# Example of an unnatural isomorphism

Can anyone give an example of an unnatural isomorphism? Or, maybe, somebody can explain why unnatural isomorphisms do not exist.

Consider two functors $F,G: {\mathcal C} \rightarrow {\mathcal D}$. We say that they are unnaturally isomorphic if $F(x)\cong G(x)$ for every object $x$ of ${\mathcal C}$ but there exists no natural isomorphism between $F$ and $G$. Any examples?

Just to clarify the air, $V$ and $V^\ast$ for finite dimensional vector spaces ain't no gud: one functor is covariant, another contravariant, so they are not even functors between the same categories. A functor should mean a covariant functor here.

For a simpler, but arguably more artificial, example than Mark's, take $\mathcal{C}$ to be the category with one object and two morphisms. Then the identity functor $\mathcal{C}\to\mathcal{C}$ is "unnaturally isomorphic" to the functor that sends both morphisms to the identity map.

• What you've written doesn't uniquely define a category. There are two such categories: either the non-identity morphism $f$ satisfies $f^2 = \text{id}$ or it satisfies $f^2 = f$. Aug 14, 2013 at 8:35
• True. Either will do. Aug 14, 2013 at 8:38
• That is very cool! Aug 14, 2013 at 10:43

If $F,G : C \to D$ are functors such that $F(x) \cong G(x)$ for every $x \in C$, I would call $F,G$ "pointwise isomorphic". You ask for examples of non-isomorphic functors which are pointwise isomorphic. There are plenty natural examples.

1. Consider the interval category $I=\{0 \to 1\}$. The category of functors $I \to C$ is isomorphic to the category of morphisms in $C$. Of course for most $C$ there are non-isomorphic morphisms in $C$ whose domain and codomain are isomorphic or even equal. For example take the identity and a constant map on a nontrivial set or space.

2. Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings; the action on morphisms is "conjugation". These functors are pointwise isomorphic, but not isomorphic (in fact between these functors there is no natural transformation at all). Actually this example (when restricted to sets of a given size) can be seen as a special case of the next one.

3. Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course for $G \neq 1$ there are non-isomorphic $G$-sets whose underlying sets are isomorphic (for example the underlying set of $G$ with the regular action and with the trivial action of $G$).

4. If $C$ denotes the category of finite abelian groups, then $\mathrm{Tor}_1^{\mathbb{Z}}$ and $\otimes_{\mathbb{Z}} : C \times C \to C$ are pointwise isomorphic (since $\mathrm{Tor}_1(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/\mathrm{gcd}(n,m) \cong \mathbb{Z}/n \otimes_{\mathbb{Z}} \mathbb{Z}/m$), but they are not isomorphic (for example since $\mathrm{Tor}_1^{\mathbb{Z}}$ is not right exact in the second or first variable).

• I think that this answer is severely underappreciated. Personally, I consider the first example here to be the most illuminating among all the examples mentioned in the answers to this question. It's a very useful exercise to explicitly write down both functors for some small sets and analyze them. The non-existence of a natural isomorphism between these functors really helps me appreciate what naturality is all about. Aug 14, 2013 at 12:00
• @OmarAntolín-Camarena: Indeed, if $G$ is a group and $X$ is a principal homogeneous space under $G$, then $G\times X$ is naturally isomorphic to $X\times X$.
– ACL
Aug 14, 2013 at 17:39
• The first example is perhaps more transparent if you consider it as a special case of the second example, where $G=S_n$ and you let it act on itself either by translation (to get Ord) or conjugation (to get Sym). Aug 14, 2013 at 18:39
• Omar, I don't think you said what you meant to say. If what you said was true, we could compose your natural iso with the first-projection functor $C \times C \to C$ to obtain a natural iso between Ord and Sym - and that's impossible. I guess you actually meant to say that there's a natural iso between the two endofunctors of $C$ defined by $X \mapsto \text{Ord}(X) \times \text{Sym}(X)$ and $X \mapsto \text{Ord}(X) \times \text{Ord}(X)$. Aug 16, 2013 at 0:29
• I have now added the most basic example which one can think of and placed it as example 1. In the comments above, example 1 refers to what is now example 2. Aug 18, 2013 at 6:51

The Universal Coefficient Theorem for, say, singular cohomology should give examples. For any abelian group $G$ and $n> 0$, the functors from spaces to abelian groups given by $$X\mapsto H^n(X;G),\qquad X\mapsto \mathrm{Ext}(H_{n-1}(X),G)\oplus\mathrm{Hom}(H_n(X),G)$$ are isomorphic, but not naturally so. See Hatcher's "Algebraic Topology", Chapter 3.1 (in particular Exercise 11 at the end of that section).

