# Deeper meaning of “canonical”?

For years I thought "canonical" for isomorphisms just means something like "obvious", "without arbitrarily choice", "simple" depending on the context, and I accepted the idea that there is no deeper meaning of "canonical" and no clear definition.

Reading about categories I noticed that "canonical" is sometimes used when there's a "natural equivalence" in the language of category theory. So my question:

Can we clearly define "canonical" by means of category theory in an abstract way, so that the situations when we use "canonical" in our "real" mathematician life are special cases of this abstract situation?

• Wikipedia discusses this here and here – Ben Grossmann Jan 12 '17 at 14:15
• Thanks for the link, especially the first is a good one. – user302982 Jan 12 '17 at 14:19
• "Canonical" is not a mathematical word, "natural" is (at least in the context of category theory). – Najib Idrissi Jan 12 '17 at 15:56
• The Wikipedia discussion fits with my understanding of the way "canonical" is often used by research mathematicians. Another example: The set of rational numbers has a canonical field structure. Since that set is just a countably infinite set you can make up lots of other field structures for it, but there is only one canonical one. – SixWingedSeraph Jan 12 '17 at 20:36
• I found now also a nice related post on mathoverflow mathoverflow.net/questions/19644/… – user302982 Jan 17 '17 at 12:18

"Natural" in the context of category theory means natural transformation. A "natural isomorphism" is just an isomorphism between functors, i.e. mutually inverse natural transformations. Sometimes we say something like "$\text{Hom}(FA,B) \cong \text{Hom}(A,UB)$ is natural in $A$ and $B$" by which we mean $(A,B)\mapsto\text{Hom}(FA,B)$ and $(A,B)\mapsto\text{Hom}(A,UB)$ are isomorphic as functors. While natural transformations are indeed often "natural" in the non-technical sense, there are a lot of them, and they can be quite arbitrary. For example, in the category of commutative monoids (or groups), $x \mapsto 42x$ is a natural transformation.

"Canonical" doesn't have a widely used technical meaning as far as I'm aware. (It does have some technical definitions, but the ones I'm aware of are not widely used.) The categorical notion that unambiguously comes closest to capturing the notion of a "canonical map" is the notion of a universal arrow. I'll use a variation on the presentation using a notion called a universal element. I describe the relation between these notions here. We say a functor $F$ is representable if $\text{Hom}(X,-) \cong F$ for a given object $X$. (Note, that in words this says $F$ is naturally isomorphic to a hom-functor.) We'll equivalently say that $X$ represents $F$. A universal element is simply the image of $id$ under this isomorphism, i.e. if $\varphi:\text{Hom}(X,-)\cong F$, then $\varphi(id)\in F(X)$ is the universal element. In the (very common, in fact necessary) case that $F$ itself is some sort of hom-functor, then the universal element will be an arrow in some category and that arrow is the universal arrow. An excellent exercise is to show that if $X'$ also represents $F$ then $X \cong X'$ and that isomorphism is unique.

The two examples of "canonical maps" from Omnomnomnom's second link are of this form. (Note, the other two examples are examples of "structure maps".) This may not be obvious, so let me spell it out.

The first example is the quotient group construction. Let $N$ be a normal subgroup of a group $G$. Define $F(H)\equiv\{f\in\text{Hom}(G,H)\mid \forall x\in N.f(x) = 0\}$. Then it is the case that $\text{Hom}(G/N,-)\cong F$. (Prove this. Don't forget naturality.) The universal element is a group homomorphism $q\in\text{Hom}(G,G/N)=F(G/N)$ for which $q(x)=0$ iff $x\in N$. In particular, any $f \in F(H)$ factors through $q$ uniquely.

The second example says there is a natural transformation $\lambda : V \to V^{**}$ natural in $V$ where $V$ is a vector space over the field $k$, and $V^* \equiv V \multimap k$ is the duality with $V\multimap W$ being the vector space of linear functions from $V$ to $W$. We have the following adjunction $$\text{Hom}(U,V\multimap W)\cong\text{Hom}(V,U\multimap W)$$ natural in $U$, $V$, and $W$. (Prove this.) Set $F_V(U)\equiv\text{Hom}(V,U\multimap k)=\text{Hom}(V,U^*)$. $\text{Hom}(-,V^*)\cong F_V$. (This looks slightly different from our definition of representability, but it's actually the same. Why?) The universal element is then a map $\lambda_V \in F_V(V^*) = \text{Hom}(V,V^{**})$. These arrows indexed by $V$ form the components of a natural transformation $\lambda : Id \to (-)^{**}$. In fact, this result can be proved generally, and the general result is a corollary to a result called parameterized representability.

• Thanks for some new aspects I can think about. – user302982 Jan 13 '17 at 10:07

IMO, a "canonical map" is one implied by context.

Language and notation is full of shorthand and abbreviations; it's far too cumbersome to say everything. For example, if we have a ring $R$ and an ideal $I$, the quotient is some ring $Q$ together with a surjective map $\pi : R \to Q$ whose kernel is $I$.

However, the usual language is just say that $Q$ is the quotient, without any mention of $\pi$. Then, if we later need $\pi$, we name it as the "canonical map" $R \to Q$.

I'm not aware of a theory that tries to capture the notion — your best bet on that, I think, would be to go over to cstheory.stackexchange.com and ask if there is a theory underpinning implicit conversions.