q) =\sup_{p\in P}\inf_{q\in Q} d_M(p,q) + \sup_{q\in Q}\inf_{p\in P}d_M(p,q) < 2\epsilon\right\}, \\
B^M_\epsilon(p) & = \left\{q\in M\ :\ d_M(p,q) <\epsilon\right\},\\
B^{\R^m}_\epsilon(0) & = \left\{x\in \R^m\ :\ d(0,x)<\epsilon\right\} \end{align*} be open balls in their respective spaces. We use $d_M$ for distance on $M$ and $d$ for distance in $\R^N$. Since $M$ is an $m$-manifold, we will work in charts in $\R^m$ when necessary.

Proposition: The stratification $f:X\to A$ is conical in the top two strata $\Ran^n(M)$ and $\Ran^{n-1}(M)$.

Proof: Let $P = \{P_1,\dots,P_n\}\in \Ran^n(M)$ and $2\epsilon = \min_{1\leqslant i <j\leqslant n}d(P_i,P_j)$. Let $Y = \emptyset$ which has a natural $(A_{>n} = \emptyset)$-stratification with $C(Y) = \{*\}$ having a natural $(A_{\geqslant n} = \{n\})$-stratification. Let $Z = B^X_\epsilon(P) = \prod_{i=1}^nB^M_\epsilon(P_i)$, for which the identity map $Z\times \{*\} \cong Z\hookrightarrow X$ is an open embedding. Hence $X$ is stratified at every $P\in \Ran^n(M)$.

Let $P = \{P_1,\dots,P_{n-1}\}\in \Ran^{n-1}(M)$ and $2\epsilon = \min_{1\leqslant i <j\leqslant n-1}d(P_i,P_j)$. Let \[ Y = \coprod_{i=1}^{n-1} \mathbf P\partial B^{\R^m}_{\epsilon/2}(0),
\hspace{1cm}
Z = B^{\R^m}_{\epsilon/2}(0), \] where $\mathbf P\partial B$ is the projectivization of the sphere, so may be viewed as a collection of unique pairs $\{\vec v, -\vec v\}$. Then the cone $C(Y)$ may be viewed as a collection of pairs $\{\vec v,t>0\}$ along with the singleton $\{0\}$, with the usual cone topology. Define a map \[ \begin{array}{r c l}
\varphi\ :\ Z\times C(Y) & \to & X, \\
(x,\vec v,t) & \mapsto & \{x+t\vec v,x-t\vec v\} ,\\
(x,0) & \mapsto & \{x\}.
\end{array} \] Note that $B^X_{\epsilon/2}(P) \subseteq \im(\varphi)\subseteq B^X_\epsilon(P)$. This map is injective as every pair of points on $M$ within an $\epsilon/2$-radius of $P_i$ is uniquely defined by their midpoint (the element of $Z$), a direction from that midpoint (the element of $Y$) and a distance from that midpoint (the cone component $t\in [0,1)$). By construction $\varphi$ is continuous and an embedding. The map takes open sets to open sets, so we have an open embedding into $X$. Hence $X$ is conically stratified at every $P\in \Ran^{n-1}(M)$. $\square$

The problem with generalizing this to $P\in \Ran^k(M)$ for all other $k$ is that an $(n-k+1)$-tuple of points has no unique midpoint. It does have a unique centroid, but it is not clear what the $[0,1)$ component of the cone should then be.

Proposition: The space $X$ is of locally singular shape.

Proof: First note that every $P\in X$ has an open neighborhood that is homemorphic to an open ball of dimension $mn$ (see Equation (1) of previous post "Attempts at proving conical stratification," 2017-10-27). Hence we may cover $X$ by contractible sets. By Remark A.4.16 of Lurie, $X$ will be of locally singular shape if every element of the cover is of singular shape. Since all elements of the cover are contractible, by Remark A.4.11 of Lurie we only need to check if the topological space $*$ is of singular shape. Finally, Example A.4.12 of Lurie gives that $*$ has singular shape. $\square$

References: Lurie (Higher algebra, Appendix A)

Attempts at proving conical stratification

This post chronicles several attempts and failures to show that $X=\Ran^{\leqslant n}(M)$ is conically stratified. Here $M$ will be a smooth, compact manifold of dimension $m$, embedded in $\R^N$ for $N\gg 0$. Recall that a stratified space $f:X\to A$ is conically stratitifed at $x\in X$ if there exist:

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

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

Let $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)=X$, and $2\epsilon = \min_{1\leqslant i<j\leqslant k}\{d(P_i,P_j)\}$.

