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    • CommentRowNumber1.
    • CommentAuthorColin Tan
    • CommentTimeJan 15th 2019

    Added the contents of the canonical isomorphism induced by some non-canonical isomorphism as coming from Lack’s proof.

    diff, v32, current

    • CommentRowNumber2.
    • CommentAuthorColin Tan
    • CommentTimeApr 1st 2019

    Added an explicit formula for the natural map from coproducts to products.

    diff, v33, current

    • CommentRowNumber3.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 6th 2019

    I consider that formula, particularly the usage of \sqcup (or rather that big chunky \coprod I see), an abuse of notation. We have a coproduct functor -\sqcup - and that takes a pair of morphisms f:abf: a \to b and cdc \to d to fg:acbdf \sqcup g: a \sqcup c \to b \sqcup d. If instead you have a pair of maps f:acf: a \to c and g:bcg: b \to c and you want a formula for the induced map abca \sqcup b \to c, then the correct formula for it would be the composite

    abfgcc cca \sqcup b \stackrel{f \sqcup g}{\to} c \sqcup c \stackrel{\nabla_c}{\to} c

    where c\nabla_c is the codiagonal map. Thus not to be notated as fg:abcf \sqcup g: a \sqcup b \to c.

    Note that this construction is dual to the following construction which takes given maps f:ab,g:acf: a \to b, g: a \to c to

    aΔ aa×af×gb×ca \stackrel{\Delta_a}{\to} a \times a \stackrel{f \times g}{\to} b \times c

    which you denote as (f,g):ab×c(f, g): a \to b \times c. Morally, it seems “unfair” to favor the product with a nice snappy notation like (f,g)(f, g) (the notation you use) while not doing the same, or otherwise spelling out a cumbersome formula, for the dual construction on the coproduct side. My own habit has been to use something like (f,g):abc(f, g): a \sqcup b \to c as notation for the coproduct construction and f,g:ab×c\langle f, g \rangle: a \to b \times c for the product construction and not play favorites. I’d be just as happy with [f,g]:abc[f, g]: a \sqcup b \to c for the coproduct.

    Following those conventions, a formula for the map c 1c 2c 1×c 2c_1 \sqcup c_2 \to c_1 \times c_2 might be

    [Id c 1,0 1,2,0 2,1,Id c 2][\langle Id_{c_1}, 0_{1, 2}\rangle, \langle 0_{2, 1}, Id_{c_2}\rangle]

    or, dually,

    [Id c 1,0 2,1],[0 1,2,Id c 2]\langle [Id_{c_1}, 0_{2, 1}], [0_{1, 2}, Id_{c_2}] \rangle

    (or some such).

    • CommentRowNumber4.
    • CommentAuthorDavidRoberts
    • CommentTimeApr 7th 2019

    Perhaps ironically, I would use parentheses for maps to products (like vector notation for functions with Euclidean codomain) and angle brackets for maps out of a coproduct (which I’ve seen used before, but I can’t recall where).

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 7th 2019

    David, that way would be fine with me too!

    • CommentRowNumber6.
    • CommentAuthorDavidRoberts
    • CommentTimeApr 7th 2019

    And yes, like Todd, I strongly disagree with the notation fgf \coprod g for maps out of a coproduct. Just like one shouldn’t write f×gf\times g for a map to a product.

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTimeApr 7th 2019

    Me three. I use (f,g)(f,g) for map into a product (analogously to the standard notation for elements of a cartesian product set, which it generalizes if you identify elements with maps out of 1) and usually [f,g][f,g] for maps out of a coproduct, calling them “pairing” and “copairing” respectively.

    • CommentRowNumber8.
    • CommentAuthorDavidRoberts
    • CommentTimeApr 7th 2019

    I also like the version Mike has for copairing, FWIW.

    • CommentRowNumber9.
    • CommentAuthorDmitri Pavlov
    • CommentTimeApr 8th 2019

    I also disagree with f∐g and I like [f,g].

