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In the book Fibre Bundles by Husemoller, universal G-bundles are introduced as bundles over a homotopy type $BG$, for which the cofunctor $[-,BG]\rightarrow k_G(-)$ is a natural isomorphism.

Contrary to this, tom Dieck defines universal G-bundles as those for which the total space EG is terminal for some homotopy category, i.e. for all numberable free G-spaces $E$ there is a G-map $E\rightarrow EG$, unique up to G-homotopy.

How does the first definition imply the second? My aim is to get a unique homotopy type to show that any total space of a universal bundle as defined in the first way is contractible without restricting to CW complexes.

I have already unsuccessfully asked this question here https://math.stackexchange.com/questions/2416706/total-spaces-of-universal-principal-g-bundles

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    $\begingroup$ How are you not satisfied with the discussion in Section 14.4 of Tom Dieck's Algebraic Topology book? This explains in detail exactly the relationship between these two perspectives. (Note that in Husemoller's book $k_G(-)$ is isomorphism classes of numerable $G$-bundles). $\endgroup$
    – Chris Schommer-Pries
    Commented Sep 6, 2017 at 15:14
  • $\begingroup$ @ChrisSchommer-Pries. Thanks for your answer. Could you specify where? In 14.4 I can only find the opposite direction, an argumentation yielding in Theorem 14.4.1. $\endgroup$
    – Max Power
    Commented Sep 6, 2017 at 15:32
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    $\begingroup$ The next theorem (thm 14.4.2) shows that there exists a bundle EG over a space BG satisfying the second defn. Hence by thm 14.1, this BG is also universal in the sense of the first definition. By the Yoneda lemma, any numerable spaces BG and BG' satisfying the first defn are homotopic. Thus if you have a space BG satisfy the first defn, then it is homotopic to the BG constructed in Thm 14.4.2, and hence also satisfies the second defn. $\endgroup$
    – Chris Schommer-Pries
    Commented Sep 6, 2017 at 15:57
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    $\begingroup$ Note also that if BG satisfies the first definition, then $id \in [BG,BG]$ corresponds to the bundle $EG \in k_G(BG)$. So you have the bundle in the first definition too. $\endgroup$
    – Chris Schommer-Pries
    Commented Sep 6, 2017 at 15:58
  • $\begingroup$ @ChrisSchommer-Pries I think I see my problem now: The Milnor construction shows that there are contractible $G$-spaces, hence the terminal objects in the appropriate category are contractible and therefore all maps into them are homotopic. $\endgroup$
    – Max Power
    Commented Sep 6, 2017 at 16:27

1 Answer 1

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There is a subtlety here that I think the classical literature doesn't deal with well. Consider the following three statements for a principle $G$--bundle $E \rightarrow B$: (a) $E$ is a terminal object in the homotopy category of (nice) free $G$--spaces. (b) $E \rightarrow B$ is the universal $G$--bundle. (c) $E$ is contractible. Then (a) $\Leftrightarrow$ (b), and (c) $\Rightarrow$ (b), by basic bundle theory as in tom Dieck's book.

It is also true that (b) $\Rightarrow$ (c), and the standard proof is as above: first one shows (e.g. using the Milnor construction) that there IS a universal bundle with contractible total space, and since universal bundles are unique, you are done. But this proof has always seemed like cheating to me, so here is a direct proof that (a) $\Rightarrow$ (c):

Observe that $E$ satisfies (a) means that $[Y,E]_G$ is a singleton for all free $G$--spaces $Y$. In particular, this is true when $Y=G \times X$, so we learn (since $[X,E] = [G \times X,E]_G$) that $[X,E]$ is a singleton for all spaces $X$. But this implies that $E$ is contractible.

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  • $\begingroup$ Thanks, but I'm still missing some point. Let $E \rightarrow G$ and $EG \rightarrow BG$ be universal. Then, I see that $id_{BG}$ corresponds to equivalences $[f] \in [B,BG]$ and $[g] \in [BG,B]$, which correspond under the functors $[B,BG] \rightarrow k_G(B)$ (resp. $[BG,B]$) to $f^*(EG)$ (resp. $g^*(E)$). So it is clear that $E$ and $EG$ correspond to each other under some bijection. But where does the homotopy equivalence between them emerge from? $\endgroup$
    – Max Power
    Commented Sep 7, 2017 at 15:10

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