Letting $EG$ be the total space of some group $G$, and $f: X \to Y$ a $G$-equivariant map with $X$ a free $G-CW$ complex, we can form the bundle $p^{\prime}:EG \times_G (X \times Y) \to EG \times_{G} X: [e,(x,y)] \to [e,x]$. The claim is that projection onto the second factor mod $G$ (i.e $\pi_2/G$) is a in fact a homotopy equivalence of bundles between $p^{\prime}$ and $p:X \times_G Y \to X/G:[x,y] \mapsto [x]$

The way I see it, there is the map $\pi_2/G:EG \times_G(X \times Y) \to X \times_G Y$ and $\pi_2/G:EG \times_G X \to X/G$ fit into a square diagram, and a bundle homotopy equivalence is saying that the classifying map $g^{\prime}$ for $p^{\prime}$ is homotopy equivalent to $(\pi_{2}/G) \circ g$, where $g$ is the classifying map for $p$.

Question $1$: Is there a better way of seeing what this means heuristically/ how does this induce a bijection of sections between the two bundles?

Question $2$: How can we prove that this is a homotopy equivalence? Since $EG$ is weakly contractible, I can see how we should be able to find some appropriate exact sequence (of fibration) to use witehead's theorem for $\pi_2/G:EG \times_G X \to X/G$ somewhere.

A reference can be found here on page 67

  • $\begingroup$ Do you have a reference for the claim? $\endgroup$ – Tyrone Nov 16 '17 at 13:20
  • $\begingroup$ @Tyrone see the edit $\endgroup$ – Andres Mejia Nov 16 '17 at 14:41
  • $\begingroup$ Do you mean that $X$ is a free $G$-complex then? $\endgroup$ – Tyrone Nov 16 '17 at 15:03
  • $\begingroup$ @Tyrone yes. My apologies. $\endgroup$ – Andres Mejia Nov 17 '17 at 5:49

Since $X$ is a free $G$-space the projection $p_X:EG\times_G X\rightarrow X/G$ is a fibration with fibre $EG$. Since $EG$ is weakly contractible this fibration is a weak equivalence, and since $X$ is a CW complex, it is in fact a homotopy equivalence. The fact that $X$ is $G$-free also means that $X\times Y$ is $G$-free, and hence the same reason applies to get a fibration $p_{X\times Y}:EG\times_G(X\times Y)\rightarrow X\times_GY$ which is also a homotopy equivalence.

Moreover the composition $EG\times_G(X\times Y)\xrightarrow{p'}EG\times_GX\xrightarrow{p_X}X/G$ factors over $p_{X\times Y}$ to give the map $p:X\times_GY\rightarrow X/G$. Putting these maps together gives a strictly commutative square diagram which is easily seen to be a pullback using the fact that the induced maps of fibres are easily verified to be homeomorphisms. The square defines a bundle map, and since $p$, $p'$ are homotopy equivalences we get that $p$, $p'$ define a homotopy equivalence of bundles.

The fact that the previously assembled square is a pullback gives you the bijection between the sections of $p$ and $p'$.

  • $\begingroup$ This is an answer that clearly deserves to be accepted, but I'm hesitant because I cannot really see the "easy" statements in the second paragraph. Also, was my notion of bundle homotopy correct? Or is it the fact that we require $p$ and $p^{\prime}$ to be homotopy equivalences that gives this? Also, your $p_x$ and is my $\pi_2/G$, correct? $\endgroup$ – Andres Mejia Nov 17 '17 at 22:50
  • $\begingroup$ Assemble the square diagram with $EG\times_G(X\times Y)$ in the top left-hand corner and $X/G$ in the bottom right-hand corner. The horizontal maps are the acyclic fibrations $p_{X\times Y}$, $p_X=\pi_2/G$. Now take a test space $E$ and maps $a:E\rightarrow X\times_GY$, $b:E\rightarrow EG\times_GX$ that are equalised by their compositions with $p'$, $p_X$, respectively. Now show that the top left-hand space $EG\times_G(X\times Y)$ satisfies the universal property of a pullback. $\endgroup$ – Tyrone Nov 18 '17 at 9:39
  • $\begingroup$ You can find a definition of a bundle homotopy (equivalence) in Rudyak's book "Thom Spectra, Orientability, and Cobordism", def. IV.1.5. To my knowledge it is not completely standard terminology. I don't think your interpretation of it is correct, though. For one thing, there is no reason that the bundles should admit classifying maps in your sense. Secondly, the bundle equivalence of the Hopf bundle $\eta:S^3\rightarrow S^2$ which is the degree $-1$ map on $S^3$ and the identity on $S^2$ gives a bundle equivalence from $-\eta$ to $\eta$, but these bundles are classified by different maps. $\endgroup$ – Tyrone Nov 18 '17 at 9:55
  • $\begingroup$ I don't have access to the book, so it would be nice if you could tell the definition you are using (I'm happy as long as it is enough to deduce a bijection between sections.) Thank you for your first comment, and for clarifying my misconception. Altogether, this was very helpful. $\endgroup$ – Andres Mejia Nov 18 '17 at 18:57
  • $\begingroup$ books.google.de/… $\endgroup$ – Tyrone Nov 18 '17 at 19:04

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