Category equivalence of sets and vector spaces It seems to me that for a field $K$, the functor $\mathbf{Set}\to\mathbf{Vec}_K$, sending a set to the free $K$-vector space on it, is a category equivalence. It is full and faithful because a linear map between vector spaces is uniquely determined by a map on bases, and it is essentially surjective because every vector space has a basis.
However I could not find any reference for this, so I doubt it is true. Could anyone clarify this for me?
 A: Your argument "because a linear map between vector spaces is uniquely determined by a map on bases" correctly shows the functor is faithful, but not full: given vector spaces with bases $(U, \{e_i\})$ and $(V,\{f_i\})$ not every linear map comes from a function of sets between $\{e_i\}$ and $\{f_i\}$.
A: There is in fact an abstract nonsense proof that $F$ is faithful but not full, just from considering the unit $\eta=$"insertion of generators" of the adjunction $ (F,U;\eta,\epsilon)$. This comes from dualizing the following result, which is Theorem $1$ in section $\mathrm{IV}.3$ in Mac Lane's Categories for the Working Mathematician.

(i) $U$ is faithful iff every component $ϵ_a$ of the counit $ϵ$ is epi.
    (ii) $U$ is full iff every $ϵ_a$ is a section.  

Since $(U,F;ϵ^{op},η^{op})$ is also an adjunction, namely $A^{op}→X^{op}$, we see that $F$ is faithful iff $η$ is monic, and $F$ is full iff $η$ is a retraction.
Since the insertion of generators as a basis into a vector space is monic but not surjective, it follows that $F$ is faithful but not full.
But you are right that $F$ is essentially surjective, as every vector space is free.
