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    • CommentRowNumber1.
    • CommentAuthorTobyBartels
    • CommentTimeAug 5th 2012
    • (edited Aug 5th 2012)

    OK, this thread is for discussing whether amalgamation should redirect to pushout or to a page on model theory such as amalgamation property or needs its own page. (It is already accepted that amalgam should redirect to pushout.)

    Zoran writes:

    I find it discuttable that “amalgamation” should redirect to pushout. Amalgam should redirect to pushout, I agree. But in combinatorial group theory and other field where amalgam is used instead of pushout, I did not see occurence of “amalgamation” (maybe I did, but not commonly). Amalgamation is not amalgam. Amalgamation is a common name for several procedures in model theory, related to amalgams. Namely there is so called “higher amalgamation” of Hrushovski which is combinatorially more involved. It does not boil down to the usual pushouts, though it has some diagrams which look like involving iterations of it.

    Tim writes:

    The terminology most often encountered in my trips into (combinatorial) group theory has been usually ’free product with amalgamated subgroup’.

    (If amalgamation seems to be its own page now, then that’s the cache bug.)

    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeAug 5th 2012

    I have seen ‘free product with amalgamation’. I think that I’ve seen ‘amalgamation’ by itself, but that may be an oversimplified memory.

    (Wikipedia says ‘free product with amalgamation’, but that’s unhelpful since no citation is given. TeX confounds us by using only the abbreviation \amalg for ‘⨿\amalg’.)

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeAug 5th 2012

    We could have amalgamation entry as disambiguation, then amalgam redirecting to pushout, then amagamation property entry and finally amalgamation in model theory talking about higher amalgamation etc..

    • CommentRowNumber4.
    • CommentAuthorTobyBartels
    • CommentTimeAug 6th 2012

    Yes; although if the higher amalgamation includes pushouts as a special case (I don’t know), then that could be at just plain amalgamation.

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeAug 6th 2012

    Amalgamation in this setup does not necessarily involve pushouts, but the existence of commutative squares of certain type, not necessarily universal. As I said, it is related, not the same.

    • CommentRowNumber6.
    • CommentAuthorTobyBartels
    • CommentTimeAug 6th 2012

    OK, then I agree with your plan.