Example of an unnatural isomorphism

Question

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.

Answer 1

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.

Answer 2

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).

Answer 3

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).

Answer 4

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.

Answer 5

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 $1,2,3\mapsto 1,2,2$).

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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.

Answer 11

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 $\eta = 0$.

Answer 12

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.)

Answer 13

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.