Talk:Natural transformation

Comment by Taku

Would you give an example of a non-natural isomorphism to the article? -- Taku 00:26, 10 November 2005 (UTC)

It looks like it has been added... Marc Harper 15:08, 26 March 2006 (UTC)

An isomorphism between a vector space and its dual V* is perhaps the commonest example of a non-natural isomorphism. Is there a sense in which it is a theorem that there is no natural isomorphism between V and V* ?Daqu 04:46, 21 February 2006 (UTC)

I removed the bit saying it was a 'counterexample'.

metterklume

It's probably not too hard to show that there can be no transformations between functors of different variance. -lethe ^{talk +} 05:25, 21 February 2006 (UTC)

example for fun

I wrote this example natural transformation for the benefit of a friend in my user space. He suggested that it was well-enough presented to warrant inclusion in this article, or else perhaps in its own article. I don't really think so; it seems to me to be too tutorial-ish to be encyclopedic. I post it on the talk page, and you can decide whether it has a place in this article (or any other).

We want to verify the equation

$\tau _{A}\circ {\mathcal {P}}(f)=f^{*}\circ \tau _{B}$

where τ_C: P(C) → 2^C is the map which sends any subset of the set C to the characteristic function on that subset, i.e.

$\tau _{C}(U)=\chi _{U},$

where χ_U is given by

$\chi _{U}(c)={\begin{cases}1&c\in U\\0&c\not \in U\end{cases}}$

for any subset U ⊆ C and any element c ∈ C. To verify the equation, let both sides act on some subset S ⊆ B. We have

${\mathcal {P}}(f)(S)=f^{-1}(S)$

by the definition of the powerset functor, and so

$\tau _{A}({\mathcal {P}}(f)(S))=\chi _{f^{-1}(S)}.$

On the right-hand side of the equation, we have

$\tau _{B}(S)=\chi _{S}$

and recall that f* is the pullback by f induced by the contravariant hom-functor; it acts on maps by multiplication on the right:

$f^{*}(\chi _{S})=\chi _{S}\circ f.$

So it remains to check the equality

$\chi _{f^{-1}(S)}=\chi _{S}\circ f.$

To verify this equation, act both maps in 2^A on an arbitrary element a ∈ A.

$\chi _{f^{-1}(S)}(a)={\begin{cases}1&a\in f^{-1}(S)\\0&a\not \in f^{-1}(S)\end{cases}}$

$(\chi _{S}\circ f)(a)={\begin{cases}1&f(a)\in S\\0&f(a)\not \in S\end{cases}}$

Since a ∈ f^–1(S) iff f(a) ∈ S, these maps are equal.

It's better than the "every group is naturally isomorphic to its opposite group"... —Preceding unsigned comment added by 130.126.108.212 (talk) 19:12, 9 November 2007 (UTC)

Sorry, but I fail to see how the above matches the definition. ${\mathcal {P}}(X)$ and $2^{X}$ are not the same category D. For the same reason, $\tau$ is not a morphism. The example is more that of the functor ${\mathcal {P}}(X)\rightarrow 2^{X}$, isn't it? -- Cobalt pen (talk) 14:04, 23 November 2015 (UTC)

natural isomorphism

I removed this paragraph from the article. I think it's wrong, but I don't know enough math to be sure of it. Paisa (talk) 02:23, 21 December 2007 (UTC)

A natural transformation η : F → G is a natural isomorphism if and only if there exists a natural transformation ε : G → F such that ηε = 1_G and εη = 1_F (where 1_F : F → F is the natural transformation assigning to every object X the identity morphism on F(X)).

