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The definitions of "natural transformation", "natural isomorphism between functors", and "natural isomorphism between objects" captures - among other things - the intuitive notion of "an isomorphism that does not depend on an arbitrary choice" (from Wikipedia). The standard example is a finite dimensional vector space $V$ being naturally isomorphic to its double dual $V$** because the isomorphism doesn't depend on the choice of basis.

What I wonder:

Does this informal notion of choice have to do with the formal notion of choice in the axiom of choice?

Is "being naturally isomorphic" somehow related to "being provably isomorphic from $ZF$ without $AC$"?

Or are these completely unrelated concepts?

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Hans-Peter Stricker
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5 Answers5

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"Naturality" in a categorical sense is much more than "not depending on choices", and, also, is essentially unrelated to issues about the Axiom of Choice.

In the example of vector spaces over a field, we can look at the category of _finite_dimensional_ vector spaces, to avoid worrying about using AxCh to find elements of the dual. The non naturality of any isomorphisms of finite-dimensional vectorspaces with their duals resides in the fact that, provably, as a not-hard exercise, there is no collection of isomorphisms $\phi_V:V\rightarrow V^*$ of isomorphisms of f.d. v.s.'s $V$ to their duals, compatible with all v.s. homs $f:V\rightarrow W$.

In contrast, the isomorphism $\phi_V:V\rightarrow V^{**}$ to the second dual, by $\phi_V(v)(\lambda)=\lambda(v)$ is compatible with all homs, as an easy exericise! This latter compatibility is the serious meaning of "naturality".

True, if capricious or random choices play a role, the chance that the outcome is natural in this sense is certainly diminished! But that aspect is not the defining property!

Edit (16 Apr '12): as alancalvitti notes, the ubiquity of adjunctions, and the naturality and sense of "naturality", and counter-examples to naive portrayals, deserve wider treatment at introductory levels. After all, this can be done with almost no serious "formal" category-theoretic overhead, and pays wonderful returns, at the very least organizing one's thinking. Distinguishing "characterization" from "construction-to-prove-existence" is related. E.g., "Why is the product topology so coarse?": to say that "it's the definition" is unhelpful; to take the categorical definition of "product" and _find_out_ what topology on the cartesian product of sets is the categorical product topology is a do-able, interesting exercise! :)

paul garrett
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  • If "adjunction arises everywhere" additional examples (and counterexamples) to natural maps should be more widely discussed. Roman gives two additional examples near the end of "Lattices and Ordered Sets". – alancalvitti Apr 16 '12 at 15:35
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Short answer: The formal definition of “natural isomorphism” is completely unrelated to any notion of choosing, let alone the axiom of choice.

Silly answer: There are natural isomorphisms which depend on arbitrary choices. For example, under the axiom of global (!) choice, the category of sets is equivalent to the category of cardinals, which is a full subcategory of the category of sets. As such, every set is “naturally” isomorphic to its cardinal! Indeed, for each set $X$, fix a bijection $\eta_X : X \to \# X$, where $\# X$ is the cardinal of $X$. (If $X = \# X$, for simplicity we require $\eta_X = \text{id}_X$. This data suffices to specify a functor $\# : \textbf{Set} \to \textbf{Card}$, which acts on arrows $f : X \to Y$ by $\# f = \eta_Y \circ f \circ {\eta_X}^{-1}$. Let $U : \textbf{Card} \hookrightarrow \textbf{Set}$ be the inclusion. Then, $\#$ is left adjoint to $U$ with counit $\eta$, since by construction $$\begin{matrix} X & \xrightarrow{\eta_X} & \# X \newline {\scriptstyle f} \big \downarrow & & \big\downarrow {\scriptstyle \# f} \newline Y & \xrightarrow{\eta_Y} & \# Y \end{matrix}$$ commutes for every arrow $f : X \to Y$, and obviously both functors are full and faithful. This shows that $\eta$ is the “natural” isomorphism we seek.

Zhen Lin
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This is far from a complete answer, but rather a remark on natural maps and the axiom of choice.