Observations

Observation 1: When $M=I = (0,1)$, the interval, we can visualize what $\Ran^{\leqslant 3}(M)$ looks like via the construction $\Ran^{\leqslant 3}(M) = (M^3\setminus \Delta_3)/ S_3$, to gain some intuition about what the Ran space looks like in general. 

A drawback is that $\dim(M)=1$, which masks the problems in higher dimensions.

Observation 2: An open neighborhood of $P\in X$ looks like
\[\coprod_{\sum \ell_i = n \atop \ell_i\in \Z_{>0}}\ \prod_{i=1}^k \Ran^{\leqslant \ell_i}(B^M_\epsilon(P_i)) = B^X_{\epsilon/2}(P) \times \coprod_{\sum \ell_i = n \atop \ell_i\in \Z_{>0}}\ \prod_{i=1}^k \Ran^{\leqslant \ell_i}(B^M_{\epsilon/2}(P_i)),\hspace{2cm} (1) \]
for $B^M_\epsilon(x) = \{y\in M\ :\ d_M(x,y)<\epsilon\}$ the open ball of radius $\epsilon$ around $x\in M$, and similarly for $P\in X$. Most attempts to prove conical stratification are based around expressing these as $Z\times C(Y)$, usually for $Z=B_{\epsilon/2}^X(P)$.

Observation 3: When $k<n$, the "steepest" direction from $P_i$ into the highest stratum of $X$ is given by $P_i$ splitting into $n-k+1$ points uniformly distributed on $\partial B^M_t(P_i)$. Hence the $[0,1)$ part of the cone (recall $C(Y)=Y\times [0,1)/\sim$) should be along $t\in [0,1)$.

Attempts

Attempt 1: Use more resrictive (but better described) AFT definition.

Ayala-Francis-Tanaka describe $C^0$ stratified spaces, a special type of stratified space. Any space that has a cover by topological manifolds is a $C^0$ stratified space, however it seems that $X$ cannot be covered by topological manifolds. Even further, each element in the cover must have the trivial stratification, and since we must have overlaps, $f:X\to A$ will have $A=\{*\}$, which is not what we want.

Attempt 2: Stratify $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$ instead.

This is more difficult, but was the original impetus, with strata defined by collecting the Vietoris-Rips complexes $VR(P,t)$ of the same type. The problem is that this space has strata next to each other of the same dimension, which does not conform to a standard definition of stratification, and so doesn't admit a conical stratification. Dimension counting and requiring an open embedding $Z\times C(Y)\hookrightarrow X$ shows this is impossible at the boundary point between two such strata.

Weinberger gives some standard stratifed space types, among them a manifold stratified space, a manifold stratified space with boundary, and a PL stratified space, but $X\times \R_{\geqslant 0}$ is none of these.

Attempt 3: Naively describe the neighborhood of $P$ as a cone. 

This is the most direct attempt to write (1) as $Z\times C(Y)$. If we say
\[ C(Y) = \underbrace{\coprod_{\sum \ell_i = n \atop \ell_i\in \Z_{>0}}\ \prod_{i=1}^k \Ran^{\leqslant \ell_i}(\partial B^M_{t}(P_i))}_{Y} \times [0,\epsilon/2) \Bigg/\sim, \]
then we miss points splitting off at different "speeds". That is, in this presentation $P_i$ can only split into points that are all the same distance away from it. Between such a collection of points and $P_i$ are points that are some closer, some the same distance away, and those are not accounted for.

Moreover, using $Z=B^X_{\epsilon/2}(P)$, leads to overcounting, and the map into $X$ would not be injective.

Attempt 4: Iterate over different number of points at common radius.

