If $X$ and $Y$ are homotopy equivalent, then are $X \times \mathbb{R}^{\infty}$ and $Y \times \mathbb{R}^{\infty}$ homeomorphic?

Question

Let $X$ and $Y$ be reasonable spaces. Since $\mathbb{R}^{\infty}$ is contractible, $$ X \times \mathbb{R}^{\infty} \cong Y \times \mathbb{R}^{\infty} \;\;\; \implies \;\;\; X \simeq Y. $$

Is the converse also true?

My vague intuition: the factors of $\mathbb{R}^{\infty}$ provide so much extra room that there will never be a geometric obstruction to producing a homeomorphism. Evidently, there is no homotopy-theoretic obstruction, so maybe the converse is true.

On the other hand, I really have no idea and could be missing something basic. For example, the plane with two punctures is homotopy equivalent to a wedge of two circles. However, I do not know about a homeomorphism $$ (\mathbb{C} - \{0, 1\}) \times \mathbb{R}^{\infty} \overset{?}{\cong} (S^1 \vee S^1) \times \mathbb{R}^{\infty}. $$

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Clarification about the meaning of $\mathbb{R}^{\infty}$ and the intent of the question

When I wrote the question, I had in mind the infinite union $\cup_n \mathbb{R}^n$ inside the product $\mathbb{R}^{\mathbb{N}}$. However, since I would like the answer to the question to be "yes," I am also interested in other versions of $\mathbb{R}^{\infty}$.

I asked the question because of an algebraic limiting construction in a paper I'm writing, and I felt that a topological version of the limit would be satisfying. The algebraic version is already working for spaces like $X \times \mathbb{C}^n$ for large-enough $n$, and converges algebraically to some limiting group, but this doesn't give too many hints about the topology I should use on $\mathbb{C}^{\infty}$, or if the limit can even be considered topologically.

My application involves singular homology, and the direct limit topology is well-suited to this application, but other choices may be as well.

Answer 1

This question is answered by two classical theorems of infinite-dimensional topology, which can be found in the books of Bessaga and Pelczynski, Chigogidze or Sakai.

Factor Theorem. For any Polish absolute neighborhood retract $X$ (= neighborhood retract of $\mathbb R^\omega$) the product $X\times\mathbb R^\omega$ is an $\ell_2$-manifold.

Classification Theorem. Two $\ell_2$-manifolds are homeomorphic if and only if they are homotopy equivalent.

So, the reasonable space in your question should read as a Polish absolute neighborhood retract. By the way, this class of spaces includes all (countable locally) finite simplicial complexes, mentioned in the answer of Igor Rivin.

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Remark on possible generalizations. Chapter IX of the book of Bessaga and Pelczynski contains generalizations of the above two theorems to manifolds modeled on normed spaces $E$, which are homeomorphic to $E^\omega$ or $E^{\omega}_0:=\{(x_n)_{n\in\omega}\in E^\omega:\exists n\in\omega\;\forall k\ge n\;(x_k=0)\}$.

Chapter 7 of Chigogidze's book contains generalizations of the Factor and Classification Theorems to manifolds modeled on uncountable products of lines.

Chapter 5 of Sakai's book contains generalizations of the Factor and Classification Theorems to manifolds modeled on the direct limits $\mathbb R^\infty$ and $Q^\infty$ of Euclidean spaces and Hilbert cubes, respectively.

So, typical generalizations of Factor and Classification Theorems look as follows:

Theorem. Let $E$ be a reasonable model space (usually it is an infinite-dimensional locally convex space with some additional properties).

$\bullet$ For any neighborhood retract $X$ of $E$ the product $X\times E$ is an $E$-manifold.

$\bullet$ Two $E$-manifolds are homeomorphic if and only if they are homotopically equivalent.

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Depending on the meaning of your model space $\mathbb R^\infty$ the meaning of a reasonable space also changes. If $\mathbb R^\infty$ is the direct limit of Euclidean spaces, then a reasonable space means an absolute neighborhood extensors which is a direct limits of finite dimensional compacta. If $\mathbb R^\infty$ is the union of the Euclidean spaces in the countable product of lines, then a reasonable space is a locally contractible space which is the countable union of finite-dimensional compact subsets.

Answer 2

I believe the answer is YES, where reasonable = finite simplicial complex, based on the work of Chapman on Hilbert Cube manifolds. Namely, I believe that it is a 1973 definition of Chapman that a Hilbert cube manifold of "type $Q$" if it is homeomorphic to $Q \backslash A,$ where $Q$ is the Hilbert cube $[0, 1]^\infty,$ and $A$ is a closed subset of $W = \{x\in Q \left| x_1 = 1\right.\}$ Now, it seems to be a theorem (of Chapman, 1972) that a manifold is of type $Q$ iff it is of the form $|Y| \times Q,$ where $Y$ is a locally finite simplicial complex, and further, two such type $Q$ manifolds are homeomorphic if and only if they are properly homotopic (clearly a homotopy induced by the homotopy of $|Y|$ is proper).

Chapman, T.A., Contractible Hilbert cube manifolds, Proc. Am. Math. Soc. 35, 254-258 (1972). ZBL0259.58004.

Chapman, T.A., On the structure of Hilbert cube manifolds, Compos. Math. 24, 329-353 (1972). ZBL0246.57005.

Answer 3

This (Brown--Cohen, A proof that simple-homotopy equivalent polyhedra are stably homeomorphic) gives a partial answer (maybe subsumed by the answer by Igor Rivin).