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

]]>Add citations to generalizations of Houston’s result

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

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

]]>Added the usual definition in an additive category

]]>Added the example of sup-lattices

]]>Added Rel as an example of category with biproducts.

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

]]>add link to Karvonen’s page

Antonin Delpeuch

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

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.

Anonymous

]]>Added a pointless definition.

]]>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 $\Sigma$-types.

]]>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 $Y \amalg_X Z$. Again, my habit is usually to write $Y +_X Z$.

]]>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.

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.

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

]]>Me three. I use $(f,g)$ 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 $[f,g]$ for maps out of a coproduct, calling them “pairing” and “copairing” respectively.

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

]]>David, that way would be fine with me too!

]]>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).

]]>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 $f: a \to b$ and $c \to d$ to $f \sqcup g: a \sqcup c \to b \sqcup d$. If instead you have a pair of maps $f: a \to c$ and $g: b \to c$ and you want a formula for the induced map $a \sqcup b \to c$, then the correct formula for it would be the composite

$a \sqcup b \stackrel{f \sqcup g}{\to} c \sqcup c \stackrel{\nabla_c}{\to} c$where $\nabla_c$ is the codiagonal map. Thus not to be notated as $f \sqcup g: a \sqcup b \to c$.

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

$a \stackrel{\Delta_a}{\to} a \times a \stackrel{f \times g}{\to} b \times c$which you denote as $(f, g): a \to b \times c$. Morally, it seems “unfair” to favor the product with a nice snappy notation like $(f, g)$ (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 $(f, g): a \sqcup b \to c$ as notation for the coproduct construction and $\langle f, g \rangle: a \to b \times c$ for the product construction and not play favorites. I’d be just as happy with $[f, g]: a \sqcup b \to c$ for the coproduct.

Following those conventions, a formula for the map $c_1 \sqcup c_2 \to c_1 \times c_2$ might be

$[\langle Id_{c_1}, 0_{1, 2}\rangle, \langle 0_{2, 1}, Id_{c_2}\rangle]$or, dually,

$\langle [Id_{c_1}, 0_{2, 1}], [0_{1, 2}, Id_{c_2}] \rangle$(or some such).

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

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

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