This came out of an attempt to fix the previous attempt. As in a previous post ("The Ran space is locally conical," 2017-10-22), let $E_\ell$ be the collection of distinct partitions of $\ell$ elements, and for $e\in E_\ell$, let $T(e)$ be the collection of distinct total orderings of $e$. A candidate for $Z\times C(Y)$ would then be

with $t_{i,0} = \epsilon$ and $t_{i,j>0}$ the chosen element of $(0,t_{i,j-1})$. The open embedding $Z\times C(Y) \to X$ would be the inclusion on the $C(Y)$ component, and would scale every factor in the $Z$ component to a neighborhood of $P_i$ of radius $t_{i,|\tau_i|}$. However, this embedding is not continuous, because a point in $\Ran^k(M)$ is next to a point in $\Ran^{n}(M)$, where $P_i$ has split off into $n-k$ points, but the radius of $B^M_\epsilon(P_i)$ in $\Ran^k(M)$ is $\epsilon$, while in $\Ran^n(M)$ it is the shortest distance from one of the new points to $P_i$.

Attempt 5: Iterate over common radii, but only "antipodal" points.
This was an attempt to fix the previous attempt and combine it with the naive description. In fact, this approach works when $k=1$ and $n=2$. Then $P = \{P_1\}$, and
\[B^M_\epsilon(P_1) \times \left.\left(\mathbf P\partial B^M_t(P_1) \times [0,1)\right)\right/\sim\]
maps into $B^X_\epsilon(P_1)$ by first scaling $[0,1)$ down to $[0,\epsilon-d_M(P,P_1))$, where $P\in B^M_\epsilon(P_1)$ is the chosen point. The object $\mathbf P\partial B^M_t(P_1)$ is the projectivization of the boundary of the open $\dim(M)$-ball of radius $t$ around $P_1$ on $M$. That is, every element in it is a pair of antipodal points on the boundary of this ball that are exactly $t\in [0,\epsilon-d_M(P,P_1))$ away from $P_1$.

This works because every pair of points in a contractible neighborhood of $P_1$ is described uniquely by a pair $(P,v)$, for $P$ the midpoint of the two points and $v$ the $\dim(M)$-vector giving the direction of the points from $P$ (this may rely on working in charts, which is fine, as $M$ is a manifold). However, trying to generalize to more than two points fails because $\ell>2$ points in general are not equally distributed on a sphere. If instead of using the "antipodal" property we take a point from which all $\ell$ points are equidistant, this point may not be in the $\epsilon$-neighborhood of $P_1$.

Possible solutions

Solution 1: Instead of a smooth manifold, let $M$ be a simplicial complex. Then $\Ran^{\leqslant n}(M)$ should also be a simplicial complex. Then it may be possible to apply a general theorem to find appropriate cones.

Solution 2: Extend the only partially successful attempt, Attempt 5. Extend by describing a point splitting off into $\ell$ pieces as a sequence of points splitting into 2 pieces. Or, extend by using the centroid of $\ell$ points instead of the midpoint.

Solution 3: Weaken definition of "conically stratifed" to exclude either open embedding condition or $A_{>f(x)}$ stratification of $Y$, though this would involve following out Lurie's proof to see what can not be concluded.

References: Lurie (Higher algebra, Appendix A), Ayala, Francis and Tanaka (Local structures on stratified spaces, Sections 2 and 3), Weinberger (The classification of topologically stratified spaces)

Exit paths, part 2

In this post we continue on a previous topic ("Exit paths, part 1," 2017-08-31) and try to define a constructible sheaf via universality. Let $X$ be an $A$-stratified space, that is, a topological space $X$ and a poset $(A,\leqslant)$ with a continuous map $f:X\to A$, where $A$ is given the upset topology relative to its ordering $\leqslant$. Recall the full subcategory $\Sing^A(X)\subseteq \Sing(X)$ of exit paths on an $A$-stratified space $X$.

Proposition: If $X\to A$ is conically stratified, $\Sing^A(X)$ is an $\infty$-category.

Briefly, a stratification $f:X\to A$ is conical if for every stratum there exists a particular embedding from a stratified cone into $X$ (see Lurie for "conical stratification" and Ayala, Francis, Tanaka for "conically smooth stratified space," which seem to be the same). We will leave confirming the described stratification as conical to a later post.

