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Are infinite dimensional simplicial complexes manifolds locally modeled on $\mathbb R^\infty=\operatorname{colim}\mathbb R^n$? If they are homotopy equivalent, are they homeomorphic?

Of course not. Part of such a complex might not be infinite dimensional: glue an interval to $\mathbb R^\infty$ by one endpoint and it stops being homogeneous. Or take two copies of $\mathbb R^\infty$ and glue them at a point and that point is special. Also, for $\alpha$ an uncountable cardinal, $\mathbb R^\alpha$, topologized as the colimit of the finite dimensional subspaces, is not locally isomorphic to $\mathbb R^\infty$ because every open subset contains an uncountable discrete subset, namely an appropriate choice of basis. But are those three obstacles the only ones?

If a simplicial complex is countable, every link* is contractible, and every simplex is the face of another, does it follow that it is locally homeomorphic to $\mathbb R^\infty$?

Is it PL isomorphic? What if we generalize to CW complexes? This leads to a second class of question. If those hypotheses are sufficient for local homeomorphism with $\mathbb R^\infty$, then apply them to the open cone on a contractible space satisfying the hypotheses. The space is now homeomorphic to $\mathbb R^\infty$. Was it already homeomorphic before applying the cone? If two such $\mathbb R^\infty$ manifolds are homotopy equivalent, are they, in fact, homeomorphic?

Is this topic in the literature? What about the special case of the product of a finite simplicial complex with $\mathbb R^\infty$? What about the product with a finite CW complex?

* The homotopy link of a point $x$ in a space $X$ is the (pro?) homotopy type of the homotopy inverse limit of the deleted neighborhoods $U-\{x\}$. This is a homeomorphism invariant of a space. Asking for the local homology $H_*(X,X-\{x\})$ to vanish is not quite sufficient for the link to be contractible because of the possibility that it is not simply connected. In a simplicial complex, the link is a specific simplicial complex, namely the union of all simplicies that are opposite $x$ in some simplex; and its PL isomorphism type is a PL invariant of the space.

Here is a potential counterexample to the PL statement. Take a closed manifold with the homology of a sphere but with nontrivial fundamental group, such as the Poincaré homology 3-sphere. The Cannon-Edwards theorem says that the double suspension is homeomorphic but not PL isomorphic to a sphere. Equivalently, the double cone is homeomorphic but not PL isomorphic to a disk. It is not PL isomorphic because a link of the link of the cone point is the homology sphere, not a real sphere. Now cross the double cone with $\mathbb R^\infty$. The result is homeomorphic to $\mathbb R^\infty$, but is it PL isomorphic? Is the cone point special? Is there some way of extracting the original fundamental group from the space, or has it been pushed off "to infinity"?

Motivation:

Is there an abelian group structure on $\mathbb C\mathbb P^\infty$? There are a couple of other models for the classifying space of $S^1$ that do have abelian group structures, such as the bar construction $BS^1$ and $FS^2$ the connected component of the free abelian group on $S^2$; we’d like to transfer them over to our favorite model. I think that we managed to cobble together a proof that $BS^1$ is PL isomorphic to $\mathbb C\mathbb P^\infty$, but what about $FS^2$? For many groups there is a model for the classifying space as the union of manifolds, but the filtration of the bar construction is only by manifolds if the group is a sphere, as fails for, eg, $\mathbb Z/3$ and $SO(3)$. But even though each $B_n\mathbb Z/3$ fails to be homogeneous, the full $B\mathbb Z/3$ is a group, thus homogeneous. So, in some sense it is a manifold. How many types of manifolds are there among infinite dimensional simplicial complexes? Perhaps they are all locally $\mathbb R^\infty$? And what about $BSO(3)$? It is neither a group nor filtered by manifolds, so it has no reason to be homogeneous. But I think it is.

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    $\begingroup$ In my answer to your earlier question I sketch a proof that $EG$ is homeomorphic (and hence $BG$ is locally homeomorphic) to $S^\infty$ for any nontrivial finite $G$. The strategy there is to show that every finite subcomplex is contained in some subcomplex homeomorphic to a PL ball, and in such a way that these balls are nested in a nice way. $\endgroup$ – Tom Goodwillie Apr 6 '16 at 12:21
  • $\begingroup$ One could also ask such a question about (realizations of) simplical sets. These are not precisely PL, right? but are a good deal more rigid than CW complexes. $\endgroup$ – Tom Goodwillie Apr 6 '16 at 12:23
  • $\begingroup$ I was going to say don't we need to assume something about skeleta being finite. But of course that condition makes no sense until you fix a cell structure; and while $S^\infty$ can be taken to have $n$-skeleton $S^n$ it is in fact homeomorphic to $\mathbb R^\infty$, which at first glance might appear to need infinitely cells in each dimension. $\endgroup$ – Tom Goodwillie Apr 6 '16 at 12:35
  • $\begingroup$ Is it clear that the link of a simplex in a PL space is well defined up to PL homeomorphism? It's certainly a standard result in the locally finite case, but in general? $\endgroup$ – Tom Goodwillie Apr 6 '16 at 12:44
  • $\begingroup$ You need not assume that every simplex is a proper face of another. This says that the link of a simplex is not empty, which follows from the link of a point being contractible. $\endgroup$ – Tom Goodwillie Apr 6 '16 at 13:43
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This question concerns manifolds modeled on the direct limit $\mathbb R^\infty$ of Euclidean spaces. The theory of such manifolds is well-developed. The most important results of this theory were obtained by Katsuro Sakai and can be found in Chapter 5 of his book. Let us mention here two results of this theory implying affirmative answers to both (topological) questions posed by Ben Wieland.

Characterization Theorem 5.4.1. A topological space $X$ is is homeomorphic to an open subspace of $\mathbb R^\infty$ if and only if any embedding $f:B\to X$ of a closed subspace $B$ of a finite-dimensional compact metrizable space $A$ can be extended to an embedding $\bar f:A\to X$.

Classification Theorem 5.5.1. Two $\mathbb R^\infty$-manifolds are homeomorphic if and only if they are homotopically equivalent.

In Sections 5.6 and 5.7 of Sakai's book one can find some results on PL-theory of $\mathbb R^\infty$-manifolds.

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