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Tue, 23 Jul 2019
Obtuse axiomatization of category theory
About ten years ago I started an article, addressed to my younger self, reviewing various books in category theory. I doubt I will ever publish this. But it contained a long, plaintive digression about Categories, Allegories by Peter Freyd and Andre Scedrov:
In light of my capsule summary of category theory, can you figure out what is going on here? Even if you already know what is supposed to be going on you may not be able to make much sense of this. What to make of the axiom that !!□(xy) = □(x(□y))!!, for example? The explanation is that Freyd has presented a version of category theory in which the objects are missing. Since every object !!X!! in a category is associated with a unique identity morphism !!{\text{id}}_X!! from !!X!! to itself, Freyd has identified each object with its identity morphism. If !!x:C\to D!!, then !!□x!! is !!{\text{id}}_C!! and !!x□!! is !!{\text{id}}_D!!. The axiom !!(□x)□ = □x!! is true because both sides are equal to !!{\text{id}}_C!!. Still, why phrase it this way? And what about that !!□(x(□y))!! thing? I guessed it was a mere technical device, similar to the one that we can use to reduce five axioms of group theory to three. Normally, one defines a group to have an identity element !!e!! such that !!ex=xe=x!! for all !!x!!, and each element !!x!! has an inverse !!x^{-1}!! such that !!xx^{-1} = x^{-1}x = e!!. But if you are trying to be clever, you can observe that it is sufficient for there to be a left identity and a left inverse:
We no longer require !!xe=x!! or !!xx^{-1}=e!!, but it turns out that you can prove these anyway, from what is left. The fact that you can discard two of the axioms is mildly interesting, but of very little practical value in group theory. I thought that probably the !!□(x(□y))!! thing was some similar bit of “cleverness”, and that perhaps by adopting this one axiom Freyd was able to reduce his list of axioms. For example, from that mysterious fourth axiom !!□(xy) = □(x(□y))!! you can conclude that !!xy!! is defined if and only if !!x(□y)!! is, and therefore, by the first axiom, that !!x□ = □y!! if and only if !!x□ = □(□y)!!, so that !!□y = □(□y)!!. So perhaps the phrasing of the axiom was chosen to allow Freyd to dispense with an additional axiom stating that !!□y = □(□y)!!. Today I tinkered with it a little bit and decided I think not. Freyd has: $$\begin{align} xy \text{ is defined if and only if } x□ & = □y \tag{1} \\ (□x)□ & = □x \tag{2} \\ (□x)x & = x \tag{3} \\ □(xy) & = □(x(□y)) \tag{4} \end{align} $$ and their duals. Also composition is associative, which I will elide. In place of 4, let's try this much more straightforward axiom: $$ □(xy) = □x\tag{$4\star$} $$ I can now show that !!1, 2, 3, 4\star!! together imply !!4!!. First, a lemma: !!□(□x) = □x!!. Axiom !!3!! say !!(□x)x = x!!, so therefore !!□((□x)x) = □x!!. By !!4\star!!, the left-hand side reduces to !!□(□x)!!, and we are done. Now I want to show !!4!!, that !!□(xy) = □(x(□y))!!. Before I can even discuss the question I first need to show that !!x(□y)!! is defined whenever !!xy!! is; that is, whenever !!x□ = □y!!. But by the lemma, !!□y=□(□y)!!, so !!x□ = □(□y)!!, which is just what we needed. At this point, !!4\star!! implies !!4!! directly: both sides of !!4!! have the form !!□(xz)!!, and !!4\star!! tells us that both are equal to !!□x!!. Conversely, !!4!! implies !!4\star!!. So why didn't Freyd use !!4\star!! instead of !!4!!? I emailed him to ask, but he's 83 so I may not get an answer. Also, knowing Freyd, there's a decent chance I won't understand the answer if I do get one. My plaintive review of this book continued:
Apparently some people like this book. I don't know why, and perhaps I never will. [Other articles in category /math] permanent link |