# A covariant functor sending every finite-dimensional vector space to its dual?

Wikipedia defines a particular map between two particular objects as an unnatural isomorphism if it is an isomorphism that cannot be extended to a natural transformation on the entire category.

Dual vector spaces provide a classic example of unnatural isomorphisms. The contravariant functor sending finite dimensional vector spaces to their duals and linear maps to the adjoints cannot be naturally isomorphic to the identity, just because it is contravariant. Furthermore, as described here, one can define a form of natural transformation between covariant and contravariant functors, and the dual functor doesn't satisfy that either.

But what is wrong with the following argument. Given a particular finite-dimensional vector space $$V$$ and an arbitrary isomorphism $$\eta_V:V\to V^*$$, chose an another arbitrary isomorphism $$\eta_W: W\to W^*$$, for every other finite dimensional space $$W\neq V$$. Define a covariant functor $$G$$ that sends each space to its dual, and sends each map $$f:W\to X$$ to $$G(f) = \eta_X \circ f \circ \eta_W^{-1}$$. This appears to be naturally isomorphic to the identity functor, since $$G(f) \circ \eta_W = \eta _X \circ f$$.

The functor $$G$$ is certainly not the standard dual functor: it is covariant, and it does not send maps to their adjoints. However it does seem like an extension of an arbitrary map between a space and its dual, to a natural transformation on the entire category. Maybe the original definition was wrong - does it even make sense to talk about an unnatural isomorphism between a single pair of objects?

Let $$D : \mathbf{fdVec}^{\mathrm{op}}_k \to \mathbf{fdVec}_k$$ be the dual space functor, defined by $$D(V) = V^*, \quad D(f : V \to W) = f^* : W^* \to V^*$$ where $$\mathbf{fdVec}_k$$ is the category of finite-dimensional vector spaces over a field $$k$$, and $$k$$-linear maps.

When we say that $$V^*$$ is 'not naturally isomorphic' to $$V$$, we mean that there is not a functor $$I : \mathbf{fdVec}^{\mathrm{op}}_k \to \mathbf{fdVec}_k$$ with $$I \cong D$$ such that $$I(V)=V$$ for all $$V$$.

So in a sense, you're answering a different question, since what you have defined is a functor $$G : \mathbf{fdVec}_k \to \mathbf{fdVec}_k$$ with $$G \cong \mathrm{id}$$ such that $$G(V)=V^*$$ for all $$V$$.

What you proved is an instance of a more general fact: given any category $$\mathcal{C}$$ and any function $$(-)^* : \mathrm{ob}(\mathcal{C}) \to \mathrm{ob}(\mathcal{C})$$ such that $$A \cong A^*$$ for all $$A \in \mathrm{ob}(\mathcal{C})$$, there is a functor $$G : \mathcal{C} \to \mathcal{C}$$ such that $$G \cong \mathrm{id}_{\mathcal{C}}$$ and $$G(A)=A^*$$ for all objects $$A$$. The proof is identical to the proof you gave.

Even more generally, given a category $$\mathcal{C}$$, a subcategory $$\mathcal{D}$$ and a function $$(-)^* : \mathrm{ob}(\mathcal{C}) \to \mathcal{ob}(\mathcal{D})$$ such that $$A \cong A^*$$ for all objects $$A$$, there is an equivalence of categories $$G : \mathcal{C} \rightleftarrows \mathcal{D} : I$$, where $$I$$ is the inclusion $$\mathcal{D} \hookrightarrow \mathcal{C}$$ and $$G(A) = A^*$$ for all objects $$A$$ of $$\mathcal{C}$$. Again, the proof is essentially identical to what you gave, and can be used to prove that every category $$\mathcal{C}$$ is equivalent to its skeleton.

To give this some more perspective: a consequence of this fact is that $$\mathbf{fdVec}_k \simeq \mathbf{Mat}_k$$, where $$\mathbf{Mat}_k$$ has $$\{ k^n : n \in \mathbb{N} \}$$ as its set of objects, with a morphism $$f : k^n \to k^m$$ being an $$m \times n$$ matrix over $$k$$. Despite all of this, it would be misleading to say that every finite-dimensional vector space over $$k$$ is 'naturally isomorphic' to $$k^n$$ for some $$n \in \mathbb{N}$$, despite the fact that there is a functor $$\mathbf{fdVec}_k \to \mathbf{fdVec}_k$$ given by $$V \mapsto k^{\mathrm{dim}(V)}$$ that is naturally isomorphic to the identity functor.

For the same reasons, despite the natural isomorphism you have defined, it would be misleading to deduce that every vector space is 'naturally isomorphic' to its dual.

• Thanks that is very helpful! Does this mean that we cannot actually ask whether a particular map between a particular pair of objects is natural? And that naturality only makes sense as a property of an entire functor? If so, the wikipedia article seems misleading. Apr 20 '19 at 15:46
• @Neuromath: Kind of, but kind of not. When we say something like "the map $\eta_V : V \to V^{**}$ is natural", implicitly what we are saying is (a) there are functors $F$ and $G$ such that $F(V) = V$ and $G(V)=V^{**}$; and (b) there is actually a family of maps $\eta_X : X \to X^{**}$, one for each $X$; and (c) these maps assemble into a natural transformation $\eta : F \to G$. So it does make sense to ask whether a map is natural, but what you're really asking is whether there exist functors and a natural transformation between those functors. Apr 20 '19 at 16:48
• Put another way, you can only make sense of 'naturality' for a family of maps, i.e. you can't just define a map $f : V \to W$ between two vector spaces and ask if it's natural, you need to define a family $$f_V : (\text{something involving } V) \to (\text{something else involving } V)$$ and ask whether that family is natural. And then to make sense of that question, you need there to be functors that send $V$ to the respective 'something' and 'something else', between which the maps $f_V$ form a natural transformation. Apr 20 '19 at 16:49
• Thanks again! However (and I believe this is just paraphrasing your answer) it seems we can't even ask if a whole family of isomorphisms $\eta_V : V \to V^*$ is natural. We have to also say what we want to happen to maps. The original example defines a family of isomorphisms from every space to its dual, which can be extended to a covariant functor naturally isomorphic to the identity. But this functor does not do what we wanted to linear maps: it does not take them to their adjoints. This is why naturality is only defined for functors, not maps or even families of maps. Apr 21 '19 at 8:59