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I've been reading the Xena Project blog, which has been loads of fun. In the linked post Kevin gives the natural isomorphism $V \to V^{\ast \ast}$ from a f.d. vector space to its dual as an example of a "canonical isomorphism":

People say that the obvious map from a vector space to its double dual is “canonical” but they could instead say that it is a natural transformation.

I think this is not a complete description. When we say that $V$ is canonically isomorphic to its double dual we don't just mean that there exists some natural isomorphism from the identity functor to the double dual functor - what we mean, and what we use in practice, is that a particularly canonical natural transformation $V \to V^{\ast \ast}$ is an isomorphism. We can multiply any such natural isomorphism by a scalar $c \neq 0, 1$ in the ground field $k$ and obtain a different natural isomorphism between these two functors, and these are not canonical!

But reading the linked post showed me I don't know precisely what I mean by "canonical" in the paragraph above. So:

Question: What, precisely, do we mean by "canonical" when we say that the usual natural transformation $V \to V^{\ast \ast}$ is the "canonical" one we want? What other "canonical" maps are canonical in the same way?

What follows are some off-the-cuff thoughts about this.

One thought is that when we write this map down we're "using as little as possible"; we're not even really using that we're working in vector spaces. If we define the dual to be the internal hom $[V, k]$ then ultimately all we're using is the evaluation map $V \otimes [V, k] \to k$ together with currying. In other words, one way to make precise what we're using is the closed monoidal structure on $\text{Vect}$. (We don't even need a closed structure if we define the dual to be the monoidal dual but I expect the internal hom to be more familiar to more mathematicians.)

So, here's what I think ought to be true: in the free closed monoidal category on an object $V$ (blithely assuming that such a thing exists - maybe I should have used monoidal duals after all, because I'm much more confident that the free monoidal category on a dualizable object exists), there ought to be a literally unique morphism $V \to [[V, 1], 1]$ where $1$ denotes the unit object. This is the "walking double dual map" and, by the universal property, reproduces all the other ones, and I think this is a candidate for what we might mean by "canonical map."

This trick of taking free categories is a really good trick actually, since it also cleanly describes the nonexistence of canonical maps. For example there is no canonical map $V \to V \otimes V$, and you could use representation theory to show that the only $GL(V)$-equivariant map is zero, but actually it's just true that in the free monoidal category on $V$ there are no maps $V \to V \otimes V$ whatsoever. Similarly there is no canonical map $V \to V^{\ast}$, and similarly you could use representation theory to show that the only $GL(V)$-equivariant map is zero, but actually it's just true that in the free closed monoidal category on $V$ (if it exists; take the free monoidal category with duals if not) there are no such maps whatsoever.

But I don't think this sort of reasoning is enough to capture other examples where two functors are naturally isomorphic and there's a particularly canonical natural isomorphism that we want in practice. For example, it's also true that de Rham cohomology $H^{\bullet}_{dR}(X, \mathbb{R})$ on smooth manifolds is canonically isomorphic to singular cohomology $H^{\bullet}(X, \mathbb{R})$ with real coefficients, and by this we don't mean just that there exists some natural isomorphism but that the particularly canonical natural transformation given by integrating differential forms over simplices is an isomorphism. We can multiply any such natural isomorphism by $c^n$ in degree $n$ where $c \in \mathbb{R} \setminus \{ 0, 1 \}$ and we'll get a different one, even one that respects cup products, and again these are not canonical (although I wonder if we could easily distinguish the usual one from the one obtained by setting $c = -1$ - do we have to choose how simplices are oriented or something like that somewhere?) But now I really don't know what I mean by that! It doesn't seem like I can pull the same trick of zooming out to a more general categorical picture. de Rham cohomology is a pretty specific functor defined in a pretty specific way. Maybe this one is a genuinely different sense of "canonical," closer to "preferred," I don't know.

Some previous discussion of "canonical" stuff on MO:

