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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeSep 25th 2012

    Since I was being asked I briefly expanded automorphism infinity-group by adding the internal version and the HoTT syntax.

    Mike, what’s the best type theory syntax for the definition of Aut(X)\mathbf{Aut}(X) via \infty-image factorization of the name of XX?

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeSep 26th 2012

    Y:TypeIsInhab(Y=X)\sum_{Y:Type} IsInhab(Y=X), or any variant depending on your chosen notation for IsInhab and identity/path types. E.g. Y:Type[Id Type(Y,X)]\sum_{Y:Type} [Id_Type(Y,X)] would be another version.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeSep 26th 2012
    • (edited Sep 26th 2012)

    Ah, thanks. I should have been able to come up with this myself.

    Can we allow ourselves to write “ Y:Type[X=Y]\sum_{Y : Type} [X = Y]”?

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeSep 26th 2012
    • (edited Sep 26th 2012)

    I have added a general remark on this to infinity-image in a new section Syntax in homotopy type theory there.

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeSep 26th 2012

    Can we allow ourselves to write “ Y:Type[X=Y]\sum_{Y : Type} [X = Y]”?

    Sure, although I would probably mention that [][-] denotes the support (and, perhaps, == the intensional identity type / path type) whenever I start using this notation on a given page or in a given paper.

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeSep 26th 2012

    OOPS! I meant [ Y:Type(X=Y)]\left[\sum_{Y : Type} (X = Y)\right]. I’ve corrected the proof at infinity-image.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeSep 26th 2012
    • (edited Sep 26th 2012)

    Hm, I was thinking that

    im (A(id A,b)A×B)A×BB im_\infty(A \stackrel{(id_A, b)}{\to} A\times B) \to A \times B \to B

    is equivalent to

    im (AbB)B. im_\infty(A \stackrel{b}{\to} B) \to B \,.

    Hm…

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeSep 27th 2012

    That’s not even true for 0-truncated objects. In that case (1,b):AA×B(1,b):A\to A\times B is already monic, so the top composite just gives you b:ABb:A\to B rather than its image.

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeSep 27th 2012
    • (edited Sep 27th 2012)

    Ah, sorry. Right, I am being stupid.

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