Let $G\text{-}PF(X)$ be the set of isomorphism classes of principal topological fibrations over the space $X$ with structural group $G$, and $G\text{-}PF\_{cw} : hCW \to Set$ the contravariant functor $X \mapsto G\text{-}PF(X)$ where $X$ is a CW-complex. Let also $G\text{-}PF\_n(X)$ be the set of isomorphism classes of *numerable* principal topological fibrations, and $G\text{-}PF\_{n} : hTop \to Set$ the contravariant functor $X \mapsto G\text{-}PF\_n(X)$.


Existence of a classifying space $BG$ is another way to ask for the functors $G\text{-}PF\_{cw}$ or $G\text{-}PF\_{n}$ to be representable. It is often said in books that the Brown's representability theorem shows that they are.


But following the definition in *nLab*, Brown's theorem apply only for contravariant functor from $hTop\_\*^c$, the homotopy category of *connected* *pointed* topological spaces, to $Set\_\*$, or from $hCW\_\*^c$, the homotopy category of *connected* *pointed* CW-complexes, to $Set\_\*$, and is false if the constraints *pointed* or *connected* are removed.


My question is : even if I assume that the two conditions of the theorem are satisfied by $G\text{-}PF\_{cw}$ and $G\text{-}PF\_n(X)$,
do we have, for $X$ and $Y$ CW-complexes
$[X,Y]\_{hCW\_\*^c} = [X,Y]\_{hCW}$ ?


Or for $X$ and $Y$ *numerable* spaces, does $[X,Y]\_{hTop\_\*^c} = [X,Y]\_{hTop}$ ?


I don't see why this should be true, and even think it is false, so how can we apply Brown's theorem here ?



Classifying spaces and Brown's representability theorem
**No**, homotopy classes of maps of unpointed spaces are not equivalent to homotopy classes of maps of pointed spaces. You need to be a little bit more clever. What follows is a spelling out of Brown's original argument in section 5.1 of [1]. I'm going to cover the statement for principal bundles of topological groups, because I'm not sure of what definition of principal fibration you are currently using (the definition I usually use makes this statement tautological, so I assume it's not it...). Hopefully you can adapt the argument at the setting you are interested in. It is a fairly general trick, analogous to move from a unreduced to a reduced cohomology theory.


Fix a topological group $G$. Then for any CW complex $X$, let $\mathrm{Bun}\_G(X)$ be the set of isomorphism classes of principal $G$-bundles $P\to X$. We are interested in showing that there is a CW complex $BG$ such that $\mathrm{Bun}\_G(X)=[X,BG]$, functorially in $X$ (the functoriality here is given by pullback of bundles).


It is clear that $\mathrm{Bun}\_G(X\amalg Y)=\mathrm{Bun}\_G(X)\times \mathrm{Bun}\_G(Y)$, so it suffices to prove the thesis when we restrict the functor to connected CW complexes. However we need somehow to consider functors from *pointed* connected spaces. The trick here is to consider a *different* functor. If $X$ is a pointed connected CW complex, we let $\mathrm{Bun}\_{G,0}(X)$ be the set of isomorphism classes of principal $G$-bundles $P$ over $X$ equipped with a point $p$ in the fiber above the basepoint of $X$. The isomorphisms are required to respect this basepoint.


It is easy now to see that $\mathrm{Bun}\_{G,0}(X)$ satisfies the hypotheses needed to apply Brown representability. It is classical that principal bundles over CW complexes satisfy the homotopy hypothesis and it is also quite easy to see that we can glue principal bundles above subcomplexes (possibly using that every subcomplex is a deformation retract of a neighbourhood). The basepoint becomes helpful when you need to verify the wedge axiom: there, to define the gluing, we need a chosen trivialization over the basepoint. But this is exactly what our additional structure provides. So to sum up, we have found a (pointed) connected space $BG$ such that for every pointed connected CW complex $X$
$$\mathrm{Bun}\_{G,0}(X)=[X,BG]\_\*\,.$$
Ah, but now you're going to complain that this was not what we wanted to do! We are not interested in $\mathrm{Bun}\_{G,0}(X)$, we are interested in $\mathrm{Bun}\_G(X)$! Thankfully there is a way to fix this.


First of all, let us note that $\pi\_1BG=\pi\_0G$. In fact
$$\pi\_1(BG)=[S^1,BG]\_\*=\mathrm{Bun}\_{G,0}(S^1)$$
and now it is an exercise to see that the latter group is canonically isomorphic to $\pi\_0G$ (hint: monodromy action). Moreover, since $BG$ is connected, we can conclude (by considering the fibration $\mathrm{Map}(X,BG)\to BG$ given by evaluation at the basepoint) that
$$[X,BG]=[X,BG]\_\*/\pi\_1BG=[X,BG]\_\*/\pi\_0G$$
and a careful look at the isomorphism will show that $\pi\_0G$ acts on $[X,BG]\_\*=\mathrm{Bun}\_{G,0}(X)$ by moving the chosen point.


Since $\mathrm{Bun}\_{G,0}(X)/\pi\_0G$ is exactly $\mathrm{Bun}\_G(X)$, we are done.


[1] *Brown, Edgar H.jun.*, [**Cohomology theories**](http://dx.doi.org/10.2307/1970209), Ann. Math. (2) 75, 467-484 (1962). [ZBL0101.40603](https://zbmath.org/?q=an:0101.40603).


