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Sep 26, 2017 at 6:55 comment added Matthias Ludewig Could somebody post this as an answer so I can accept? If anybody has a definite reference for these statements, I would greatly appreciate this!
Sep 26, 2017 at 3:06 comment added Nicholas Kuhn Of course, then there are fun examples like $X = \mathbb RP^{\infty}$ and $Y = S$ (Segal conjecture for $\mathbb Z/2$) to mess with your head! (The mapping spectrum is still $-1$-connected.)
Sep 26, 2017 at 1:00 comment added Tyler Lawson Just to tag along with Dylan's comments, if you want bounds on the nonzero groups of the function spectrum you most easily get them from bounds on the homotopy groups of Y and the cohomology groups of X. Connectivity can give you a bound on the cohomology of X but coconnectivity can't.
Sep 25, 2017 at 21:11 comment added Dylan Wilson Yeah all your statements are okay with mapping space replacing [,]. (Obligatory comment about using derived mapping spaces if you're in some model category.)
Sep 25, 2017 at 21:10 comment added Dylan Wilson The mapping spectrum will usually not be n-connective period. After all, the mapping spectrum into the sphere flips your spectrum upside down... or more readily: mapping spectra out of a sphere will give you desuspensions of the target, which are evidently less connective than they started.
Sep 25, 2017 at 21:08 comment added Matthias Ludewig Sorry, I corrected my mistake. So the mapping spectrum between $X$ and $Y$ will generally not be connective unless both $X$ and $Y$ are connective? But he equalities do hold if I consider the mapping space?
Sep 25, 2017 at 21:06 history edited Matthias Ludewig CC BY-SA 3.0
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Sep 25, 2017 at 21:04 comment added Dylan Wilson Yes, no (you've got it backwards), yes, and no (because mapping spectra know about maps from desuspensions of the source.) EDIT: Now you've got the second thing the right way round, so yes to that too.
Sep 25, 2017 at 20:57 history asked Matthias Ludewig CC BY-SA 3.0