It is possible without the axiom of choice to have a vector space which is not finitely generated and isomorphic to its double dual $[1]$ (we can find ones which are isomorphic to their first dual, and we can find those which are not isomorphic to their first dual but still isomorphic to the second dual).

If you think about it, assuming the axiom of choice, the natural map is always injective. This is not the case without the axiom of choice. It is possible to have nontrivial vector spaces that have no non-zero functionals, so the natural map is constantly zero $[2,3]$.

So the axiom of choice actually plays a role in restricting the class of vector spaces which are naturally isomorphic to their double duals (the axiom of choice giveth and the axiom of choice taketh).

Further reading (MathOverflow):

  1. Does the fact that this vector space is not isomorphic to its double-dual require choice?
  2. What’s an example of a space that needs the Hahn-Banach Theorem?
  3. Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?
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Asaf Karagila
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    The dual always has the zero element, AC or no AC :) – Mariano Suárez-Álvarez Jul 04 '11 at 16:55
  • @Mariano: I was trying to act like a cool algebraist and pretend that ${0}$ is an empty space... but very well. :-) – Asaf Karagila Jul 04 '11 at 17:03
  • @Asaf: you seem to be missing here the fact that no infinite-dimensional vector space is isomorphic to its second dual. Moreover, for finite-dimensional $V$, neither the natural isomorphism between $V$ and $V^{* }$ nor the unnatural isomorphism between $V$ and $V^$ requires the Axiom of Choice. – Pete L. Clark Jul 06 '11 at 06:16
  • @Pete: I am aware for the lack of need for choice in finite dimensions. As for the infinite dimensional case, I was sure that $\ell_2$ is isomorphic to its dual, and thus to its double dual. – Asaf Karagila Jul 06 '11 at 06:28
  • @Asaf: you are thinking of the set of continuous linear functionals on a topological vector space, i.e., you are working in a different category. (And I am pretty sure that this isomorphism does not use the Axiom of Choice either.) – Pete L. Clark Jul 06 '11 at 06:40
  • Also, could you please fix the bit about ${0}$ being the empty space? This is simply a false statement and therefore potentially confusing to readers. – Pete L. Clark Jul 06 '11 at 06:44
  • @Pete: I never claimed that the isomorphism requires choice. I said that the existence of the dual may require choice. As for the empty space, I was always taught that the empty vector space is the trivial space with only the zero vector in it. – Asaf Karagila Jul 06 '11 at 06:47
  • @Asaf: But the existence of the dual space certainly does not require choice, as Mariano pointed out. As for your second sentence: I think you know well more than enough set theory to see that a set which contains the zero vector is not empty, so this is statement is false (whether you were taught this or not). Perhaps you are thinking of the fact that the empty set is a basis for the zero vector space? – Pete L. Clark Jul 06 '11 at 06:50
  • @Pete: You might be surprised but it is very easy to construct a model of $ZF$ with a vector space without any functionals except the zero functional. I am guessing that there is some nomenclature confusion here, I was taught that the empty space is not the empty set, it is the space with no nontrivial vectors. – Asaf Karagila Jul 06 '11 at 06:53
  • @Asaf: no, I know this: my point (I'm sincerely sorry to harp on something so trivial) is that the zero functional counts as a linear functional. Moreover, your convention is leading you to make false statements as in "There is no second dual": again, every vector space has a dual, which might be the zero vector space. I'm sorry to say that it is starting to seem like you have been taught something incorrect. Can you supply a reference where this "empty vector space" terminology is used? – Pete L. Clark Jul 06 '11 at 07:06
  • Let me rephrase some of what I said before and add a more positive question: if one assumes AC (as is typical in algebra), then a(n abstract) vector space $V$ over a field $k$ is isomorphic to its second dual iff $V$ is finite-dimensional. If one does not assume AC then still every finite-dimensional vector space is isomorphic to its second dual. But without AC is there perhaps an infinite-dimensional vector space $V$ such that the canonical injection $V \rightarrow V^{* *}$ is an isomorphism? – Pete L. Clark Jul 06 '11 at 07:13
  • @Pete: As for the terminology, I will have to direct you back in time and learn to speak Hebrew. I agree that the terminology is troublesome and will try to correct myself from here on end. As for the dual, I see your point. However if the dual is trivial, the second dual is either trivial or not a vector space (since every functional is continuous on ${0}$) and either way it is not isomorphic to the original space. – Asaf Karagila Jul 06 '11 at 07:15
  • @Pete: I do not know an answer for that, since I do not know the proof for the equivalence you gave I cannot speak on how the axiom of choice might be relevant to that. I guess that my limited knowledge in algebra causes me to confuse between the notions of dual. It is still unclear to me. I will try to remedy that, and if not I will post a question. In the meantime I will delete the answer until such time that I can rectify its content. – Asaf Karagila Jul 06 '11 at 07:18
  • @Asaf: well, if the convention only exists in Hebrew then...(I feel it would be obnoxious to spell this out: you see where I'm going). However, I don't recall saying that I don't speak Hebrew: if you have a Hebrew reference, I would be interested. – Pete L. Clark Jul 06 '11 at 07:23
  • Anyway, I'm sorry to pick at you repeatedly. If it helps, it's for a good cause. :) – Pete L. Clark Jul 06 '11 at 07:24
  • @Pete: I will have to drag you back in time and have you sit with me in class. :-) – Asaf Karagila Jul 06 '11 at 07:44
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    @Pete: Turns out that I wasn't that far from the truth when I thought about $\ell_2$. In Solovay's model its topological dual is actually its algebraic dual. :-) – Asaf Karagila Apr 16 '12 at 21:56
  • @Pete: Whether or not it's still interesting, I've put the arguments about self-duality in a short note online. http://karagila.org/2012/the-axiom-of-choice-and-self-dual-vector-spaces/ – Asaf Karagila Oct 27 '12 at 23:26
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The categorical meaning of "natural" means that your definition is "invariant by morphism" (e.g. change of coordinates in vector spaces). This is a very strong property which implies the definition is somehow canonical, but not necessarily obvious.

