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Let $F(M,k)=\{(x_1,\cdots,x_k)\mid x_1\cdots,x_k\in M,x_i\neq x_j, \text{ for } i\neq j \}$. It is known that $F(\mathbb{R}^\infty,k)$ is contractible for each $k$.

My question: is $F(S^\infty,k)$ contractible for each $k$? For what kind of space $X$ can we have $F(X,k)$ is contractible for each $k$?

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    $\begingroup$ $F(M,k)$ is contractible if $M$ is a contractible infinite-dimensional manifold. That's not an if and only if statement, but for many purposes it's close to one. $\endgroup$
    – Ryan Budney
    Commented Mar 3, 2015 at 5:04

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The space $S^\infty$ is actually homeomorphic to $\mathbb{R}^\infty$. To see this, put \begin{align*} B_n &= \{x\in\mathbb{R}^\infty\::\: \|x\|\leq n, x_k = 0\text{ for } k \geq n\} \\ C_n &= \{x\in S^\infty\::\: x_n \leq 1/2,\; , x_k = 0\text{ for } k > n\}. \end{align*} Then $\mathbb{R}^\infty$ is the colimit of the spaces $B_n$, and $S^\infty$ is the colimit of the spaces $C_n$. Moreover, $B_n$ and $C_n$ are both homeomorphic to the $n$-ball, and they are contained in the interiors of $B_{n+1}$ and $C_{n+1}$ respectively. One can cook up compatible homeomorphisms $f_n\:B_n\to C_n$, and pass to the colimit to obtain a homeomorphism $\mathbb{R}^\infty\to S^\infty$.

One can prove along similar lines that various other standard contractible spaces are also homeomorphic to $\mathbb{R}^\infty$, for example $\mathbb{R}^\infty\setminus A$ (for any finite subset $A$), or $F(\mathbb{R}^\infty\setminus A,n)$, or the Stiefel manifold $V_n(\mathbb{R}^\infty)$. However, the linear isometry space $L(\mathbb{R}^\infty,\mathbb{R}^\infty)$ is homeomorphic to $\prod_{i=0}^\infty\mathbb{R}^\infty$, which is different.

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I believe each of these arguments will work.

Argument 1: Consider $S^n \subset \Bbb R^{n+1} \subset S^{n+1}$, where the last inclusion is given by the upper hemisphere (which is homeomorphic to $\Bbb R^{n+1}$. This gives rise to a string of inclusions

$$ \cdots \subset F(S^n,k) \subset F( \Bbb R^{n+1},k) \subset F(S^{n+1};k) \subset F(\Bbb R^{n+2},k)\subset \cdots $$ The colimit this system is identified with both $F(S^\infty,k)$ and $F(\Bbb R^\infty,k)$.

Argument 2: The projection onto the last coordinate defines a Hurewicz fibration $$ F(S^\infty,k) \to S^\infty $$ whose fiber is identified with $F(\Bbb R^\infty,k)$. But $S^\infty$ is contractible, which shows that the fibration is trivializable. In particular, one has a homotopy equivalence $$ F(S^\infty;k) \simeq F(\Bbb R^\infty,k)\times S^\infty\, . $$ Now use the fact that $F(\Bbb R^\infty,k)$ is contractible.

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  • $\begingroup$ In Argument 1, the inclusions $\mathbb{R}^n\to\mathbb{R}^{n+1}$ you obtain are not the standard inclusions, and so it is not obvious that their colimit is $\mathbb{R}^\infty$. I believe it still is, though I haven't checked all the details (I think you can identify the $\mathbb{R}^n$ appearing in your direct system with the ball of radius $n$ around the origin inside the standard $\mathbb{R}^n\subset\mathbb{R}^\infty$). $\endgroup$
    – Eric Wofsey
    Commented Mar 3, 2015 at 7:41
  • $\begingroup$ You're correct of course. But I think you can replace the euclidean space with $D^n_+$ (the upper hemisphere). Then $F(D^n_+,k)$ becomes weakly contractible as $n$-tends to infinity by a general position argument. Argument 1 will at then at least imply that $F(S^\infty,k)$ is weakly contractible. $\endgroup$
    – John Klein
    Commented Mar 3, 2015 at 21:40

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