Post #1441646
2026-04-01 21:57 UTC
Is A₀ necessarily contractible? Be careful not to think I'm asking a very similar-sounding but distinct question: I know if the sequence was ordered the other way around, with A₀ = 0, A₁ = ΣA₀, A₂ = ΣA₁, etc. then the colimit of the sequence ("the infinite-dimensional sphere") would be contractible.
Replies (2)
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@jcreed@mastodon.social 2026-04-01 21:57
If it's possible for A₀ to not be trivial, then I have a hunch it ought to behave kind of like a "negative point"; I think for any function B → C I can naturally construct a map from a colimit of C many copies of A₀ to B many copies of A₀.
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@ncf@types.pl 2026-04-01 22:46
@jcreed at least yes assuming Whitehead's principle, since A₀ is n-connected for all n.