This proposition, given as part of Theorem A.6.4 in Lurie, has a very long proof, so is not repeated here. Lurie actually proves that the natural functor $\Sing^A(X)\to N(A)$ described below is a (inner) fibration, which implies the unique lifting property of $\Sing^A(X)$ via the unique lifting property of $N(A)$ (and we already know nerves are $\infty$-categories).

Example: The nerve of a poset is an $\infty$-category. Being a nerve, it is already immediate, but it is worthwhile to consider the actual construction. For example, if $A = \{a\leqslant b\leqslant c \leqslant d\}$ is the poset with the ordering $\leqslant$, then the pieces $N(A)_i$ are as below.

It is immediate that every 3-horn can only be filled in one unique way (as there is only one element of $N(A)_3$), as well as that every 2-horn can be filled in one unique way (as every sequence of two composable morphisms appears as a horn of exactly one element of $N(A)_2$).

In Appendix A.9 of Higher Algebra, Lurie says that there is an equivalence of categories \[(A\text{-constructible sheaves on }X)   \cong   \left[(A\text{-exit paths on }X),\mathcal S\right],\] given that $X$ is conically stratified, and for $\mathcal S$ the $\infty$-category of spaces (equivalently $N(Kan)$, the nerve of all the simplicial sets that are Kan complexes). So, instead of trying to define  a particular constructible sheaf on $X = \Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, (as in previous posts "Stratifying correctly," 2017-09-17 and "A constructible sheaf over the Ran space," 2017-06-24) we will try to make a functor that takes an exit path of $X$ and gives back a space.

Fix $n\in \Z_{>0}$ and set $X = \Ran^{\leqslant n}\times \R_{\geqslant 0}$. Let $SC$ be the category of simplicial complexes and simplicial maps, with $SC_n$ the full subcategory of simplicial complexes with at most $n$ vertices. There is a map
\[\begin{array}{r c l}
g\ :\ X & \to & SC_n \\
(P,t) & \mapsto & VR(P,t),
\end{array}\]
allowing us to say
\[X = \bigcup_{S\in SC_n}g^{-1}(S).\]
Here we consider that two elements $P_i,P_j\in P$ give an edge of $VR(P,t)$ whenever $t>d(P_i,P_j)$ (this is chosen instead of $t\geqslant d(P_i,P_j)$ so that the boundaries of the strata ``facing downward," with respect to the poset ordering, are open). Now we define a stratifying poset $A$ for $X$.

Definition: Let $A = \{a_S\ :\ S\in SC_n\}$ and define a relation $\leqslant$ on $A$ by
\[ \left(a_S\leqslant a_T\right)\ \ \Longleftarrow\ \ \left(
\begin{array}{c}
\exists\ \sigma\in \Sing(X)_1\ \text{such that}\\
g(\sigma(0))=S,\ g(\sigma(t>0))=T.
\end{array}
\right)\]
Let $(A,\leqslant)$ be the poset generated by relations of the type given above.

We claim that $f:X\to A$ given by $f(P,t)=a_{g(P,t)}$ is a stratifying map, that is, continuous in the upset topology on $A$. To see this, take the open set $U_S = \{a_T\in A\ :\ a_S\leqslant a_T\}$ in the basis of the upset topology of $A$, for any $S\in SC_n$, and consider $x\in f^{-1}(U_S)$. If for all $\epsilon>0$ we have $B_X(x,\epsilon)\cap f^{-1}(U_S)^C\neq \emptyset$, then there exists $T_\epsilon\in SC_n$ with $B_X(x,\epsilon)\cap f^{-1}(a_{T_\epsilon})\neq\emptyset$, for $S\not\leqslant T_\epsilon$ (as $T_\epsilon\not\in U_S$). This means there exists $\sigma\in \Sing(X)_1$ with $\sigma(0)=x$ and $\sigma(t>0)\in f^{-1}(a_{T_\epsilon})$, which in turn implies $S\leqslant T_\epsilon$, a contradiction. Hence $f$ is continuous, so $f:X\to A$ is a stratification.

