Drying categories

by Fred Ross Last updated: February 27, 2009

Category theory is rife with commutative diagrams, such as the "naturality square",


These diagrams make me vaguely uncomfortable. I don't know what to do with them. I asked myself how to make them look like simple things such as commutative (a\\cdot b = b\\cdot a) or associative (a \\cdot (b \\cdot c) = (a \\cdot b) \\cdot c) laws; about which John Baez commented, "Alas, 1-dimensional representations of commutative diagrams are sort of like dried roses..." So here is my recipe for drying commutative diagrams:

First, I give function application and composition its own symbol. E.W. Dijkstra used (.), but I'm going to generate an infinite stack of such symbols in a moment, so we'll call it (\\overset{1}{.}). I'm vaguely distressed by using the same symbol for composing morphisms and applying them to objects, but consideration of the type of its arguments makes it unambiguous. Diagram (1) above is "composition is associative" for morphisms f, g acting on object x,

(f\\, \\overset{1}{.}\\, g)\\; \\overset{1}{.}\\; x = f\\; \\overset{1}{.}\\; (g\\, \\overset{1}{.}\\, x)

Along with existence of identity morphisms, that defines a category. Applying a functor Fto either object or morphism gets the symbol (\\overset{2}{.}). A functor Fis defined by

F\\; \\overset{2}{.}\\; (f\\, \\overset{1}{.}\\, x) = (F\\, \\overset{2}{.}\\, f)\\; \\overset{1}{.}\\; (F\\, \\overset{2}\\,{.} x)

The requirement of mapping identities to identities follows from this with fset to an identity element. Functors preserve associativity of functions (proof: apply the equation above to everything in sight in F\\; \\overset{2}{.}\\; (f\\, \\overset{1}{.}\\, g\\, \\overset{1}{.}\\, x)) and identity morphisms (proof: (F\\, \\overset{2}{.}\\, id) \\; \\overset{1}{.} \\; (F\\, \\overset{2}{.}\\, x) = F \\;\\overset{2}{.} \\; (id\\, \\overset{1}{.}\\, x) = F\\, \\overset{2}{.}\\, x).

Natural transformations are maps between functors, and I write (\\overset{3}{.}) for applying them. It's convenient to say that (\\overset{n}{.}) has a higher precedence than (\\overset{m}{.}) when n > m. A natural transformation Q obeys

Q\\; \\overset{3}{.} (F\\, \\overset{2}{.}\\, k) = Q\\, \\overset{3}{.}\\, F\\: \\overset{2}{.} \\: k

where k is object or morphism. This reproduces the naturality square (proof: apply Q to F\\; \\overset{2}{.}\\; (f\\, \\overset{1}{.}\\, x)).

I can continue and define

H\\; \\overset{4}{.} (Q\\, \\overset{3}{.}\\, F) = H\\, \\overset{3}{.}\\, Q\\: \\overset{2}{.} \\: F

Z\\; \\overset{5}{.} (H\\, \\overset{4}{.}\\, Q) = Z\\, \\overset{5}{.}\\, H\\: \\overset{4}{.} \\: Q

and onwards. At any stage, the algebra continues to look like functors and natural transformations, though on more and more involved categories.

Fred Ross
February 2009
Lausanne, Switzerland