• I "almost knew" this one but I did not know how to show unnaturality. Aug 14, 2013 at 10:45
• Beautiful example. Aug 14, 2013 at 13:04

Your non-example of vector spaces and their duals can be souped up to a real example.

Let $C$ be the groupoid of finite-dimensional vector spaces and linear isomorphisms. Then there are two obvious functors $C \to C^{op}$: the linear dual, and the natural isomorphism $C \stackrel \sim \to C^{op}$ that one has for any groupoid. These functors are unnaturally isomorphic.

• Specifically, the second functor takes any map to its inverse. Aug 14, 2013 at 18:34
• If we take the groupoid of finite dimensional real Hilbert spaces and linear isomorphisms, these functors are unnaturally isomorphic. But, if we restrict to orthogonal maps instead of linear ones, the inner product provides a natural isomorphism between them. Apr 8, 2016 at 15:52

The geometric realization of a simplicial set and the geoemetric realization of its barycentric subdivision are always homeomorphic. However there cannot be a natural isomorphism between these two functors. (Look at the diagram of simplicial sets $\Delta^1 \leftarrow \Delta^2\rightarrow \Delta^1$. The maps are induced by $1,2,3 \mapsto 1,1,2$ and $1,2,3\mapsto 1,2,2$).

• This example is beautifully different from most of the others. The others give examples of clearly non-natural families of isos. This one feels much more like it ought to be natural — even with the counterexample in front of me, I find it hard to disbelieve that these functors are naturally isomorphic. Oct 26, 2015 at 21:38
• Very beatiful!! Jun 15, 2022 at 20:30

Take $C = BG$ for some group $G$ and take $D = \text{Set}$. A functor $BG \to \text{Set}$ is a $G$-set. Two $G$-sets are unnaturally isomorphic iff they have the same cardinality, and it's easy to find two $G$-sets of the same cardinality which are not isomorphic as $G$-sets, e.g. find a group with two non-conjugate subgroups of the same index.

• Perhaps a better construction of nonisomorphic G-sets of the same cardinality, that works for any nontrivial group G, is to take G acting on itself by multiplication and G acting on itself trivially. Aug 14, 2013 at 9:21

Here's a nice example that recently came up in an MSE question. Let $k$ be a field and let $Vect$ be the category of $k$-vector spaces and $Aff$ be the category of $k$-affine spaces. Every vector space is an affine space, giving a forgetful functor $F:Vect\to Aff$. On the other hand, every affine space has an associated vector space of the same dimension (the vector space of formal differences), giving a functor $G:Aff\to Vect$. The composition $GF:Vect\to Vect$ is naturally isomorphic to the identity. The composition $FG:Aff\to Aff$, on the other hand, is only unnaturally isomorphic to the identity: it takes every affine space to another affine space of the same dimension, but this cannot be made compatible with morphisms.

The structure theorem for finitely generated abelian groups furnishes for each $$A$$ an isomorphism $$A\cong T(A)\oplus \frac{A}{T(A)}$$ where $$T$$ is torsion. This is a family of pointwise isomorphisms between $$1_{\mathsf{Ab}_\text{f.g}}$$ and the functor $$T\oplus \frac 1T$$.

Claim. These functors are not naturally isomorphic. In particular, the isomorphisms of the structure theorem are not natural.

Proof. The endomorphism monoid of the identity functor is the multiplicative monoid $$\mathbb Z$$. This can be seen by looking at naturality squares mapping out of $$\mathbb Z$$ and using its universal property as the free abelian group on a single generator. On the other hand, the functor $$T\oplus \frac 1T$$ admits a nilpotent endomorphism $$T\oplus \frac 1T\overset{ \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix} }{\longrightarrow}T\oplus \frac 1T$$ where $$\alpha:\frac 1T\Rightarrow T$$ is given componentwise by $$\frac{A}{T(A)}\to T(A)\oplus \frac{A}{T(A)}\to T(A)$$. Thus $$1,T\oplus \frac 1T$$ have non-isomorphic endomorphism monoids whence they are themselves non-isomorphic functors.

