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12:52 AM
Let $f : A \to B$ be a based fibration. Suppose $\Omega f$ has a section $\Gamma$. How do I show that $\mu \circ (\Gamma \times \Omega i) : \Omega B \times \Omega \mathsf{fib}(f) \to \Omega A$, where $i$ is the inclusion of the fiber and $\mu$ loop composition, is an equivalence?

I tried constructing an inverse $(f \times 0) \circ \Delta$ but to show that this is a section it seems that I need to also know that $\Gamma$ is a retraction. I don't think I have the right map but I can't think of any other one.
Assume we are working in some higher topos.
I've seen this claimed in a few places, most recently here: sciencedirect.com/science/article/pii/S0166864108003647
The author there cites the following: [2] B. Eckmann, P.J. Hilton, Operators and cooperators in homotopy theory, Math. Ann. 141 (1960) 1–21. I don't have good access to this article so it would be very helpful if somebody could point me to a solution!
 
 
3 hours later…
4:00 AM
@AliCaglayan it’s a map of objects over \Omega B which is an equivalence on each fiber so it’s an equivalence.
 
4:15 AM
@DylanWilson Ahhh I see it now of course! Thank you for your help!
 
 
2 hours later…
6:20 AM
can you do paragraphs
apparently not without copy/paste sorry. just testing.
 
 
3 hours later…
9:43 AM
@DylanWilson Yes, I think you're right. :) To be honest, I was hoping that such an $n$ can be fixed for all ring spectra and perhaps that an appropriate "triangulated category of modules" is already defined for $A_{n}$-algebras, even in the case they don't lift to an $A_{\infty}$-algebra. Surely, this is too much to ask..?
 

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