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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 7th 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 7th 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.

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