• I don’t understand what $\alpha$ is. But $T\oplus1/T$ is not faithful (e.g. it takes both morphisms $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$ to the same thing), which is another way to see this functor is not isomorphic to the identity. Nov 5, 2021 at 16:38

I gave a more elaborate example to the Universal Coefficient splitting being non natural in my paper Cohomology with chains as coefficients'', Proc. London Math. Soc. (3) 14 (1964), 545-565, available here. It is proved there that for chain complexes $K,L$ which are free and are zero below dimension $0$, there is an isomorphism for any abelian group $G$

$$H^*( K \otimes L, G) \cong H^*(K, H^*(L,G))$$

which can be chosen to be natural with respect to maps of $K$ but not with regard to maps of $L$, nor in Example 3.2 maps of $G$. The naturality with respect to maps of $K$ is useful to recover R. Thom's determination of the weak homotopy type of the function space $K(G,n)^Y$ and further to determine $k^Y$ where $k$ is a cohomology operation (see the paper On Kunneth suspensions'', Proc. Camb. Phil. Soc. 60 (1964) 713-720, available here.

It seems that the functor on the category infinite sets that adds one disjoint point * to any set is not naturally isomorphic to the identity functor.

• Why? Can you add details or proof? Nov 26, 2018 at 23:43
• It looks like there isn't even a natural transformation $\phi\colon (-)\sqcup 1 \Rightarrow \mathrm{id}$, since it would imply that every function in the category of infinite sets should preserve the basepoints that every set $X$ gets from $1 \to X\sqcup 1 \xrightarrow{\phi_X} X$, and this is not true. Nov 27, 2018 at 4:28

Here is an example of unnaturally isomorphic functors for which there does not exist any non-trivial natural transformation between them.

Let $\mathcal{C} = \mathbb{N}^{\mathrm{op}}$, $\mathcal{D} = \mathcal{Ab}$, and consider $F, G : \mathcal{C} \rightarrow \mathcal{D}$ defined by $$F(n) = G(n) = \mathbb{Z} \quad\text{for all } n \in \mathbb{N},$$ $$F(m \le n)(x) = 2^{n-m} x,\quad G(m \le n)(x) = 3^{n-m} x \quad\text{for all } x \in \mathbb{Z}.$$

Suppose $\eta = \{ \eta_n : F(n) \rightarrow G(n) \}_{n \in \mathbb{N}}$ is a natural transformation. Then for any $n$ we have that $\eta_0 (2^n x) = 3^n \eta_n (x)$, so $2^n \eta_0 (x) = 3^n \eta_n(x)$. But then $3^n \mid \eta_0 (x)$ for all $n$, which implies that $\eta_0(x) = 0$, and so $\eta = 0$.

• I do not think $3^n \mid \eta_0 (x)$ for all $n$ is clearly absurd: take $\eta_0 = 0$. In fact, there is always the zero natural transformation between two functors into abelian groups. Aug 14, 2013 at 9:20
• You're right, I forgot to say non-zero, I'll edit that. Aug 14, 2013 at 9:31

I'm pretty sure one can also categorify the fact that for ordinary complex representations of finite groups, number of irreducible representations = number of conjugacy classes. As in this closely related question, one has a bijection (which categorifies to a pointwise isomorphism) but not a natural one.

(The two functors I'm thinking of here are contravariant functors from the category of finite groups to the category of $k$-linear categories. The first is $F_1(G) = \mathrm{rep}_{\mathbb{C}}(G)$. The second, $F_2$, takes a finite group $G$ to the $k$-linear category freely generated by the conjugacy classes of $G$. I'd have to think about how to define $F_2$ on morphisms, but I don't think there's any choice of definition of $F_2$ on morphisms that will make $F_1$ and $F_2$ naturally isomorphic, despite the fact that they are pointwise isomorphic.)

• There's no ambiguity on how to define $F_2$ for isomorphisms of groups, and already in that case $F_1$ and $F_2$ cannot be naturally isomorphic; see the answers to mathoverflow.net/questions/21606/…. Aug 14, 2013 at 19:29
• As explained in the closely related question that you linked to, it probably makes most sense to think of one functor as covariant and the other contravariant. If you buy this, then the OP has ruled out this type of example by fiat. Jun 26, 2019 at 2:14
• @TimothyChow: If we only consider these on the category of groups & isomorphisms, then we can take $F_2$ to be contravariant as well. Yet even in that situation, I think the answers on the question linked to by Eric rule out a natural isomorphism of these functors. Jun 26, 2019 at 16:12

Although there are already too many answers, let me just add the observation that one of the real motivations for The General Theory of Natural equivlances, was to understand the distinction between the fact that a finite dimensional vector space is isomorphic to its dual space, but naturally isomorphic to its second dual.

• Which is undoubtedly the reason why this example is mentioned in the question itself. Oct 14, 2013 at 19:30