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Stream: learning: questions

Topic: About Natural Isomorphisms

Suraaj K S (Oct 10 2024 at 21:35):

I'm sure this is a pretty basic question, but here it goes.

I was watching some CT lectures, and there was a segment where they had to prove two functors F: C → Set and G: C → Set to be isomorphic. The proof went a little bit like follows:

• Let X ∈ C.
  + We can see that FX ≡ GX.
• Therefore F≡G

( I think that the exact functors were Hom(AxB, _) and Hom(A,_) x Hom(B,_) or something similar).

I think that, that argument demonstrates: ∀X∈C: FX≡GX.
I don't think that for arbitrary functors F and G, this implies that F≡G. For example, I think that I can construct categories where ∀X∈C: FX≡GX, but F and G are not isomorphic.

However, when the target category is Set, does this implication hold?

Nathanael Arkor (Oct 10 2024 at 21:39):

You also have to show that the isomorphisms between each $FX$ and $GX$ are natural. This is what it means for two functors to be isomorphic: that there is a natural isomorphism between them.

Kevin Carlson (Oct 10 2024 at 22:04):

And no, objectwise isomorphic functors need not be naturally isomorphic, even in case the target category is Set. The most famous example of an unnatural isomorphism, that between a finite-dimensional vector space and its dual, is such a case, if you just add an underlying-set functor.

Mike Shulman (Oct 10 2024 at 22:28):

With that said, there's a reason natural transformations are called "natural": in many, many, cases, when you define some transformation in a "natural" way (in the informal sense), it turns out to be natural in the formal sense as well. And so, mathematicians frequently omit to verify naturality explicitly because it is "obvious".

Kevin Carlson (Oct 11 2024 at 00:52):

Right, that's presumably what happened in this lecture. "Natural" examples of unnatural isomorphisms are rare!

Kevin Carlson (Oct 11 2024 at 00:55):

In the particular case, where the functors must have been something more like Hom(A+B,-) and Hom(A,-) x Hom(B,-) for the claim to be correct, the point is that the way your mind notices that Hom(A+B,X) is isomorphic to Hom(A,X) x Hom(B,X) is by observing a specific isomorphism, not a random one, and the specific isomorphism you've observed happens to be natural, which semi-justifies not discussing the naturality.

Max New (Oct 23 2024 at 12:56):

Kevin Carlson said:

In the particular case, where the functors must have been something more like Hom(A+B,-) and Hom(A,-) x Hom(B,-) for the claim to be correct, the point is that the way your mind notices that Hom(A+B,X) is isomorphic to Hom(A,X) x Hom(B,X) is by observing a specific isomorphism, not a random one, and the specific isomorphism you've observed happens to be natural, which semi-justifies not discussing the naturality.

To add to this example, any time you are showing that a functor $F : C \to \textrm{Set}$ is representable by constructing a natural isomorphism $C(c, -) \cong F$ for some $c$, you don't need to check any naturality conditions. By the Yoneda lemma, the natural transformation $C(c,-) \to F$ always arises as the functorial action of $F$ on a "universal element" of $F(c)$, and such a functorial action is natural. So as long as you present that morphism by an action on a universal element, there is no actual manual verification necessary

Max New (Oct 23 2024 at 12:58):

And if it's not clear you don't have to check that the inverse $F \to C(c,-)$ is natural because inverses of natural transformations are natural. This is one of my favorite practical motivations for the Yoneda lemma, without something like it category theory really would have an insane proliferation of side-conditions that would all need to be manually verified.

John Baez (Oct 23 2024 at 16:16):

Kevin Carlson said:

The most famous example of an unnatural isomorphism, that between a finite-dimensional vector space and its dual....

By the way, the unfortunate thing about this claimed example is that taking the dual is not a functor $\mathsf{FinVect} \to \mathsf{FinVect}$, so the idea of a natural isomorphism between this functor and the identity $1: \mathsf{FinVect} \to \mathsf{FinVect}$ doesn't even parse!

Taking the dual gives a functor $\ast: \mathsf{FinVect} \to \mathsf{FinVect}^{\text{op}}.$

I owe this observation to my friend Bill Schmitt.

Mike Shulman (Oct 23 2024 at 16:29):

One of my favorite examples of an unnatural isomorphism where the statement of naturality would typecheck is between an [[affine space]] $A$ and the underlying affine space of the vector space of displacements in $A$. This is important because if the isomorphism were natural, the categories of vector spaces and affine spaces would be equivalent, which they're not. (The other isomorphism, between a vector space $V$ and the vector space of displacements in the underlying affine space of $V$, is natural.) You can also postcompose it with an underlying-set functor to get a pair of unnaturally isomorphic functors to Set.

A simpler version of the same thing is between a [[heap]] and the underlying heap of its structure group.

Kevin Carlson (Oct 23 2024 at 16:59):

Ah, that's a really nice example, Mike! So nice, you (or demons in the Zulip server) posted it wice :)

Mike Shulman (Oct 23 2024 at 17:16):

Begone, Zulip demons; I banish you with the arcane Delete button!

Suraaj K S (Oct 26 2024 at 01:21):

Max New said:

inverses of natural transformations

Should it be natural isomorphisms here? I don't think all natural transformation have an inverse..

Suraaj K S (Oct 26 2024 at 01:22):

Max New said:

that morphism

Here, this morphism is the component of the natural transformation, right?

Max New (Oct 28 2024 at 00:40):

Suraaj K S said:

Max New said:

inverses of natural transformations

Should it be natural isomorphisms here? I don't think all natural transformation have an inverse..

The property I'm saying is that if you have a natural transformation $\alpha : F \Rightarrow G$ where $F,G : C \to D$ and for every $c$, $\alpha_c : Fc \to G c$ has an inverse, then those inverses assemble into a natural transformation $G\Rightarrow F$

Max New (Oct 28 2024 at 15:10):

Suraaj K S said:

Max New said:

that morphism

Here, this morphism is the component of the natural transformation, right?

To be totally precise, to construct a natural transformation $C(c,-) \to F$ it is sufficient to define an element $e \in F(c)$ and define the transformation $\alpha_{c'}(f : C(c,c')) = F(f)(e)$. You can check once and for all such an $\alpha$ is natural, and the Yoneda lemma says that in fact all natural transformations $\beta : C(c,-) \to F$ arise in this way by picking $e$ to be $\beta_{c}(\textrm{id})$. The Yoneda lemma is that $\beta_{c'}(f) = F(f)(\beta_{c}(\textrm{id}))$.