I don't have CWM handy, but Mac Lane and Moerdijk in Sheaves and Geometry in Logic on page 13 call a natural transformation η : F → G a natural isomorphism if it is componentwise an isomophism, that is, if η_X is an isomorphism for each object X. (This is the definition currently in the article.) The Mac Lane–Moerdijk definition is equivalent to the one above. If there exists a natural transformation ε as above, then the compositional identities imply that η_X and ε_X are inverse morphisms. Thus η is componentwise an isomorphism, hence a natural isomorphism. Conversely, given a natural isomorphism η, define ε by the rule ε_X := η_X^–1 for each object X. One can verify that this definition yields a natural transformation satisfying the compositional identities above. I did not find an explicit comment in Mac Lane–Moerdijk acknowledging this equivalent definition. Michael Slone (talk) 04:09, 21 December 2007 (UTC)

Actually, I think that the definition

A natural transformation η : F → G is a natural isomorphism if and only if there exists a natural transformation ε : G → F such that ηε = 1_G and εη = 1_F (where 1_F : F → F is the natural transformation assigning to every object X the identity morphism on F(X)).

is better than the one that requires each η_X to be an isomorphism, as it gives the right generalisations to terms such as natural retraction, for which the two mechanisms of definition give non-equivalent concepts. I.e. a natural retraction should be a natural transformation with a right-inverse natural transformation, not a natural transformation η for which each η_X is a retraction (and which retractions may not necessarily fit together to make a right inverse natural transformation, unlike the case with isomorphisms). If nobody objects I may therefore update the article with this definition, pointing out the equivalence with the other definition for isomorphisms, and indicating the generalisations to retraction, coretraction, etc. Rfs2 (talk) 09:38, 7 October 2011 (UTC)

Natural map vs transformation

I think there's an informal notion of natural map that's not a natural transformation. I posted a section in Talk:Eilenberg-Steenrod axioms. Can someone explain how exactly the connecting homomorphism is a transformation? Money is tight (talk) 05:50, 12 January 2011 (UTC)

opposite group of abelian group

For all abelian groups X, Y and Z we have a group isomorphism

Hom(X ⊗ Y, Z) → Hom(X, Hom(Y, Z)). These isomorphisms are "natural" in the sense that they define a natural transformation between the two involved functors Ab × Ab^op × Ab^op → Ab.

What is the opposite group of an abelian group? In line with the first example, it should be the group itself with "reversed order" operation, i.e., the abelian group itself without modification to its group operation. But then the "op" markings are unecessary and misleading. --217.253.231.134 (talk) 09:14, 23 January 2012 (UTC)

Is dual really a counterexample?

Admittedly, this is far from my expertise; I am hoping someone will enlighten me. I looked at the MacLane and Birkhoff book, secion 6.4, and didn't see any detailed discussion of the assertion that V and V* are not naturally isomorphic.

I thought the reason we say that V and V** are naturally isomorphic is that if C is the category of finite-dimensional vector spaces over K, then there's a functor F: C --> C satisfying two properties: (1) F(V) = V** for all V in C (2) F is naturally isomorphic to the identity functor.

Why does the following not work: define, for each V, an isomorphism phi_V: V --> V*. Now define a functor F: C --> C by F(V) = V*, and for a linear map psi: V --> W, F(psi) = phi_W o psi o phi_V^-1. I think this is a functor, and (1) F(V) = V* for all V in C (2) F is naturally isomorphic to the identity functor, via the system of isomorphisms that we've chosen.

So why do we not say V and V* are naturally isomorphic? My procedure "requires choices," so it may be aesthetically unappealing, but what's the rigorous mathematical reason it's not allowed? Kier07 (talk) 22:14, 23 April 2012 (UTC)

• What you're missing is that the functor F isn't the dual-space functor; if psi: V --> W, then F(psi): V* --> W*; but the dual-space functor instead gives a psi*: W* --> V*. By contrast, the double-dual-space functor G has G(psi) = psi**: V** --> W**.

