Let $X = S^1 \sqcup S^1 \sqcup S^1 \sqcup \cdots \sqcup S^1$, where there are $m$ disjoint circles. Let $Y \subset X^n$ be the set of $n$ ordered points on $X$ such that there is at least one point on each component. Note that each connected component of $Y$ is isomorphic to $(S^1)^n$. Let the group $G = (S^1)^m$ act on $X$, where the $k$-th factor of $G$ rotates the $k$-th component of $X$. So $G$ acts on $Y$. Let $Z = Y/G$. Each connected component of $Z$ is isomorphic to $(S^1)^{n-m}$ so, if $n>m$, we have $\chi(Z) = 0$.

I now present a regular CW structure on $Z$ so that the number of $k-m$ cells is $S(n,k) c(k,m)$, proving your identity. A given cell of $Z$ will correspond to those arrangements of points on $X$ which lie in a given cyclic order. So, to specify a cell of $Z$, we need to say (1) which points are equal to each other and (2) how to arrange those piles of points around circles. The number of ways to group $n$ points into $k$ equality classes is $S(n,k)$ and the number of ways to arrange those $k$ classes around $m$ circles is $c(k,m)$. We now need to see that this is a triangulation.

So, fix a partition of $[n]$ into $k$ blocks, and an arrangement of those blocks around $m$ circles. Let $U^{\circ}$ be the set of points in $Z$ with that configuration, and let $U$ be the closure of $U$. We need to see that $U^{\circ}$ is the interior of a ball and there is a continuous map from the closed ball to $Z$ extending the inclusion of $U^{\circ}$.

Choose coordinates on $U^{\circ}$ to be the angles between the blocks. Let the number of blocks on the $i$-th circle be $c_i$, so we'll write $\theta_1^i$, $\theta_2^i$, ..., $\theta_{c_i}^i$ for the coordinates coming from the $i$-th circle. So $\sum_{i=1}^m c_i = k$ and $\sum_{r=1}^{c_i} \theta_r^i = 2 \pi$ for each $i$.

Then $$U^{\circ} = \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i >0 {\Large \}}$$. Clearly, $U^{\circ}$ is the interior of the ball $$U := \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i \geq 0 {\Large \}}$$

It is also easy to build a map $U \to Z$ extending the inclusion of $U^{\circ}$. It is not quite an injection: Whenever there is an $i$ such that every $\theta^i_r$ is either $0$ or $2 \pi$, the various points which are formed by moving the position of the $2 \pi$ will be identified.

Let $X = S^1 \sqcup S^1 \sqcup S^1 \sqcup \cdots \sqcup S^1$, where there are $m$ disjoint circles. Let $Y \subset X^n$ be the set of $n$ ordered points on $X$ such that there is at least one point on each component. Note that each connected component of $Y$ is isomorphic to $(S^1)^n$. Let the group $G = (S^1)^m$ act on $X$, where the $k$-th factor of $G$ rotates the $k$-th component of $X$. So $G$ acts on $Y$. Also, let $S_m$ act by permuting the factors. Let $Z = S_m\backslash Y/G$. Each component of $Y/G$ is isomorphic to $(S^1)^{n-m}$. The $S_m$ action permutes the components, so the quotient is the union of a smaller number of copies of $(S^1)^{n-m}$. So we see that $\chi(Z)=0$ for $n-m>0$.

I now present a regular CW structure on $Z$ so that the number of $k-m$ cells is $S(n,k) c(k,m)$, proving your identity. A given cell of $Z$ will correspond to those arrangements of points on $X$ which lie in a given cyclic order. So, to specify a cell of $Z$, we need to say (1) which points are equal to each other and (2) how to arrange those piles of points around circles. The number of ways to group $n$ points into $k$ equality classes is $S(n,k)$ and the number of ways to arrange those $k$ classes around $m$ circles is $c(k,m)$. We now need to see that this is a triangulation.

So, fix a partition of $[n]$ into $k$ blocks, and an arrangement of those blocks around $m$ circles. Let $U^{\circ}$ be the set of points in $Z$ with that configuration, and let $U$ be the closure of $U$. We need to see that $U^{\circ}$ is the interior of a ball and there is a continuous map from the closed ball to $Z$ extending the inclusion of $U^{\circ}$.

Choose coordinates on $U^{\circ}$ to be the angles between the blocks. Let the number of blocks on the $i$-th circle be $c_i$, so we'll write $\theta_1^i$, $\theta_2^i$, ..., $\theta_{c_i}^i$ for the coordinates coming from the $i$-th circle. So $\sum_{i=1}^m c_i = k$ and $\sum_{r=1}^{c_i} \theta_r^i = 2 \pi$ for each $i$.

