Perspectives on the Ran space

This post combines the finite subset approach with the mapping space approach of the Ran space, in the context of stratifications. The goal is to understand the colimit construction of the Ran space, as that leads to more powerful results.

Topology

Let $X,Y$ be topological spaces.

Definition: The mapping space of $X$ with respect to $Y$ is the topological space $X^Y = \{f:Y\to X$ continuous$\}$. The topology on $X^Y$ is the compact-open topology which has as basis finite intersections of sets
$$\{f\in X^Y\ :\ f(K)\subseteq U\}, \hspace{3cm} (1)$$
for all $K\subseteq Y$ compact and all $U\subseteq X$ open.

Now fix a positive integer $n$.

Definition: The Ran space of $X$ is the space $\Ran^{\leqslant n}(X) = \{P\subseteq X\ :\ 0<|P|\leqslant n\}$. The topology on $\Ran^{\leqslant n}(X)$ is the coarsest which contains
$$\left\{P\in \Ran^{\leqslant n}(X)\ :\ P\subseteq \bigcup_{i=1}^k U_i,\ P\cap U_i\neq \emptyset\ \forall\ i \right\} \hspace{3cm} (2)$$
as open sets, for all nonempty finite collection of parwise disjoint open sets $\{U_i\}_{i=1}^k$ in $X$.

From now on, we let $I$ be a set of size $n$ and $M$ be a compact, smooth, connected $m$-manifold. There is a natural map
$$\begin{array}{r c l} \varphi\ :\ M^I & \to & \Ran^{\leqslant n}(M), \\ (f:I\to M) & \mapsto & f(I). \end{array}$$
This map is surjective, and for $n>1$, is not injective.

Proposition 1: The map $\varphi$ is continuous and an open map.

Proof: For continuity, take an open set $U\subseteq \Ran^{\leqslant n}(M)$ as in (2) and consider $\varphi^{-1}(U)$. We use the fact that $\{*\}\subset I$ is a compact (in fact open and closed) subset of $I$ and that all the $U_i$ are open, as is their union. Observe that
$$\begin{align*}\varphi^{-1}(U) & = \left\{f\in M^I \ :\ f(I)\subset \bigcup_{i=1}^k U_i,\ f(I)\cap U_i\neq \emptyset\ \forall\ i \right\} \\ & = \left\{f\in M^I\ :\ f(I)\subset \bigcup_{i=1}^k U_i\right\} \cap \bigcap_{i=1}^k \left\{f\in M^I\ :\ f(*\in I) \in U_i\right\},\end{align*}$$
which is a finite intersection of sets of the type (1), and so $\varphi^{-1}(U)$ is open in $M^I$.

For openness, take an open set $V$ as in (1), so $V = \bigcap_{i=1}^k \{f\in M^I\ :\ f(K)\subseteq U_i\}$ for different subsets $K\subseteq I$. By Lemma 1, we may assume that the $U_i$ are pairwise disjoint. For each $U_i$, let $\{U_{i,j}\}_{j=1}^\infty$ be a sequence of increasing open sets in $U_i$ such that $U_{i,j}\subseteq U_{i,j+1}$ and $U_{i,j}\xrightarrow{\ j\to\infty\ } U_i$. Then
$$\varphi (V) = \underbrace{\left\{P\in M\ :\ P\subset \bigcup_{i=1}^k U_i,\ P\cap U_i\neq \emptyset\ \forall\ i\right\}}_{f\in M^I\text{\ with image completely in the\ }U_i} \cup \underbrace{\bigcap_{i=1}^k \bigcup_{j=1}^\infty \left\{ P\in M\ :\ P\subset U_{i,j}\cup \left(\overline{U_{i,j}}\right)^c,\ P\cap U_{i,j}\neq \emptyset,\ P \subset \left(\overline{U_{i,j}}\right)^c \neq \emptyset\right\}}_{f\in M^I\text{\ with image partially in the\ }U_i}.$$
Note that $U_{i,j}$ and $\left(\overline{U_{i,j}}\right)^c$, the complement of the closure of $U_{i,j}$ are both open and disjoint in $M$. Since infinite unions and finite intersections of elements in the topology are also open, we have that $\varphi(V)$ is open in $\Ran^{\leqslant n}(M)$. $\square$

The above proposition says that we may talk equivalently about the compact-open topology on $M^I$ and the Ran space topology on $\Ran^{\leqslant n}(M)$. Viewing the Ran space as a function space allows for more general terminology to be applied.

