Let $\mathbb{F}$ be a field. The category of matrices $\mathbf{Mat}$ has $\mathbb{N}_0$ as class of objects and $\mathrm{hom}(n,m)=\mathbb{F}^{n\times m}$ for $n,m\in\mathbb{N}_0$, with $\mathrm{id}_n=\mathbf{1}_n$ and $B\circ A=A\cdot B$ with the usual matrix multiplication whenever defined. The category of finite-dimensional vector spaces $\mathbf{FinVect}$ has $\mathbb{F}$-vector spaces as objects and linear maps as morphisms, where composition is the usual composition of functions and the identity morphism is the identity map. The categories $\mathbf{Mat}$ and $\mathbf{FinVect}$ are equivalent.

There is a contravariant duality functor $(-)^\ast\colon\mathbf{FinVect}\rightarrow\mathbf{FinVect}$ with $V^\ast=\mathrm{Hom}(V,\mathbb{F})$ for any vector space $V$ and $f^\ast\colon W^\ast\rightarrow V^\ast,\,g\mapsto g\circ f$ for any linear map $f\colon V\rightarrow W$. There also is a contravariant transpose functor $(-)^t\colon\mathbf{Mat}\rightarrow\mathbf{Mat}$ where $n^t=n$ for $n\in\mathbb{N}_0$ and $A^t$ is the usual transpose of a matrix.

These two contravariant functors on equivalent categories are related by the fact that for finite-dimensional vector spaces $V,W$ with bases $\mathscr{B},\mathscr{C}$ respectively, we have that $$(T^{\mathscr{B}}_{\mathscr{C}}(f))^t=T^{\mathscr{C}^\ast}_{\mathscr{B}^\ast}(f^\ast),$$ where $T$ denotes the transformation matrix for the respective bases.

Is there a way to describe this relationship of functors in category-theoretical terms?


1 Answer 1


There's an obvious functor $\text{Mat} \to \text{FinVect}$ which is an equivalence, but no obvious choice of inverse to this functor; choosing an inverse amounts to choosing, for each finite-dimensional vector space (note that there is more than a set's worth of such things!), a basis of that vector space. Not a fun operation to perform; we need something like the axiom of global choice to do it.

What can be done without making such choices is the following: there is an intermediate category which I'll just call $\text{Base}$, the category of finite-dimensional vector spaces together with a choice of basis (morphisms are still just linear transformations). There is an obvious functor $F : \text{Base} \to \text{FinVect}$ given by forgetting the basis, and an obvious functor $M : \text{Base} \to \text{Mat}$ given by writing linear transformations with respect to a basis. Both of these functors are equivalences, but again the functor $F : \text{Base} \to \text{FinVect}$ can't really be inverted obviously. See also anafunctor.

$\text{Base}$ has a dualization endofunctor $D : \text{Base} \to \text{Base}$ which takes duals of linear transformations and sends a pair $(V, B)$ of a vector space and a basis to the pair $(V^{\ast}, B^{\ast})$ of the dual vector space and the dual basis. Now what we can say is that the functors $F$ and $M$ intertwine this dualization functor and the other dualization functors, in the sense that we have not just natural isomorphisms but equalities

$$F \circ D = (-)^{\ast} \circ F : \text{Base} \to \text{FinVect}$$ $$M \circ D = (-)^T \circ M : \text{Base} \to \text{Mat}.$$


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