Let $\xi=(E,p,B)$ and $\omega=(EG,\pi,BG)$ be principal $G$-bundles. Further, let $\omega$ be universal. According to Husemoller that means $[-,BG]\rightarrow k_G(-)$ is a natural isomorphy, where $k_G(X)$ are the isomorphy classes of principal $G$-bundles over $X$. If $\xi$ is universal, it is clear that $B$ and $BG$ are homotopy equivalent.

In contrast to this, tom Dieck defines universality in terms of the total spaces, i.e. $EG \rightarrow BG$ is defined as terminal object in the appropriate homotopy category of principal $G$-bundles. This means that $\omega$ is universal if for all $\xi$ there is a principal $G$-bundle map $E\rightarrow EG$, unique up to homotopy.

Does the first definition imply the second and are they equivalent?

  • $\begingroup$ In your first paragraph, what does $\xi$ have to do with the rest of what you write? Why is it clear that $B$ and $BG$ are homotopy equivalent. Also, it seems that the only difference between Husemoller and tom Dieck is that the first gives a bijection $[-,BG]\rightarrow k_G(-)$ and the second gives a bijection $k_G(-)\rightarrow [-,BG]$. $\endgroup$ – Jason DeVito Sep 4 '17 at 18:21
  • $\begingroup$ Sorry, in my first version $\xi$ was meant to be universal, it is corrected now. My question is: how do we get a map $E \rightarrow EG$, unique up to homotopy, from the first definition? In the end, I would like to be able to prove that maps between universal total spaces are homotopy equivalences without considering CW-complexes and weak equivalence. $\endgroup$ – Max Power Sep 4 '17 at 18:31
  • $\begingroup$ I'm not sure if this is answering your question..., but, Given a bundle $\xi$, by Husemoller's definition, there is a (homotopy class of ) map $f:B\rightarrow BG$ for which $f^\ast(EG)\cong E$. Note that $f^\ast(EG)$ is, by definition, a subset of $EG\times B$, so projection onto the first factor gives a map $f^\ast(EG)\rightarrow EG$. Then the composition $E\cong f^\ast(EG)\rightarrow EG$ is the map you want, I think. $\endgroup$ – Jason DeVito Sep 5 '17 at 2:11
  • $\begingroup$ Yes, but is it unique over f up to homotopy? If $f,g $ are homotopic, then one can construct these maps you've mentioned. But is there a unique homotopy between them over the homotopy between $f$ and $g$? $\endgroup$ – Max Power Sep 5 '17 at 6:28
  • $\begingroup$ Sorry for being slow - I understand your question now! (Unfortunately, I do not know the answer. I mean, they must be equivalent because at the end of the day, $EG$ ends up being contractible. But I don't know how to prove it using Husemoller's definition.) $\endgroup$ – Jason DeVito Sep 5 '17 at 13:40

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