Although using the axiom of choice is likely to lead to non-natural constructions, not all non-natural constructions use this axiom, even if there might be some "choice" involved in the definition ("choice" in the colloquial sense does not always imply actual use of the axiom). For example, to embed any vector space in its dual, you need a basis (and thus the axiom of choice). But to do that for a finite-dimensional vector space, the axiom of choice is not needed. The construction is never natural.

Finally, it may also very well happen that a construction using the axiom of choice turns out to be natural (I'll admit I don't know of any example).

Joel Cohen
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I'm no expert but I think they're different. My understanding is that "natural isomorphism" has always been a little vaguely defined but the concept is fairly intuitive.

It's a case where there is clearly a "best" or simplest isomorphism. Like the Chinese Remainder Theorem perhaps. Cmn ~ Cm x Cn where m, n are relatively prime. There may be more than one isomorphism but one stands out as the most natural or most obvious, I guess.

Generally the "natural" isomorphisms are easily and unambiguously provable. It's recognising that such a connection exists that's the real art. Funnily enough I find the axiom of choice much vaguer (and have never been a fan).

I'm probably way off...

karnok
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  • I always did understand "natural isomorphism" as strictly defined (just like the axiom of choice) - where is there any vagueness? – Hans-Peter Stricker Jul 04 '11 at 15:10
  • I could be wrong of course, but I only mean the natural part. Isomorphism is strict enough. But natural isn't. There is a distinction - I think natural isomorphisms are especially used when there's a highish degree of generalisation. If it's a one-off, you'd rarely call it natural. Hope that makes sense (and is roughly accurate)... – karnok Jul 04 '11 at 15:21
  • What about the definitions I linked to? – Hans-Peter Stricker Jul 04 '11 at 15:28
  • Sorry I'm a moron, don't know that much about category theory. I think "natural isomorphism" gets used elsewhere in maths and I got confused... My bad! – karnok Jul 04 '11 at 15:35
  • "Whoever is free from sin..." ;-) – Hans-Peter Stricker Jul 04 '11 at 15:40