    Additionally, ∐ (\coprod) instead of ⊔ (\sqcup) is a wrong symbol to use for binary coproducts, just like ⨁ (\bigoplus) instead of ⊕ (\oplus) is a wrong symbol for binary direct sums.

    • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeApr 8th 2019

    Note that in addition to \coprod (\coprod) and \sqcup (\sqcup) there is also ⨿\amalg (\amalg). I’ve never been clear on whether there is an intended semantic distinction between \sqcup and ⨿\amalg, but both seem to be sized as binary operators.

    • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 8th 2019
    • (edited Apr 8th 2019)

    Following this discussion, and unless there are serious objections, some time soon I am going to start making edits at a number of pages, getting around to biproduct after clearing away some underbrush elsewhere. I’ve made a start at product, and following Dmitri’s observation, I would like to edit away in coproduct the big chunky \coprod as a binary operator, replacing it with \sqcup.

    (Personally, in my own work I prefer to use \sum over \coprod, and ++ over \sqcup, but it would be close to impossible to get universal agreement on such things. However, some mention of these alternatives should be made.)

    I would guess that people like ’\amalg’ for pushout notation like Y⨿ XZY \amalg_X Z. Again, my habit is usually to write Y+ XZY +_X Z.

    • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeApr 8th 2019

    Thanks Todd!

    I use ++ sometimes too, but particularly in extensive categories where the coproduct really is a “disjoint union” I often find \sqcup or ⨿\amalg easier for me personally to parse, and I also sometimes find that a ++ isn’t quite strong enough to comfortably support subscripts as used for pushouts (particularly when the subscript gets longer than a single letter). And when both a category and its internal type theory are in play, I sometimes find it’s useful to distinguish between \coprod for the external coproducts and \sum for the internal Σ\Sigma-types.

    • CommentRowNumber13.
    • CommentAuthorDmitri Pavlov
    • CommentTimeMar 31st 2020

    Added a pointless definition.

    diff, v34, current

  1. I have added several more elementary and perhaps more enlightening examples. I think the example that was there was meant as an entry for the more general context of “ambidexterity”, and it somehow ended up being the only example for this very elementary concept.


    diff, v38, current

    • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeJun 29th 2021

    added more hyperlinks to keywords in these examples (Ab, derived category, stable homotopy category, exact functor)

    diff, v39, current

  2. add link to Karvonen’s page

    Antonin Delpeuch

    diff, v42, current

    • CommentRowNumber17.
    • CommentAuthorJ-B Vienney
    • CommentTimeNov 24th 2022
    • (edited Nov 24th 2022)

    Put examples of categories with binary products and coproducts which are not equal and don’t give biproducts.

    diff, v45, current

    • CommentRowNumber18.
    • CommentAuthorJ-B Vienney
    • CommentTimeNov 24th 2022

    Added Rel as an example of category with biproducts.

    diff, v45, current

    • CommentRowNumber19.
    • CommentAuthorJ-B Vienney
    • CommentTimeNov 24th 2022

    Added the example of sup-lattices

    diff, v46, current

    • CommentRowNumber20.
    • CommentAuthorJ-B Vienney
    • CommentTimeFeb 5th 2023

    Added the usual definition in an additive category

    diff, v47, current

    • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeFeb 5th 2023

    turned the three Definition-sections into subsections of a single Definition-section

    diff, v48, current

    • CommentRowNumber22.
    • CommentAuthormattecapu
    • CommentTimeAug 25th 2023

    Added remark that semiadditive categories are (co)cartesian objects in the 2-category of (co)cartesian categories and (co)product-preserving functors.

    diff, v49, current

    • CommentRowNumber23.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 23rd 2024

    Add citations to generalizations of Houston’s result

    diff, v52, current

    • CommentRowNumber24.
    • CommentAuthorJ-B Vienney
    • CommentTimeMar 6th 2024

    Added the definition of a biproduct of n3n \ge 3 objects in an additive category which requires one more equation than in the case n=2n=2 (namely i k;p l=0i_k;p_l=0 if klk \neq l, which is automatically verified for n=2n=2).

    diff, v53, current