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

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 $$(\mathbb{C} - \{0, 1\}) \times \mathbb{R}^{\infty} \overset{?}{\cong} (S^1 \vee S^1) \times \mathbb{R}^{\infty}.$$

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.

• (a) An earlier start is even to determine for which contractible $X$ is $X\times\mathbb{R}^\infty$ homeomorphic to $\mathbb{R}^\infty$. (b) I have several options in mind for "reasonable", maybe you could elaborate. Is every compact metrizable space "reasonable"? (c) you could have defined $\mathbb{R}^\infty$; I guess you mean the inductive limit topology on a countable increasing union rather than something like a separable Hilbert space, but I'm not sure.
– YCor
Feb 19, 2018 at 21:40
• @YCor (a) Yes, a good point (b) + (c) I am hoping that someone with expertise will be able to fill in the details by making good choices for "reasonable" and $\mathbb{R}^{\infty}$. If the statement is false, then I expect it will be false in some robust way that cannot be corrected by changing technical details. If it is true, then I expect that it will require careful consideration of the issues you raise! Feb 19, 2018 at 21:51
• Related: mathoverflow.net/questions/269413/… (see the comments) Feb 20, 2018 at 2:15
• What about $X= [0,1]^I, Y = \{pt\}$, for a very large $I$ (something like $\mathbb{R}$) ? I assume that such an $X$ won't be "reasonable" but then perhaps you should be clearer on what you call "reasonable" Feb 20, 2018 at 9:19
• @JohnWiltshire-Gordon Please explain what do you have in mind writing $\mathbb R^\infty$. Is it the countable product of lines or the direct limit of Euclidean spaces? Feb 20, 2018 at 17:18

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.

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.

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.

• The OP might have intended $\mathbb R^\infty$ to mean $\mathrm{colim}_{n\to\infty}\mathbb R^n$. I'm guessing that, in your answer, $\mathbb R^\omega$ refers to the infinite product of $\mathbb R$'s, with the product topology. Could you please comment on the difference? Feb 20, 2018 at 16:45
• The theory of manifolds modeled on direct limits of $\mathbb R^n$ also is well-developed and has similar factor and classification theorems. The author of the question should explain what he had in mind under $\mathbb R^\infty$. If necessary I can add to my answer the necessary information on $\mathbb R^\infty$-manifolds, too. Feb 20, 2018 at 16:50
• "$colim$" stands for colimit and a "direct limit" is just a "directed colimit", that is, the colimit of a system indexed by a directed set. So here he means the direct limit. Feb 20, 2018 at 17:48
• @SimoneVirili Thank you for the explanation. After reading Wikipedia I realised that colimits (in the sense of Andre Henriques) are just direct limits and have deleted my comment. But this does not explain the meaning of $\mathbb R^\infty$ in OP. Feb 20, 2018 at 17:55
• Yes, I cannot explain the meaning of $\mathbb R^{\infty}$. Colimits are a more general concept than direct limits, but yes, they are defined by the same universal property. The fact is that the "indexing category" or "indexing set" is completely general for colimits, while it has to be directed for direct limits. In this particular case, the indexing set is the natural numbers, so it is clearly directed and the two concepts coincide. (For example, a pushout is a colimit but not a direct limit) Feb 20, 2018 at 18:11

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.

• The type $Q$ manifolds considered in Chapman's paper and in your answer are contractible, so $X$ and $Y$ would also need to be contractible. Feb 20, 2018 at 5:57
• @IanAgol Yes, I was free-associating and answered Yves' more specific question, however the general question (at least for simple homotopy type) is answered in a 1974 paper of Chapman, I will add the reference when more awake. Feb 20, 2018 at 12:38
• I saw that too from the references in the paper linked in the other answer. But Chapmans paper addresses a different question, since crossing with the Hilbert cube gives something different than crossing with $R^\infty$. Feb 20, 2018 at 14:44
• @IanAgol you mean the closed hilbert cube? Feb 20, 2018 at 19:41
• Yes, the compact Hilbert cube defined in Chapman's paper that you reference. Feb 20, 2018 at 19:55

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).