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)

The Ran space is locally conical

In this post we show that every point in the Ran space $\Ran^{\leqslant n}(M)$, for $M$ a compact, smooth embedded manifold, is the base of a cone in $\Ran^{\leqslant n}(M)$. Let $\dim(M) = m$ and let $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)$. We write $d(x,y)$ for distance in Euclidean space $\R^N$ where $M$ is embedded, and $d_M(x,y)$ for distance on the embedded manifold $M$ (note $d\leqslant d_M$). Define the following objects:
\begin{align*}
N_\epsilon(x) & = \{z\in M\ :\ d_M(x,z)<\epsilon\}, \\
E_n & = \{\text{distinct partitions of an unlabeled set of $n$ elements}\}, \\
T(e) & = \{\text{distinct total orderings of }e\in E_n\}.
\end{align*}
We write $\tau=(\tau_1<\cdots<\tau_{|\tau|})$ for an element $\tau\in T(e)$.

Example: Let $n=4$, so then
\[
E_4 = \Big\{\{\{*\},\{*\},\{*\},\{*\}\},\hspace{10pt}
\{\{*,*\},\{*\},\{*\}\},\hspace{10pt}
\{\{*,*\},\{*,*\}\},\hspace{10pt}
\{\{*,*,*\},\{*\}\},\hspace{10pt}
\{\{*,*,*,*\}\}\Big\}.
\]
By stacking the $*$ on top of one another to indicate containment in a single set, and for order increasing from left to right, we have the following distinct total orderings for every element of $E_4$.

Set $\epsilon = \min_{1\leqslant i<j\leqslant k}\{d(P_i,P_j)\}$, $t_0\in(0,\epsilon/2)$, and $t_{j>0}\in (0,t_{j-1})$. By construction, the object
\begin{align*}
C_P & = \{P\} \cup \coprod_{\genfrac{}{}{0pt}{}{\sum \ell_i=n-k}{\ell_i \in \Z_{>0}}}\ \prod_{i=1}^k\  \coprod_{\genfrac{}{}{0pt}{}{\tau\in T(e)}{e\in E_{\ell_i}}}\  \prod_{j=1}^{|\tau|} \Ran^{|\tau_j|}\left(\partial N_{t_j}(P_i)\right) \times (0,t_{j-1}) \\
& = \left.\coprod_{\genfrac{}{}{0pt}{}{\sum \ell_i=n-k}{\ell_i \in \Z_{>0}}}\ \prod_{i=1}^k\ \coprod_{\genfrac{}{}{0pt}{}{\tau\in T(e)}{e\in E_{\ell_i}}}\left( \Ran^{|\tau_1|} (\partial N_{t_0}(P_i))\times \prod_{j=2}^{|\tau|} \Ran^{|\tau_j|}\left(\partial N_{t_j}(P_i)\right) \times (0,t_{j-1})\right) \times [0,\epsilon/2)\right/\sim
\end{align*}
is an open cone based at $P$ sitting inside $\Ran^{\leqslant n}(M)$. Here $\sim$ is the equivalence relation of all elements with $t_0=0$, with $[0,\epsilon/2)\owns t_0$ representing the unit interval in the usual cone construction. Moreover, given the point-counting stratification $f:\Ran^{\leqslant n}(M)\to A$, there is a natural stratification $g:C_p\to A_{\geqslant f(P)}$, with $P\in C_P$ the only element mapping to $f(P)$ under $g$.

The next step is to show that $P$ has an open neighborhood in $\Ran^{\leqslant n}(M)$ that is the image of an open embedding $Z\times C_P$, for some topological space $Z$. The obvious choice $Z = \prod_{i=1}^k N_{\epsilon/2}(P_i)$ does not work, because we double count points in higher strata, so we do not have an embedding.

Basic topological constructions

 Preliminary exam prep

Let $X,Y$ be topological spaces based at $x_0,y_0$, respectively, and $I=[0,1]$ the unit interval.
\[
\begin{array}{r r c l}
\text{cone} & CX & = & X\times I / X\times \{0\} \\[5pt]
\text{suspension} & \Sigma X & = & X\times I / X\times \{0\}, X\times \{1\}\\[5pt]
\text{reduced suspension} & \widetilde\Sigma X & = & X\times I/X\times\{0\}, X\times \{0\}, \{x_0\}\times I \\[5pt]
\text{wedge} & X\vee Y & = & X\sqcup Y / \{x_0\} \sim \{y_0\} \\[5pt]
\text{smash} & X\wedge Y & = & X\times Y / X\times \{y_0\}, \{x_0\}\times Y \\[5pt]
\text{join} & X * Y & = & X\times Y \times I \left/\begin{array}{l l}
X\times \{y\}\times \{0\} & \forall\ y\in Y \\
\{x\}\times Y \times \{1\} & \forall\ x\in X
\end{array}\right. \\[5pt]
\text{connected sum} & X \# Y & = & (X\setminus D^n_X)\sqcup (Y\setminus D^n_Y) / \partial D^n_X \sim \partial D^n_Y
\end{array}
\]
In the last description, $X$ and $Y$ are assumed to be $n$-manifolds, with $D^n_X$ a closed $n$-dimensional disk in $X$ (similarly for $Y$). The quotient identification may also be made via some non-trivial map. In fact, only the interior of each $n$-disk is removed from $X$ and $Y$, so that the quotient makes sense.

Remark: Some of the above constructions may be expressed in terms of others, for example
\[
X\wedge Y = X\times Y / X\vee Y,
\hspace{1cm}
X*Y = \Sigma(X\wedge Y).
\]
The first is clear by viewing $X = X\times \{y_0\}$ and $Y = \{x_0\}\times Y$ as sitting inside $X\times Y$. The second is clear by letting $X\times \{y\}\times \{0\}$ be identified to $\{x_0\}\times\{y\}\times \{0\}$ for every $y\in Y$, and analogously with $Y$.

Example: Here are some of the constructions above applied to some common spaces.
\begin{align*}
CX & \simeq \text{pt} & \Sigma S^n & = S^{n+1} & S^n \wedge S^m & = S^{n+m}\\
\Sigma X & = S^1 \wedge X & S^n * S^m & = S^{n+m+1}\end{align*}
Remark: We may also calculate the homology of the new spaces in terms of the old ones.
\[
\begin{array}{r c l l}
\widetilde H_k(CX) & = & 0 & \text{via homotopy} \\
\widetilde H_k(\Sigma X) & = & \widetilde H_{k-1}(X) & \text{via Mayer--Vietoris} \\
\widetilde H_k(X\vee Y) & = & \widetilde H_k(X)\oplus \widetilde H_k(Y) & \text{via Mayer--Vietoris}\\
\widetilde H_k(X\wedge S^\ell) & = & \widetilde H_{k-\ell}(X) & \text{via Kunneth} \\
\widetilde H_k(X\# Y) & = & \widetilde H_k(X) \oplus \widetilde H_k(Y) & \text{via Mayer--Vietoris and relative homology}
\end{array}
\]
The last equality holds for $k<n-1$, for $M$ and $N$ both $n$-manifolds, and for $k=n-1$ when at least one of them is orientable.

References: Hatcher (Algebraic Topology, Chapters 0, 2)