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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?

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Is there an argument involving inverse limits? For example, if $H$ is the union of an increasing sequence of compact subspaces $X_i$, does $[H,G]$ surject onto $\lim_{\leftarrow}[X_i,G]$? –  Mark Grant Jun 12 '11 at 1:22

1 Answer 1

up vote 8 down vote accepted

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.

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