Timeline for What is $TC(\Sigma^\infty \Omega X)$?

10 events

when toggle format	what		by	license	comment
Feb 13, 2019 at 7:49	history	edited	Dan Petersen	CC BY-SA 4.0	deleted 7 characters in body
Dec 2, 2009 at 18:45	comment	added	Reid Barton		Yes, my thoughts exactly (that's the behavior I want).
Dec 2, 2009 at 18:30	comment	added	Tyler Lawson		Right, that makes sense and I see where you're going now. I believe you're correct and that mapping object does map to $TC$. It seems, however, like the mapping object is p-equivalent to $X$ itself and $TC$ is p-local, so this map may just factor through evaluation at the basepoint?
Dec 2, 2009 at 17:18	comment	added	Reid Barton		Oops, where that error message is was just supposed to be $(\Lambda X)^C \to \Lambda X$.
Dec 2, 2009 at 17:17	comment	added	Reid Barton		Thanks, this is a great explanation. For my "space-level $TC$", I was imagining using the $c$th power map $(\Lambda X)^C → \Lambda X$ for $R$ (which is an equivalence of spaces), so $TR(\Omega X, p)$ would just be $\Lambda X$ with $F$ acting as the $p$th power map. Then $TC(\Omega X, p)$ would be $\mathrm{Map}(B\mathbb{Z}[1/p], X)$, I think. That maps to $\Omega^{\infty} TC(\Sigma^\infty \Omega X, p)$, right?
Dec 2, 2009 at 3:46	vote	accept	Reid Barton
Dec 2, 2009 at 3:46	vote	accept	Reid Barton
Dec 2, 2009 at 3:46
Dec 2, 2009 at 3:43	history	edited	Tyler Lawson	CC BY-SA 2.5	dumb mistake
Dec 2, 2009 at 3:08	history	edited	Tyler Lawson	CC BY-SA 2.5	elaboration
Dec 1, 2009 at 5:46	history	answered	Tyler Lawson	CC BY-SA 2.5