Not signed in (Sign In)

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

  • Sign in using OpenID

Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

Welcome to nForum
If you want to take part in these discussions either sign in now (if you have an account), apply for one now (if you don't).
    • CommentRowNumber1.
    • CommentAuthorYimingXu
    • CommentTimeFeb 17th 2022
    • (edited Feb 17th 2022)

    Hi, All. I am reading through McLarty’s axiomatization of category of categories, and here is a theorem that I cannot understand. It might be my silly or have forgotten some basics, but it has annoyed me for several hours.

    The paper “Axiomatizing category of categories” is here

    On page 1245, Theorem 5 states:

    Let ff and gg be composable arrows of any category AA. If gg is an identity arrow then g Af=fg\circ_A f = f

    And the proof says:

    Theorem 4 plus chasing through the pushout for g Afg \circ_A f.

    I think to be able to use Theorem 4, we need the unique arrow induced by the pair f,gf,g to factorize through the one induced by id 2id_2 and 0! 20 \circ !_2. if such factorization does not exist, I do not see any chance of applying theorem 4.

    I think there is some condition that pushout arrow induced by of faf \circ a and gbg \circ b factors though the pushout arrow induced by aa and bb, and the case described in Theorem 5 of the paper satisfies the condition. If there is a condition, what is it? (I agree with the intuition that if two pairs of things have the same shape, then their pushout should have a “same fragment”, my trouble is to figure out how it formally happens.)

    Thanks for any hint and attempt to help!

    • CommentRowNumber2.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 17th 2022

    May I suggest editing your post to use markdown formatting for the long url? Like this [link text](

    • CommentRowNumber3.
    • CommentAuthorYimingXu
    • CommentTimeFeb 17th 2022

    Edited! Thanks for the suggestion!

    • CommentRowNumber4.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 18th 2022

    I think one place to start is with the universal case, namely taking the category AA to be 3\mathbf{3}.

    • CommentRowNumber5.
    • CommentAuthorYimingXu
    • CommentTimeFeb 18th 2022
    • (edited Feb 18th 2022)

    I think there are only 6 maps 232\to 3, so I think the case AA is 33 can be shown using case-by-case argument.

    It seems to me the claim is:

    If Theorem 3 holds for AA is 33, then it for all AA’s. I believe this is true if: 1. 33 is indeed universal, in the following sense. 2.Theorem 5 holds for 3.

    But it is not clear to me that the case for AA is 33 is a universal case. Expand the “sense of universal”, it means:

    For every composable pair f:2A,g:2Af:2\to A,g:2\to A such that gg factors through 11, there is a pair of composable arrows f 0:23,g 0:23f_0:2\to 3,g_0:2\to 3 such that g0g0 factors through 11, and exists arrow t 0:3At_0:3\to A such that t 0f 0=ft_0\circ f_0 = f and t 0g 0=gt_0\circ g_0 = g. I am struggling on proving this.

    Thanks for correcting me if my sense of “universal” is wrong here!

    Update: I just realised that I have not used the fact that 33 retracts to 22, I will retry tomorrow.

    • CommentRowNumber6.
    • CommentAuthorYimingXu
    • CommentTimeFeb 20th 2022

    I have convinced myself that 33 is indeed universal. Thank you very much for the useful hint!

    • CommentRowNumber7.
    • CommentAuthorDavidRoberts
    • CommentTimeFeb 21st 2022

    Excellent, glad it helped!