Let $X$ be a topological space and $A$ a subspace of $X$. Given $k\geq 2$, let the unordered configuration space be $$ B(X,k)=\{(x_1,x_2,\cdots,x_k)\in X^k\mid x_i\neq x_j \text{ for any } i\neq j\} $$ and the relative unordered configuration space be $$ B(X,A;k)=\{(x_1,x_2,\cdots,x_k)\in X^k\mid x_i\neq x_j \text{ for any } i\neq j,\text{ and } x_t\in A\text{ for some } 1\leq t\leq k\}. $$

Question: I want to express $B(X,A;k)$ in terms of $B(Y,i)$ 's for some space $Y$.

I have tried: For $j=1,2,\cdots, k$, first choose $j$ points in $A$, then choose $k-j$ points in $X\setminus A$. Then we have $$ B(X,A;k)=\bigsqcup_{j=1}^k B(A,j)\times B(X\setminus A,k-j). $$ However, this expression seems wrong since it is a non-connected disjoint union... Where is the problem?

  • $\begingroup$ “Where is the problem?” You have said the problem yourself: this is disconnected (whenever $k > 1$ and both $A$ and $X \setminus A$ are non-empty), while $B(X,A;k)$ will be connected whenever $X$ is. What exactly do you mean by “expressible in terms of”, though, and why do you expect this should be possible? One option would be to take the image of the map from your disjoint union into $B(X,k)$, if this fits your goal of “expressible in terms of”. $\endgroup$ Commented Sep 15, 2015 at 13:56

1 Answer 1


You have to map $B(A,j)\times B(X\setminus A,k-j)$ to $B(X,A;k)$ by the "union" map $\phi:2^X\times 2^X \to 2^X$ defined as $\phi(C,D)=C\cup D$. Also define $B(Y,n)$ for $n\ge 0$, not just $n\ge 2$. Then $B(X,A;k)=\cup_{j=1}^k \phi(B(A,j)\times B(X\setminus A,k-j))=\cup_{j=1}^k\{\{x_1,\cdots,x_k\}:\{x_1,\cdots,x_j\}\in B(A,j),\{x_{j+1},\cdots,x_k\}\in B(X\setminus A,k-j)\}$

There's no topology here, only sets...

  • $\begingroup$ Thanks, Prof. Duchon! I want to know $H^*(B(X,A;k)$. If I decompose in such a way, how could I study the cohomology? $\endgroup$
    – QSR
    Commented Sep 15, 2015 at 14:34
  • $\begingroup$ That's another question... $\endgroup$ Commented Sep 15, 2015 at 15:07

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