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Coming off of this discussion, I'm wondering what the term "canonical" really means. In that thread, many suggested category theory as a way to formalize the concept of what "canonical" means, using the precise term "natural" (and, many suggested that the two were not the same thing). Beyond its formal equivalent in category theory, the word natural seems to have nothing other than an intuitive or even "theological" meaning.

However, I was wondering if there is some way to define the notion of canonical by using formal logic.

Here's my own idea: After all, when we choose something we have to use some sort of logical procedure. The notion of canonicity then might mean that there actually exists some logical way to pick out a particular element, morphism, etc. When there's no canonical choice, it might mean that there is no logical way to pick out one choice over another.

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David Corwin
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    Might, but not always. – Will Sawin Jun 04 '12 at 20:54
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    There are two obviously logical ways of picking a shoe from a pair, neither seems canonical to me... – François G. Dorais Jun 04 '12 at 21:20
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    I am more interested in the canonical definition of formal logic. – Sniper Clown Jun 04 '12 at 21:55
  • As far as i am concerned, a way to get an object from another, is 'canonical' if it is given by an "application" between the two class of object (application is between brackets only because thats between class and not between set and hence you have to have a precise notion off "class" to give a precise meaning to this definition). Natural mean that this application can be made into a functor. – Simon Henry Jun 04 '12 at 22:33
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    François makes an excellent point. To push it further: suppose I hand you a set X with two distinguished elements, x and y, and I ask you to choose an element of X. Depending on how you want to use words, you might say that there is no canonical choice, or that there are two canonical choices. But I guess everyone would agree that even if the rules "pick x" and "pick y" aren't completely canonical, they are in some sense more canonical than "pick an arbitrary element of X". – Tom Leinster Jun 05 '12 at 13:42
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    @Tom: Okay, I think that is getting into why I think we can get around Francois's point. YES, we can go and pick one, but there is no logical way to distinguish the two. Here's a better example: there is no canonical root of $x^2+1$ (but when we talk about $i$, we pretend there is one). In other terms, there is really no logical way to pick one over the other. You can pretend, i.e. use the fact that there exist two of them, and then just say "let $i$ be one of them." But there is actual no logical way to specify which one we are talking about. – David Corwin Jun 14 '12 at 02:37
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    Another way to describe this: There is no surefire way to tell the two apart. Imagine you're handed the two roots of the equation. It's like being handed two identical balls. Let's say that the person who handed to you knows which is which (say, for example, they have different names). Then you have no way of telling! – David Corwin Jun 14 '12 at 02:47
  • Here's another way to explain it: If we are given a set with a distinguished point, then yes, there is a canonical choice, even if, in a sense, we're forcing that choice. If we are given a set containing x and y, then by labeling them, we are making them distinguished.

    Thus, in Francois's example with the shoes, unless we're told which shoe is left and which is right (which is just like having distinguished points), there's no actual way of distinguishing the two shoes. Maybe "logical way" or "distinguishing" isn't the right phrase?

     – David Corwin Jun 15 '12 at 05:42
  • I am not a logician, but I agree with @DavidCorwin here. It's clear that there are lots of logical ways to pick one shoe over the other, but that seems to be missing the point. It's like working with objects versus pointed objects. However, this seems like a tacitly illogical state of affairs, in that by asking to pick one "over" the other, you're assigning value in some way, which seems a little non-mathematical. – Jonathan Beardsley Aug 24 '14 at 19:33
  • @François: Right. But if choosing the right shoe is not right, then you are left with choosing the left shoe, and that's just not right. Therefore the right shoe is the right choice, and therefore the canonical choice! – Asaf Karagila Oct 23 '14 at 21:35
  • As Gauss said, philosophy oscillates between banality and nonsense--there is never nothing in between. – Włodzimierz Holsztyński Oct 23 '14 at 21:39
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    An algorithmic perspective on the question might be interesting: If we think of a choice as being provided by an algorithm, then if there is an obviously simplest/least complex algorithm, it would make the canonical choice. In the case of two shoes, there are two equally simple algorithms. In the case of a finite vector space and its double dual, the identity isomorphism is the simplest to construct. – usul Oct 25 '14 at 03:04
  • Usul's idea is somewhat similar to what I had in mind. – David Corwin Nov 07 '14 at 02:48

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It might be of interest to recall how Bourbaki uses the word "canonical" -- though admittedly this is far from formal logic. When a notion is first defined, for instance the homomorphism from a group to a quotient group, Bourbaki says "this homomorphism is called canonical". This allows him to talk about "the canonical homomorphism from $G$ to $G/H$" without any ambiguity whenever this situation occurs.

So he does not define the term canonical, but only certain canonical maps, sets, objects ...

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The term 'canonical' is very general and any attempt to create a "precise notion" may be questioned and can never be proved as such. However, there is a canonical candidate, which at least concerns canonical morphisms for constructs and other structured sets, which has a clear logical character.

First, every mathematical structure on a set is determined by relations between sets. For group structures there is a primary relation $r$ that is a function $G\times G \overset {r}\longrightarrow G$, $(x,y)\underline{r} z \Leftrightarrow z=x\centerdot y$ and some secondary relations which are conditions on $r$ (associativity, unit element and inverses). For a topological space the primary relation could be $2^X \overset{r}\longrightarrow X$, where $2^X$ is the set of all subsets of $X$ and $M\underline{r}x$ is the relation $x\in \bar{M}$ (the closure of $M$).

Whenever the main relation of a mathematical structure is given on the form $F(X)\rightarrow X$, for a functor $F$ in the category Rel (where sets are objects and binary relations are morphisms) which maps morphisms $X\overset {f}\longrightarrow Y$ on $F(X)\overset {F(f)}\longrightarrow F(Y)$, it is possible to define morphisms between the structures as (in general non commuting) diagrams: $\require{AMScd}$ \begin{CD} F(X) @>F(f)>> F(Y)\\ @VrV V @VVsV\\ X @>>f> Y \end{CD} such that

(1) $\quad \rho\underline{F(f)}\sigma \Rightarrow (\rho\underline{r}x\Rightarrow\sigma\underline{s}f(x))$.

This condition on $f$ gives the canonical morphisms to every construct that provide canonical morphisms.

Example: If $F$ is the (contravariant) functor defined as $2^X\overset{2^f}\longrightarrow 2^Y$, where $M\underline{2^f}M'\Leftrightarrow M=f^{-1}(M')$ and $r,s$ are defined as above, then due to (1):

$M=f^{-1}(M')\Rightarrow (x\in \bar{M}\Rightarrow f(x)\in \bar{M}')$, so $x\in \overline{f^{-1}(M')}\Rightarrow f(x)\in \bar{M}'$. (Continuity).

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