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1. • CommentRowNumber1.
   • CommentAuthorColin Tan
   • CommentTimeJan 15th 2019
   • PermaLink
   Author: Colin Tan
   Format: MarkdownItexAdded the contents of the canonical isomorphism induced by some non-canonical isomorphism as coming from Lack's proof. <a href="https://ncatlab.org/nlab/revision/diff/biproduct/32">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/32">v32</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

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

   diff, v32, current

2. • CommentRowNumber2.
   • CommentAuthorColin Tan
   • CommentTimeApr 1st 2019
   • PermaLink
   Author: Colin Tan
   Format: MarkdownItexAdded an explicit formula for the natural map from coproducts to products. <a href="https://ncatlab.org/nlab/revision/diff/biproduct/33">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/33">v33</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

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

   diff, v33, current

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeApr 6th 2019
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexI 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).

   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).

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeApr 7th 2019
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexPerhaps 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).

   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).

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeApr 7th 2019
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexDavid, that way would be fine with me too!

   David, that way would be fine with me too!

6. • CommentRowNumber6.
   • CommentAuthorDavidRoberts
   • CommentTimeApr 7th 2019
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexAnd 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.

   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.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeApr 7th 2019
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexMe 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.

   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.

8. • CommentRowNumber8.
   • CommentAuthorDavidRoberts
   • CommentTimeApr 7th 2019
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexI also like the version Mike has for copairing, FWIW.

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

9. • CommentRowNumber9.
   • CommentAuthorDmitri Pavlov
   • CommentTimeApr 8th 2019
   • PermaLink
   Author: Dmitri Pavlov
   Format: MarkdownItexI 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 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.

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeApr 8th 2019
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexNote 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.

    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.

11. • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 8th 2019
    • (edited Apr 8th 2019)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexFollowing 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 $Y +_X Z$.

    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 $Y +_X Z$.

12. • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeApr 8th 2019
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexThanks 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.

    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.

13. • CommentRowNumber13.
    • CommentAuthorDmitri Pavlov
    • CommentTimeMar 31st 2020
    • PermaLink
    Author: Dmitri Pavlov
    Format: MarkdownItexAdded a pointless definition. <a href="https://ncatlab.org/nlab/revision/diff/biproduct/34">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/34">v34</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    Added a pointless definition.

    diff, v34, current

14. • CommentRowNumber14.
    • CommentAuthornLab edit announcer
    • CommentTimeJun 29th 2021
    • PermaLink
    Author: nLab edit announcer
    Format: MarkdownItexI 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 <a href="https://ncatlab.org/nlab/revision/diff/biproduct/38">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/38">v38</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    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

    diff, v38, current

15. • CommentRowNumber15.
    • CommentAuthorUrs
    • CommentTimeJun 29th 2021
    • PermaLink
    Author: Urs
    Format: MarkdownItexadded more hyperlinks to keywords in these examples (*[[Ab]]*, *[[derived category]]*, *[[stable homotopy category]]*, *[[exact functor]]*) <a href="https://ncatlab.org/nlab/revision/diff/biproduct/39">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/39">v39</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

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

    diff, v39, current

16. • CommentRowNumber16.
    • CommentAuthornLab edit announcer
    • CommentTimeJun 18th 2022
    • PermaLink
    Author: nLab edit announcer
    Format: MarkdownItexadd link to Karvonen's page Antonin Delpeuch <a href="https://ncatlab.org/nlab/revision/diff/biproduct/42">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/42">v42</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    add link to Karvonen’s page

    Antonin Delpeuch

    diff, v42, current

17. • CommentRowNumber17.
    • CommentAuthorJ-B Vienney
    • CommentTimeNov 24th 2022
    • (edited Nov 24th 2022)
    • PermaLink
    Author: J-B Vienney
    Format: MarkdownItexPut examples of categories with binary products and coproducts which are not equal and don't give biproducts. <a href="https://ncatlab.org/nlab/revision/diff/biproduct/45">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/45">v45</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

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

    diff, v45, current

18. • CommentRowNumber18.
    • CommentAuthorJ-B Vienney
    • CommentTimeNov 24th 2022
    • PermaLink
    Author: J-B Vienney
    Format: MarkdownItexAdded [[Rel]] as an example of category with biproducts. <a href="https://ncatlab.org/nlab/revision/diff/biproduct/45">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/45">v45</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    Added Rel as an example of category with biproducts.

    diff, v45, current

19. • CommentRowNumber19.
    • CommentAuthorJ-B Vienney
    • CommentTimeNov 24th 2022
    • PermaLink
    Author: J-B Vienney
    Format: MarkdownItexAdded the example of sup-lattices <a href="https://ncatlab.org/nlab/revision/diff/biproduct/46">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/46">v46</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    Added the example of sup-lattices

    diff, v46, current

20. • CommentRowNumber20.
    • CommentAuthorJ-B Vienney
    • CommentTimeFeb 5th 2023
    • PermaLink
    Author: J-B Vienney
    Format: MarkdownItexAdded the usual definition in an additive category <a href="https://ncatlab.org/nlab/revision/diff/biproduct/47">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/47">v47</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    Added the usual definition in an additive category

    diff, v47, current

21. • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeFeb 5th 2023
    • PermaLink
    Author: Urs
    Format: MarkdownItexturned the three Definition-sections into subsections of a single Definition-section <a href="https://ncatlab.org/nlab/revision/diff/biproduct/48">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/48">v48</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

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

    diff, v48, current

22. • CommentRowNumber22.
    • CommentAuthormattecapu
    • CommentTimeAug 25th 2023
    • PermaLink
    Author: mattecapu
    Format: MarkdownItexAdded remark that semiadditive categories are (co)cartesian objects in the 2-category of (co)cartesian categories and (co)product-preserving functors. <a href="https://ncatlab.org/nlab/revision/diff/biproduct/49">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/49">v49</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    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

23. • CommentRowNumber23.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 23rd 2024
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexAdd citations to generalizations of Houston's result <a href="https://ncatlab.org/nlab/revision/diff/biproduct/52">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/52">v52</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    Add citations to generalizations of Houston’s result

    diff, v52, current

24. • CommentRowNumber24.
    • CommentAuthorJ-B Vienney
    • CommentTimeMar 6th 2024
    • PermaLink
    Author: J-B Vienney
    Format: MarkdownItexAdded 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 $n=2$). <a href="https://ncatlab.org/nlab/revision/diff/biproduct/53">diff</a>, <a href="https://ncatlab.org/nlab/revision/biproduct/53">v53</a>, <a href="https://ncatlab.org/nlab/show/biproduct">current</a>

    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 $n=2$).

    diff, v53, current