As all morphisms in $\Sing(X)$ are compsitions of the face maps $s_i$ and degenracy maps $d_i$, so are all morphisms in $\Sing^A(X)$. There is a natural functor $F:\Sing^A(X)\to N(A)$ defined in the following way:
\[\begin{array}{r r c l}
%%
%% L1
%%
\text{objects} & \left(
\begin{array}{c}
\sigma:|\Delta^k|\to X \\
a_0\leqslant \cdots \leqslant a_k\subseteq A \\
f(\sigma(t_0,\dots,t_i\neq 0,0,\dots,0)) = a_i
\end{array}
\right) & \mapsto & \left( a_0\to\cdots\to a_k\in N(A)_k\right) \\[20pt]
%%
%% L2
%%
\text{face maps} & \left(
\begin{array}{c}
\left(
\begin{array}{c}
\sigma:|\Delta^k|\to X \\ a_0\leqslant \cdots \leqslant a_k\subseteq A
\end{array}
\right)\\[10pt]
\downarrow \\[10pt]
\left(
\begin{array}{c}
\tau:|\Delta^{k+1}|\to X \\ a_0\leqslant \cdots \leqslant a_i\leqslant a_i\leqslant \cdots a_k\subseteq A
\end{array}
\right)
\end{array}
\right) & \mapsto &
\left(\begin{array}{c}
\left(a_0\to\cdots \to a_k\right)\\[10pt]
\downarrow\\[10pt]
\left(a_0\to\cdots \to a_i\xrightarrow{\text{id}}a_i\to\cdots \to a_k\right)
\end{array}\right)\\[40pt]
%%
%% L3
%%
\text{degeneracy maps} & \left(
\begin{array}{c}
\left(
\begin{array}{c}
\sigma:|\Delta^k|\to X \\ a_0\leqslant \cdots \leqslant a_k\subseteq A
\end{array}
\right)\\[10pt]
\downarrow \\[10pt]
\left(
\begin{array}{c}
\tau:|\Delta^{k-1}|\to X \\ a_0\leqslant \cdots \leqslant a_{i-1}\leqslant a_{i+1}\leqslant \cdots a_k\subseteq A
\end{array}
\right)
\end{array}
\right) & \mapsto &
\left(\begin{array}{c}
\left(a_0\to\cdots \to a_k\right)\\[10pt]
\downarrow\\[10pt]
\left(a_0\to\cdots \to a_{i-1}\xrightarrow{\circ}a_{i+1}\to\cdots \to a_k\right)
\end{array}\right)
\end{array}\]
As all maps in $\Sing^A(X)$ are generated by compositions of face and degeneracy maps, this completely defines $F$. Naturality of $F$ follows precisely because of this.

A poset (which can be viewed as a directed simple graph) may be naturally viewed as a 1-dimensional simplicial set, moreover an $\infty$-category (by virtue of being a \emph{simple} graph, with no multi-edges or loops). Hence there is a natural map, the inclusion, that takes $N(A)$ into $N(\mathcal Kan) = \mathcal S$.  Finally, Construction A.9.2 of Lurie describes a map that takes a functor from $A$-exit paths into spaces and gives back an $A$-constructible sheaf over $X$, which Theorem A.9.3 shows to be an equivalence, given the following conditions:

• $X$ is paracompact,
• $X$ is locally of singular shape,
• the $A$-stratification of $X$ is conical, and
• $A$ satisfies the ascending chain condition.

The first condition is satisfied as both $\Ran^{\leqslant n}(M)$ and $\R_{\geqslant 0}$ are locally compact and second countable. The last condition is satisfied because $A$ is a finite poset. We already mentioned that the conical property will be checked later, as will the singular shape property. Unfortunately, Lurie gives a definition of singular shape only for $\infty$-topoi, so some work must be done to translate this into our simpler setting. However, in the introduction to Appendix A, Lurie says that if $X$ is "sufficiently nice" and we assume some "mild assumptions" about $A$, then the described categorical equivalence follows, so it seems there is hope that everything will work out well in the end.

References: Stacks Project, Lurie (Higher algebra, Appendix A), Ayala, Francis and Tanaka (Local structures on stratified spaces, Sections 2 and 3)