distinct multiple points in a space with at least one point lying in a subspace

Question

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?

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...