Lemma 1: Let $U_i\subseteq M$ be open, for $i=1,\dots,k$. Then $\bigcap_{i=1}^k\{f\in M^I\ :\ f(K)\subseteq U_i\}$ may be written as a union of intersections $\bigcap_{j=1}^\ell \{f\in M^I\ :\ f(K)\subseteq V_j\}$ with the $V_j$ open, pairwise disjoint, and $\ell\leqslant k$.

Proof: It suffices to prove this in the case $k=2$. Let $U,V\subseteq M$ open and suppose than $U\cap V\neq \emptyset$. Note that $U\setminus V$ and $V\setminus U$ are separated (that is, $(U\setminus V) \cap \overline{V\setminus U} = \emptyset$ and $(V\setminus U)\cap \overline{U\setminus V} = \emptyset$), and since $\R^N$ is a completely normal space (equivalently, satisfies the $T5$ axiom), there exist disjoint open sets $A,B$ with $U\setminus V\subseteq A$ and $U\setminus V\subseteq B$. So for $A' = A\cap (U\cup V)$ and $B' = B\cap (U\cup V)$, we have
$$\begin{align*} \{f\in M^I\ &:\ f(K)\subseteq U\} \cap \{f\in M^I\ :\ f(K)\subseteq V\} \\ & = \left(\{f\in M^I\ :\ f(K)\subseteq U\setminus V\} \cap \{f\in M^I\ :\ f(K)\subseteq V\setminus U\}\right) \cup \{f\in M^I\ :\ f(K)\subseteq U\cap V \} \\ & = \left(\{f\in M^I\ :\ f(K)\subseteq A'\} \cap \{f\in M^I\ :\ f(K)\subseteq B'\}\right) \cup \{f\in M^I\ :\ f(K)\subseteq U\cap V \}, \end{align*}$$
for $A',B',U\cap V$ open, and $A'\cap B' = \emptyset$.  $\square$

Note that in the last calculation of the proof, the intersection of sets in the second line is smaller than the intersection of sets in the last line (as $U\setminus V \subsetneq A$ and $V\setminus U\subsetneq B$). However, all the extra ones in the third line appear in the set $\{f\in M^I\ :\ f(K)\subseteq U\cap V\}$.

Stratifications

Now we compare stratifications on $M^I$ and $\Ran^{\leqslant n}(M)$. As before, $I$ is a set of size $n$.

Corollary: An image-constant $A$-stratification on $M^I$ is equivalent to an $A$-stratification on $\Ran^{\leqslant n}(M)$.

This follows from Proposition 1. By image-constant we mean if $\alpha,\beta\in M^I$ have the same image (that is, $\alpha(I)=\beta(I)$), then $\alpha,\beta$ are sent to the same element of $A$.

Proof: If we start with a continuous map $f:M^I\to A$, setting $g(P) = f(I\to M)$ whenever $(I\to M) \in \varphi^{-1}(P)$ is continuous, as $\varphi(f^{-1}(U))$ is open, by continuity of $f$ and openness of $\varphi$. The assignment $g(P) = f(I\to M)$ whenever $(I\to M) \in \varphi^{-1}(P)$ is well defined, as the stratification is image-constant, so any continuous map from $M^I$ must send every element of $\varphi^{-1}(P)$ to the same place.

Conversely, if we start with a continuous map $g:\Ran^{\leqslant n}(M)\to A$, setting $f(I\to M) = g(\varphi(I\to M))$ is continuous, as $\varphi^{-1}(g^{-1}(U))$ is open, by continuity of $g$ and continuity of $\varphi$. This map is image-constant, as $\varphi(\alpha:I\to M) = \alpha(I)$. $\square$

Next we consider a particular stratification of $M^I$, adapted from Example 3.5.17 of Ayala-Francis-Tanaka, simplified with $P=\{*\}$. That is, the example begins with a stratified space $M\to P$ and proceeds to construct another stratification $M^I\to P'$, but we only consider the trivial stratification $M\to \{*\}$.