Then $$U^{\circ} = \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i >0 {\Large \}}$$. Clearly, $U^{\circ}$ is the interior of the ball $$U := \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i \geq 0 {\Large \}}$$

It is also easy to build a map $U \to Z$ extending the inclusion of $U^{\circ}$. It is not quite an injection: Whenever there is an $i$ such that every $\theta^i_r$ is either $0$ or $2 \pi$, the various points which are formed by moving the position of the $2 \pi$ will be identified.

Let $X = S^1 \sqcup S^1 \sqcup S^1 \sqcup \cdots \sqcup S^1$, where there are $m$ disjoint circles. Let $Y \subset X^n$ be the set of $n$ ordered points on $X$ such that there is at least one point on each component. Note that each connected component of $Y$ is isomorphic to $(S^1)^n$. Let the group $G = (S^1)^m$ act on $X$, where the $k$-th factor of $G$ rotates the $k$-th component of $X$. So $G$ acts on $Y$. Let $Z = Y/G$. Each connected component of $Z$ is isomorphic to $(S^1)^{n-m}$ so, if $n>m$, we have $\chi(Z) = 0$.

I now present a regular CW structure on $Z$ so that the number of $k-m$ cells is $S(n,k) c(k,m)$, proving your identity. A given cell of $Z$ will correspond to those arrangements of points on $X$ which lie in a given cyclic order. So, to specify a cell of $Z$, we need to say (1) which points are equal to each other and (2) how to arrange those piles of points around circles. The number of ways to group $n$ points into $k$ equality classes is $S(n,k)$ and the number of ways to arrange those $k$ classes around $m$ circles is $c(k,m)$. We now need to see that this is a triangulation.

So, fix a partition of $[n]$ into $k$ blocks, and an arrangement of those blocks around $m$ circles. Let $U^{\circ}$ be the set of points in $Z$ with that configuration, and let $U$ be the closure of $U$. We need to see that $(U, U^{\circ})$ is a $k-m$ ball and its interior.

Choose coordinates on $U$ to be the angles between the blocks. Let the number of blocks on the $i$-th circle be $c_i$, so we'll write $\theta_1^i$, $\theta_2^i$, ..., $\theta_{c_i}^i$ for the coordinates coming from the $i$-th circle. So $\sum_{i=1}^m c_i = k$ and $\sum_{r=1}^{c_i} \theta_r^i = 2 \pi$ for each $i$.

Then $$U = \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i \geq 0 {\Large \}}$$ $$U^{\circ} = \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i >0 {\Large \}}$$.

I leave it to you to see that this is a $k-m$ ball and its interior.

Let $X = S^1 \sqcup S^1 \sqcup S^1 \sqcup \cdots \sqcup S^1$, where there are $m$ disjoint circles. Let $Y \subset X^n$ be the set of $n$ ordered points on $X$ such that there is at least one point on each component. Note that each connected component of $Y$ is isomorphic to $(S^1)^n$. Let the group $G = (S^1)^m$ act on $X$, where the $k$-th factor of $G$ rotates the $k$-th component of $X$. So $G$ acts on $Y$. Let $Z = Y/G$. Each connected component of $Z$ is isomorphic to $(S^1)^{n-m}$ so, if $n>m$, we have $\chi(Z) = 0$.

I now present a regular CW structure on $Z$ so that the number of $k-m$ cells is $S(n,k) c(k,m)$, proving your identity. A given cell of $Z$ will correspond to those arrangements of points on $X$ which lie in a given cyclic order. So, to specify a cell of $Z$, we need to say (1) which points are equal to each other and (2) how to arrange those piles of points around circles. The number of ways to group $n$ points into $k$ equality classes is $S(n,k)$ and the number of ways to arrange those $k$ classes around $m$ circles is $c(k,m)$. We now need to see that this is a triangulation.

So, fix a partition of $[n]$ into $k$ blocks, and an arrangement of those blocks around $m$ circles. Let $U^{\circ}$ be the set of points in $Z$ with that configuration, and let $U$ be the closure of $U$. We need to see that $U^{\circ}$ is the interior of a ball and there is a continuous map from the closed ball to $Z$ extending the inclusion of $U^{\circ}$.

Choose coordinates on $U^{\circ}$ to be the angles between the blocks. Let the number of blocks on the $i$-th circle be $c_i$, so we'll write $\theta_1^i$, $\theta_2^i$, ..., $\theta_{c_i}^i$ for the coordinates coming from the $i$-th circle. So $\sum_{i=1}^m c_i = k$ and $\sum_{r=1}^{c_i} \theta_r^i = 2 \pi$ for each $i$.

Then $$U^{\circ} = \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i >0 {\Large \}}$$. Clearly, $U^{\circ}$ is the interior of the ball $$U := \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i \geq 0 {\Large \}}$$

It is also easy to build a map $U \to Z$ extending the inclusion of $U^{\circ}$. It is not quite an injection: Whenever there is an $i$ such that every $\theta^i_r$ is either $0$ or $2 \pi$, the various points which are formed by moving the position of the $2 \pi$ will be identified.