Suppose I have an $H$-space $H$ and a topological group $G$, such that for *compact* spaces $X$ there is a natural equivalence of group valued functors


$[X, H] \to [X, G]$


Now $H$ is a (non-finite) CW-complex (in my case it is $BU\_{\otimes}$), but $G$ is some really huge not even locally compact space. Can I somehow deduce from this, that $H$ and $G$ are weakly homotopy equivalent? 


This would of course follow, if I had a map $H \to G$ inducing the natural equivalence, right? But I only have the natural transformation above. Can I deduce the existence of a map $H \to G$ from this somehow? 



Weak homotopy equivalence of $H$-spaces
If $H$ and $G$ are spaces having abelian fundamental groups (for all basepoints), and if there is a natural isomorphism between the sets $[X,H]$ and $[X,G]$ for finite CW complexes $X$, then $H$ and $G$ are weakly homotopy equivalent.


Proof: Wlog $H$ and $G$ are CW complexes. $H$ is the direct limit of all of its finite subcomplexes $H\_i$. Inductively find maps $H\_i\to G$ inducing the given natural maps $[X,H\_i]\to [X,G]$ for finite CW $X$, choosing the maps to be compatible so that they yield a map $H\to G$. This map induces the given bijection from $[X,H]=lim\_i[X,H\_i]$ to $[X,G]$. 


It's not always true that a map $H\to G$ is a weak homotopy equivalence if it induces bijections on $[X,-]$ for finite CW (or compact) spaces, but it is true if the fundamental groups are abelian.


