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The unpointed version is easy: the model $X = EG \times X \to (EG \times X)/G = X^{un}_{hG}$ is a fibration with fiber $G$. But when we go pointed, $X = EG_+ \wedge X \to (EG_+ \wedge X) / G = X_{hG}$ is no longer a fibration: its fiber changes from $G$ over non-basepoints to $\ast$ over the basepoint.

Of course, these two cases are still closely related. So perhaps there's some hope of understanding the pointed case.

Questions: Let $G$ be a finite group (or perhaps something a bit more general), and let $X$ be a pointed $G$-space.

  1. Is there a good way to understand the homotopy fiber $F$ of the map $X \to X_{hG}$ (where these are pointed homotopy orbits)?

  2. In particular, does the homotopy fiber sequence $F \to X \to X_{hG}$ deloop to a homotopy fiber sequence $X \to X_{hG} \to BF$ (as it does in the unpointed case)?

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    $\begingroup$ this pointed thing $X_{*hG}$ is the cofiber of $BG \to X_{hG}$. It follows that the fiber you're looking for sits in a fiber sequence with $G$ and the fiber of $X_{hG} \to X_{hG}/BG$... but the fiber of that map is pretty gross. My guess is that there's no straightforward description of this fiber... but I'd be happy to be proven wrong! $\endgroup$ – Dylan Wilson Jun 13 '19 at 12:41
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Here is a special case which gives a partial answer:

(i). Suppose $G$ acts in a homotopically trivial way on $X$. This means that there is a trivial $G$-space $Y$ and a pair of $G$-equivariant maps $X \overset\sim\leftarrow X' \overset\sim\to Y$ each which is a weak homotopy equivalence of underlying spaces.

(ii). Assume further that $X \simeq \Sigma Z$ has the homotopy type of a suspension of a based space $Z$ (note: $Z$ does not need to have a $G$ action).

Using (i), the reduced homotopy orbits $X_{hG}$ is identified with the cofiber of the inclusion $\ast \times BG \to X\times BG$ and the latter is just $X\wedge (BG_+)$.

Using (ii), since $X$ is a suspension, $X\wedge (BG_+) \simeq X \vee (X\wedge BG)$.

Set $U = X\wedge BG$. The map $X\to X_{hG}$ with respect to this identification corresponds to the inclusion $X\to X \vee U$. So we have to identify the homotopy fiber of the latter.

It is straightforward to check that this homotopy fiber is the loop space of the homotopy fiber of the first summand projection $X\vee U \to X$. But the homotopy fiber of the latter is just $U\wedge(\Omega X_+)$. So the homotopy fiber of the map $X\to X_{hG}$ in our special case is then $$ \Omega (X\wedge BG \wedge (\Omega X_+)) \simeq \Omega ((X\wedge BG) \vee (X\wedge BG \wedge \Omega X) ) $$ Answering your question (1) in this instance.

Speculation on your Question (2): For the fiber sequence $F\to X\to X_{hG}$ to be induced by a map $X_{hG} \to BF$ in our example, you would need to know that the fiber sequence $U\wedge (\Omega X)_+ \to U \vee X \to X$ is fiber homotopically trivial. The latter is equivalent to finding a retraction to the map $U\wedge (\Omega X)_+ \to U \vee X$. Denote the latter map by $\iota$.

This seems unlikely to be true in general: a retraction to $\iota$ is given by a pair of maps $a:U\to U\wedge (\Omega X)_+$ and $b:X\to U\wedge (\Omega X)_+$ such that $\iota \circ a: U \to U\vee X$ and $\iota\circ b:X \to U \vee X$ are the summand inclusions. The map $a$ is clear, but there is no obvious candidate for $b$.

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  • $\begingroup$ Wow, thanks! If this special case is any indication, this fiber is probably fairly complicated in general. $\endgroup$ – Tim Campion Jun 13 '19 at 16:10
  • $\begingroup$ Yes, I bet it is very complicated. $\endgroup$ – John Klein Jun 13 '19 at 16:43

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