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Jun 30, 2014 at 20:43 comment added Ronnie Brown I don not know details of the Joyal-Tierney model of $3$-types, but crossed squares as models of $3$-types were published by Loday in 1982, and were used in a paper of mine and Nick Gilbert, published in 1988, and in which the model was related to automorphisms of crossed modules. Advantages of crossed squares are: they are a strict model; are values of a homotopically defined functor; colimits are commonly directly related to geometry and algebra; they extend in a natural way to all dimensions; the concept applies to other algebraic structures than groups, e.g. Lie algebras.
Jun 30, 2014 at 20:33 comment added Ronnie Brown I agree with Todd $(\infty,1)$-groupoids are not really algebraic, and neither are $k$-invariants. As models of $2$-types, crossed modules (over groupoids) seem pretty good, as their limits and colimits, where the latter correspond to many gluings of $2$-types, are good for examples and calculations. These gluings are also related to the convenient compositions which you get in cubical theories. These (partial) compositions are often neglected, but seem to me more "algebraic" than the Kan condition.
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Jun 14, 2014 at 16:52
Jun 14, 2014 at 3:38 comment added user62675 @ToddTrimble Thanks for the detailed explanation. I was quite interested in a generalization of the HH; thanks for giving me an interesting idea to pursue!
Jun 13, 2014 at 17:20 comment added Todd Trimble @HiroLeeTanaka I definitely agree with you, as your remark is very much in line with the homotopy type theory philosophy that equations between operations can be replaced by contractible spaces of paths between operations.
Jun 13, 2014 at 16:08 comment added Hiro Lee Tanaka ... A Kan complex generalizes this to a collection of things (simplices) with operations (face and degeneracy maps) satisfying some property. I think the whole point of homotopical algebra is that a "property" (like inverses existing, associativity) formerly written via equalities are replaced by properties expressed via existence of homotopies. So I think Kan complexes fit exactly the bill of the philosophy in Philippe Gaucher's comment: Representing homotopy types purely algebraically.
Jun 13, 2014 at 16:05 comment added Hiro Lee Tanaka Re: "Kan complexes... are not, strictly speaking, algebraic." I couldn't agree more, if the word "strictly" means exactly the "strictness" of equations. But if we grant ourselves the philosophy that algebra does not need to be strict, I rather think Kan complexes are precisely the algebraic replacement of spaces: A group is one set together with with an operation satisfying some properties, likewise a groupoid is two collections of things (objects and homs) with an operation satisfying some properties. ...
Jun 13, 2014 at 14:32 history answered Todd Trimble CC BY-SA 3.0