Qiaochu Yuan
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    Here's a fun sort-of-related question I was thinking about: does the point, as a $0$-dimensional manifold, have a canonical orientation? – Qiaochu Yuan Sep 13 '20 at 05:38
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    I think this question, though closed, is probably also relevant; https://mathoverflow.net/questions/345136/is-a-vector-space-naturally-isomorphic-to-its-dual#comment863420_345136 – user44191 Sep 13 '20 at 06:19
  • -1 f.t.a.'f.d.' – bof Sep 13 '20 at 14:55
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    A point does not have a canonical orientation, nothing stops the global choice of $+1$ or $-1$, and I think this hits to the heart of your question. There are "preferred choices" of what is to be the "canonical" object and it depends on context (I see this is also stated in one of your MO links). If I want to know the total number of points in my hand, the canonical sign is $+1$ based on the context "minus means 'taking away'". – Chris Gerig Sep 13 '20 at 17:16
  • Perhaps folks can clarify the definition of orientation they have in mind. The tangent space to a point is 0-dimensional, and a 0-dimensional vector space has a unique ordered basis, so in this sense there is a canonical orientation. – Julian Rosen Sep 13 '20 at 17:29
  • @Julian: that definition would give a unique orientation, which is not right. You can define an orientation to be a choice of generator for $H^n(-, \mathbb{Z})$ and then something funny happens when $n=0$ - this group is arguably canonically isomorphic to $\mathbb{Z}$ so has a distinguished generator, namely $1$ (distinguished abstractly by being the multiplicative unit). – Qiaochu Yuan Sep 13 '20 at 17:34
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    Possibly also related: https://mathoverflow.net/questions/244131/nuances-regarding-naturality – Steven Landsburg Sep 13 '20 at 18:09
  • To me, the essence of the "canonicity" that we care about is the fact that the construction is "defined the same way for every object", and this is something that the concept of a natural transformation nicely packages if you look at it the right way. – Alex Mine Sep 13 '20 at 18:28
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    @Alex: I think the standard analysis of the double dual isomorphism $V \to V^{\ast \ast}$ does not capture the sense in which it is "defined the same way for every object." When we talk about it as a natural transformation we don't capture e.g. that we are using the same map over every field, in a way which is compatible with extension of scalars. On the other hand the free category argument I used implies this. Note that this is a strictly stronger statement with concrete mathematical consequences: e.g. it implies that the double dual map is compatible with twisting by automorphisms of $k$. – Qiaochu Yuan Sep 13 '20 at 18:32
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    A card-carrying structuralist confessed that she secretly believed in sets. The priest replied, "Your sins are forgiven. To be honest, I secretly believe in urelements." Okay, dumb joke, but I'm having trouble seeing why this isn't a purely philosophical question. Does any substantive mathematical question hinge on it? – Timothy Chow Sep 13 '20 at 19:42
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    @Timothy: most concretely I am hoping that there is some nice category-theoretic thing to say here that is strictly stronger than just that there is some natural isomorphism of functors. For example, as I mentioned in the previous comment, the double dual map $V \to V^{\ast \ast}$ is compatible with extension of scalars, and this has concrete mathematical substance (e.g. it would be relevant to understanding the interaction of taking the double dual with Galois descent). There is at least one more kind of functoriality to incorporate, in the underlying field $k$ itself. – Qiaochu Yuan Sep 13 '20 at 20:21
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    @QiaochuYuan : I guess what I mean is, if when trying to solve a problem, you have a good mathematical reason to distinguish between different "natural isomorphisms," then the context should suggest what categories to consider. But it sounds like you're trying to pre-emptively pin down the word "canonical" before a practical need arises. That sounds to me like premature theorizing. Or maybe I'm still not understanding the motivation behind your question? – Timothy Chow Sep 13 '20 at 20:34
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    @Timothy: the linked Xena project blog post gives some context. Kevin gives the example that Grothendieck freely calls two different localizations $R[1/f], R[1/g]$ "equal" if they are canonically isomorphic, e.g. if $g^2 = f^3$, and describes some trouble he ran into trying to formalize arguments about localizations in Lean involving showing that a bunch of diagrams commuted. That example has me curious about the whole "canonical isomorphism" thing as a whole, although I agree that this is perhaps zooming out too much. – Qiaochu Yuan Sep 13 '20 at 20:39
  • There is this beautiful text https://perso.univ-rennes1.fr/matthieu.romagny/notes/canonicite.pdf by Mathieu Romagny which shows that there is no functorial isomorphism between a space and its double dual, which might be relevant. – Filippo Alberto Edoardo Sep 15 '20 at 09:26
  • Regarding the second paragraph after the fold, it's definitely true that the free closed monoidal category on a category exists. – Todd Trimble Nov 01 '20 at 23:56
  • @FilippoAlbertoEdoardo It looks to me that Romagny is talking about something else, about single duals. – Todd Trimble Nov 02 '20 at 00:05
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    A silly remark : there are canonical maps that aren't canonical isomorphisms; and whether or not they are isomorphisms depends on the base (for instance $2: M\to M$ is certainly defined canonically). Now if you're working over a field of char. $\neq 2$, would you say "the canonical isomorphism $2: M\to M$", or "the canonical map $2: M\to M$, which is an isomorphism" ? – Maxime Ramzi Nov 02 '20 at 13:10

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