Let $\mathcal{V}$ be the category of vector spaces with finite dimension. Show that it exists a small category $\mathcal{C}$ which is equivalent to $\mathcal{V}$.
If we take the category $\operatorname{Mat}_K=\mathcal{C}$, with the naturals $0,1,2,\dotso$ as objects and where a morphism $n\to m$ is a $m\times n$ matrix over a fixed field $K$ are.
The identity $n\to n$ is the identity matrix and the composition of morphisms given by the multiplication of matrices, which has all the required properties.
By definition I need two functors $F:\operatorname{Mat}_K\to\mathcal{V}$ and $G:\mathcal{V}\to\operatorname{Mat}_K$ and two natural transformations $\tau:F\circ G\simeq \operatorname{id}_\mathcal{V}$, $\eta: G\circ F\simeq \operatorname{id}_{\operatorname{Mat}_K}$, such that for every object $M\in\operatorname{Ob}(\operatorname{Mat}_K)$ and $V\in\operatorname{Ob}(\mathcal{V})$ the morphism $\tau_M: F\circ G(M)\to M$, and $\eta_V: G\circ F(V)\to V$ are isomorphisms.
I tried this:
$F:\mathcal{C}\to\mathcal{V}$, for $n\in\mathbb{N}$ let $F(n):=K^n$ and for a $m\times n$ matrix $A$ let $F(A):K^n\to K^m$ with $x\mapsto Ax$.
This is indeed a functor, since $A(Bx)=(AB)x$ and we have the identity matrix $I$ as neutral element: $Ix=x$
But now I would have to construct a functor $G:\mathcal{V}\to \mathcal{C}$, but I doubt that this is so easy.
Can you give me a hint?
Best Answer
Construction of functor from category to one of its skeletons.
Let $\mathcal C$ be a category.
On objects we have the relation of "being isomorphic" which is evidently an equivalence relation.
From every equivalence class choose a representative.
If $R\subseteq\mathsf{Ob}(\mathcal C)$ denotes this collection of representatives then let $\mathcal S$ be the full subcategory that has $R$ as its collection of objects.
Then $\mathcal S$ is a so-called skeleton of $\mathcal C$.
For object $c$ in $\mathcal C$ let $F(c)$ be the representative of $c$ which is an object of $\mathcal S$. Then $F$ can be looked at as the object function of a functor. To make it indeed a functor we must find for every arrow $f:c\to d$ in $\mathcal C$ an arrow $F(f):F(c)\to F(d)$ in an appropriate way.
For this choose for every object $c$ an isomorphism $i_c:c\to F(c)$.
Then prescribe $F$ on arrows by stating that $f:c\to d$ is mapped by $F$ to $i_d\circ f\circ i_c^{-1}:F(c)\to F(d)$.
In my (favorite) notation: $$\left[c\stackrel{f}{\longrightarrow}d\right]\stackrel{F}{\mapsto}\left[F\left(c\right)\stackrel{i_{d}\circ f\circ i_{c}^{-1}}{\longrightarrow}F\left(d\right)\right]$$
Then it is not difficult to verify that $F$ respects identities and composition whence is a functor.
Actually $F$ is an equivalence and $\mathcal C$ and $\mathcal S$ are equivalent categories.
This can be applied in your case and it remains to prove that there $\mathcal S$ is small.