Does $\mathbb C\mathbb P^\infty$ have a (commutative) group structure? More specifically, is it homeomorphic to $FS^2$, (the connected component of) the free commutative group on $S^2$?

$\mathbb C\mathbb P^\infty$ is well known to have the homotopy type of the classifying space of a commutative group, $\mathbb C^\times$. One model for the classifying space is the bar construction, which is a functor from groups to spaces and is product preserving, so the classifying space of a commutative group is again a commutative group (since the group operation is a homomorphism iff the group is a commutative). The Dold-Thom theorem says that the component of the free commutative group on $S^n$ is a $K(\mathbb Z,n)$. So those are two fairly nice group models. But is either actually homeomorphic to the space we started with, $\mathbb C\mathbb P^\infty$?

By $\mathbb C\mathbb P^\infty$ I mean the colimit of the $\mathbb C\mathbb P^n$ under the closed inclusions. Similarly, I topologize the free commutative group on a space $X$ as the colimit of the inclusions of the subset of words of length $n$, which is topologized as a quotient of $(X\times\{\pm1\}\cup\{*\})^n$, that is, sequences of $n$ letters, each of which may be a generator, the inverse of a generator, or the identity element. So $\mathbb C\mathbb P^\infty$, $BS^1$, and $FS^2$ are CW complexes with finite skeleta. Are they all locally modeled on $\mathbb R^\infty$? Do such infinite dimensional manifolds have homeomorphism types determined by their homotopy types?

Appendix: Motivation

In the question, I invoked the Dold-Thom theorem to motivate the particular model $FS^2$ under consideration. But this case is easy and should motivate the general theorem, rather than vice versa. So I will suggest a proof. Moreover, I will motivate the choice of space, how we get from $\mathbb C\mathbb P^\infty$ to $FS^2=F\mathbb C\mathbb P^1$. We want to make a model for $B\mathbb C^\times$ that is as nice as possible. If we identify $\mathbb C^\infty=\mathbb C[x]$, it is a ring, so its multiplication makes $\mathbb C\mathbb P^\infty=\mathbb P(\mathbb C[x])$ into an topological monoid. That’s a pretty nice structure; it just lacks inverses. So we should consider $\mathbb P(\mathbb C(x))$. For any reasonable topology on $\mathbb C(x)$, the vector space is contractible, as is the complement of the origin; and the multiplication on the ring is continuous. Thus $\mathbb P(\mathbb C(x))$ is a topological monoid homotopy equivalent to $\mathbb C\mathbb P^\infty$. This is better than the previous model because every element has an inverse. However, the inversion map is continuous only for some topologies. For counterexample, the finest possible topology is given by making the topological vector space the colimit of its finite dimensional subspaces (a colimit indexed by an uncountable filtering poset). But then the inverse is not continuous: the copy of $\mathbb C$ given by $y\mapsto x-y$ has inverses a set of linearly independent vectors, thus discrete.

Turning away from topological vector spaces, how else can we put a topology on $P(\mathbb C(x))$? A polynomial is determined (up to scale) by its zeros and a rational function by its zeros and poles. Then we can think of $\mathbb P(\mathbb C[x])$ as something like the free commutative monoid on $S^2=\mathbb C\mathbb P^1$ (specifically, the free monoid modulo the point at infinity). Similarly, $\mathbb P(\mathbb C(x))$ is the component of the free commutative group on $\mathbb C\mathbb P^1$, that is, $F\mathbb C\mathbb P^1$. To put the zeros and poles on the same footing, we need the topology described from the beginning.

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