# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• 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.

• CommentRowNumber2.
• CommentAuthorColin Tan
• CommentTimeApr 1st 2019

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

• 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 $f: a \to b$ 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).

• 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 $f \coprod g$ for maps out of a coproduct. Just like one shouldn’t write $f\times g$ for a map to a product.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeApr 7th 2019

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.

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

• 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 $\Sigma$-types.

• CommentRowNumber13.
• CommentAuthorDmitri Pavlov
• CommentTimeMar 31st 2020

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.

Anonymous

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeJun 29th 2021

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