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  1. Page created, but author did not leave any comments.

    Anonymous

    v1, current

  2. added info about the circle type

    Anonymous

    diff, v5, current

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJun 8th 2022

    fixed some grammar and touched the wording in the Definition-section

    diff, v6, current

  3. Adding link to Jordan curve

    Anonymous

    diff, v8, current

  4. Adding coequalizer definition of the circle type

    Anonymous

    diff, v9, current

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeJan 30th 2023

    tried to bring some logical order back into the list of references.

    diff, v11, current

    • CommentRowNumber7.
    • CommentAuthorGuest
    • CommentTimeMar 1st 2023

    This page says:

    Its induction principle says that for any P:S 1TypeP:S^1\to Type equipped with a point base:P(base)base' : P(base) and a dependent path loop:base=baseloop':base'= base', there is f: (x:S 1)P(x)f:\prod_{(x:S^1)} P(x) such that:

    Should it instead say loop:tr P loop(base)=baseloop' : tr^{loop}_P(base') = base', or loop:base= P loopbaseloop' : base' =^{loop}_{P} base', or “dependent path loop:base=baseloop':base'= base' over looploop”, or something like that? Does it matter?

    Adrian

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeMar 2nd 2023
    • (edited Mar 2nd 2023)

    I think that’s right, there is a transport involved. That’s what the link dependent path is referring to. For the moment i have added (here) the missing = loop=_{loop}-subscript and a pointer to UFP13, p. 177.

    Ideally I would like to polish up the whole presentation (which is due to the notorious “Anonymous”, in revision 5) but not now.

    diff, v12, current

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeMar 2nd 2023

    On whether it matters: I expected it does. While I haven’t tried to write out a formal argument, a quick idea goes as follows:

    In the higher induction principle of the suspension type SS 0\mathrm{S}S^0 there is certainly such dependent identifications involved, namely along the two “meridian” paths from the “north pole” to the “south pole”: by the general rules of higher inductive types, but also because here it would not even type-check otherwise. But then for the evident map S 1SS 0S^1 \to \mathrm{S}S^0 to be an equivalence, there must be a corresponding dependent identification also in the induction principle of S 1S^1.

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