Obscure wording in definition

a natural transformation η from F to G associates to every object X in C a morphism ηX : F(X) → G(X) between objects of D, called the component of η at X

Who or what is called "component"? Are objects called the "component", or are all morphisms called "component"? Perhaps it is a synonym of "natural transformation" itself? Sorry, I'm not familiar with the term, so I just cannot make it up from the description. A general request (if you will edit the description). Can you please avoid sentences with long chains of descriptive clauses like this? These make not the easiest to understand content particularly unintelligible. 109.65.104.130 (talk) 14:07, 30 September 2013 (UTC)

The components are morphisms. I agree that the wording is ambiguous, though it's not obvious how to best fix it. I've updated it in https://wiki.riteme.site/w/index.php?title=Natural_transformation&diff=610072841&oldid=609455728 at the cost of more verbosity, but I think it helps clarify matters (see "Style — Towards clarity and grace" for an analysis of the tradeoffs). --Blaisorblade (talk) 14:02, 25 May 2014 (UTC)

Definitions of horizontal composition

The article contains no definition of horizontal composition, nor does it state the interchange law. I think there are two definitions — for instance they're given in this blog post, though in form of Haskell code and one of them is incorrectly called vert, for vertical composition. I think they're equivalent by naturality, but I've not seen a proof. In more category-theoretic term, the two definition of $a\circ _{0}b:G'F'\to GF$ (with $a:G'\to G,b:F'\to F$ natural transformations) would be

$a\circ _{0}b:G'F'\to GF$

$(a\circ _{0}b)_{X}=a_{FX}\circ G'(b_{X})=G(b_{X})\circ a_{F'X}$

And the interchange law (with $\circ _{0},\circ _{1}$ being respectively horizontal and vertical composition, as in 2-category) — see the diagram in the linked post:

$(a\circ _{1}b)\circ _{0}(a'\circ _{1}b')=(a\circ _{0}a')\circ _{1}(b\circ _{0}b')$

But I am not entirely sure whether the above are correct, much less a source for them. (I've typechecked them in Haskell, https://github.com/Blaisorblade/CategoryTheory/blob/master/code.hs#L22, and Haskell's type system tends to guarantee that a function with the right type has the right behavior — the type of natural transformation between f and g is forall a. f a -> g a, and it is all is needed to enforce naturality, for instance). --Blaisorblade (talk) 14:36, 25 May 2014 (UTC)

Section "Unnatural Isomorphism" is misleading

I am learning category theory and found the section "Unnatural isomorphism" misleading. Because I am new to the subject I hesitate to label it inaccurate, or to remove the section. However I really hope someone can fix it, because I think others will be misled also.

The problem is the sentence

Conversely, a particular map between particular objects may be called an unnatural isomorphism (or "this isomorphism is not natural") if the map cannot be extended to a natural transformation on the entire category.

As commented above by Kier07, you can define a covariant functor that sends each finite-dimensional vector space to its dual, and which is naturally isomorphic to the identity. This functor is not the dual space functor, because it does not send linear maps to their adjoints. However, by the definition quoted above, an arbitrary isomorphism between a space and its dual should not be called unnatural.

It seems that naturalness can only be defined for functors, and this is indeed what is in all the textbooks. This section, perhaps trying to make things easier for beginners, appears to say you can define naturalness for single maps. I believe this is incorrect.

The references to MacLane & Birkhoff 1999, §VI.4 are also inappropriate - that section does not go into anything like this level of detail. And it does not define unnatural this way.

See also discussion here.

--Kenneth.harris (talk) 09:29, 21 April 2019 (UTC)

I agree with the above completely. In fact, suppose in some category we have a map F defined on objects of C such that X is isomorphic to F(X) for all X, and we arbitrarily pick isomorphisms eta_X : X --> F(X) for all X. Then there is a unique way to define F on morphisms so that it becomes a functor and so that eta is a natural transformation from I to F. Indeed, the definition of naturality demands F(f) = eta_Y f eta_X^-1 whenever f : X --> Y. This shows that there is no such thing as an unnatural isomorphism in the sense proposed.

While the section may gave some vague idea of what mathematicians mean when they say natural, mathematically it is wrong, and therefore more harmful than useful. Frankly it should be deleted. I would do so myself but I am not an expert in category theory and maybe something can be salvaged. --Sean Eberhard (talk) 00:08, 27 February 2022 (UTC)