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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