Definition: Given $M$ and $I$, let the poset $\mathcal P(I)$ of coincidences on $I$ be the set of equivalence relations on $I$, ordered by reverse set inclusion. Let $f_I:M^I\to \mathcal P(I)$ be the natural stratification that takes a map $\alpha: I\to M$ to the equivalence relation on $I$ describing which elements of $I$ coincide in the image of $\alpha$.

Example: An element of $\mathcal P(I)$ is a subset of $I\times I$ always containing $(a,a)$ for every $a\in I$ (reflexivity), and satisfying the symmetry and transitivity conditions. For example, if $|I|=3$ or 4, then $\mathcal P(I)$ is ordered as in the diagrams below, with order increasing from left to right. We simplify things by writing $[x_1,\dots,x_k]$ for the collection $(x_i,x_j)$ of all $i\neq j$ (the equivalence class).

To check that the map $f_I:M^I\to \mathcal P(I)$ is continuous, we first note that an element $U_{[x_1],\dots,[x_k]}$ in the basis of the upwards-directed topology on $\mathcal P(I)$ contains images of $\alpha\in M^I$ whose images have at most the elements of each equivalence class $[x_i]$ coinciding. Hence
$$f_I^{-1}(U_{[x_1],\dots,[x_k]}) = \bigcup_{U_1,\dots,U_k\subseteq M \atop \text{open, disjoint}}\ \bigcap_{i=1}^k \left\{\alpha\in M^I\ :\ \alpha(K = \{x\in [x_i]\}) \subseteq U_i\right\},$$
which is an open set in the compact-open topolgy on $M^I$.

The Ran space as a colimit

Beilinson-Drinfeld (Section 3.4) and Ayala-Francis-Tanaka (Section 3.7) describe the Ran space as a colimit, the former of a functor into topological spaces, the latter of a functor into stratified spaces. See Mac Lane for a full treatment of colimits. Both BD and AFT use the category $\Fin^{surj,\leqslant n}$ of finite sets and surjections, that is,
$$\begin{align*}\Obj(\Fin^{surj,\leqslant n}) & = \{I\in \Obj(\Set)\ :\ 0<|I|\leqslant n\}, \\ \Hom_{\Fin^{surj,\leqslant n}}(I,J) & = \begin{cases} \emptyset, & \text{\ if\ } |I|<|J|, \\ \left\{\text{surjections\ }I\to J\right\}, & \text{\ if\ } |I|\geqslant |J|. \end{cases}\end{align*}$$
AFT uses more involved terminology, with "conically smooth" stratified spaces instead of just poset-stratified. They use a category $\Strat$, which for our purposes we may define as
$$\begin{align*} \Obj(\Strat) & = \{\text{poset-stratified topological spaces }X\xrightarrow{ f } A\}, \\ \Hom_{\Strat}(X\xrightarrow{ f } A,Y\xrightarrow{ g } B) & = \{(\mu\in \Hom_{\Top}(X,Y), \nu \in \Hom_{\Set}(A,B)\ :\ g\circ \mu = \nu\circ f\}. \end{align*}$$

Remark:  There is a natural functor $\mathcal F_M:(\Fin^{surj,\leqslant n})^{op} \to \Top$, given by $I\mapsto M^I$. A surjection $s:I\to J$ induces a map $M^J\to M^I$, with $(f:J\to M)\mapsto (f\circ s :I\to M)$. BD use this to declare that $\Ran^{\leqslant n}(M) = \colim(\mathcal F_M)$.

Remark: There is also a natural functor $\mathcal G_M:(\Fin^{surj,\leqslant n})^{op} \to \Strat$, given by $I\mapsto (M^I\to \mathcal P(I))$. AFT use this to declare that $(\Ran^{\leqslant n}(M)\to \{1,\dots,n\}) = \colim(\mathcal G_M)$.

The construction of AFT is even more general, as they consider the Ran space of an already stratified space. Here we use their result for $M\to \{*\}$ trivially stratified.

References: Ayala, Francis, and Tanaka (Local structures on stratified spaces, Sections 3.5 and 3.7), Beilinson and Drinfeld (Chiral algebras, Section 3.4), Mac Lane (Categories for the working mathematician, Chapter III.3)