Post #1816228
2026-02-23 21:15 UTC
Replies (4)
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@antoinechambertloir@mathstodon.xyz 2026-02-23 21:17
@MartinEscardo @dwarn I like this, because it does not seem clear to me that once algebra is done within hott, every basic algebraic structure should a priori be a set. Maybe it's because higher groups are much more complicated than classical groups, but still.
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@ToucanIan@mathstodon.xyz 2026-02-25 02:11
@MartinEscardo thanks for this! I tried and failed to do the very same thing. @dwarn
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@de_Jong_Tom@mathstodon.xyz 2026-02-25 10:47
@MartinEscardo @dwarn To really understand it, I've been working on my own retelling. If I manage to complete it, I'll share it here. Thanks to both of you for sharing!
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@jakub_et_al@mathstodon.xyz 2026-02-28 15:56
@MartinEscardo @dwarn This thread forced me to finally learn some agda, and try to write some proofs. I am attempting to reprove another statement from Taylor's paper. If a topological space admits a majority operation ๐ satisfying ๐(๐ฅ,๐ฅ,๐ฆ)=๐(๐ฅ,๐ฆ,๐ฅ)=๐(๐ฆ,๐ฆ,๐ฅ)=๐ฅ, then it is weakly-equivalent to a discrete set. Naturally, for homotopy types this would translate to three equations like ๐โ : ฮ (๐ฅ,๐ฆ : ๐ด) ๐(๐ฆ,๐ฅ,๐ฅ)=๐ฅ ๐โ : ฮ (๐ฅ,๐ฆ : ๐ด) ๐(๐ฅ,๐ฆ,๐ฅ)=๐ฅ ๐โ : ฮ (๐ฅ,๐ฆ : ๐ด) ๐(๐ฅ,๐ฅ,๐ฆ)=๐ฅ I managed to make ๐ act idempotently on the loop space, but encountered an interesting problem: There are a few ways how to prove that ๐ is idempotent, and it is not clear which choice to make! I would need something like ฮ (๐ฅ:๐ด) ๐โ(๐ฅ,๐ฅ) = ๐โ(๐ฅ,๐ฅ). But maybe this is a consequence of what we mean when we write the above majority identities.