Category of sets equivalent to category of finite dimensional vector spaces? Suppose char(K) = 0. Prove or disprove: The category of sets is equivalent to the category of finite dimensional K-vector spaces.
I've only just begun to learn about category theory, so I'm not entirely sure how to approach this task. By the type of question I'd imagine that they are not equivalent, which would make sense to me, since sets carry far less structure than vector spaces. 
In this case I'd have to prove that there can't be any equivalence F from the sets into the category of vector spaces. How would I go about that? 
Any help would be appreciated!
 A: In the category of finite dimensional vector spaces, the initial object and the terminal object are equal: it's the zero-dimensional vector space $0$. Thus the initial and the terminal object are isomorphic (yes, equal implies isomorphic...!). But in the category of sets, the initial object $\varnothing$ is not isomorphic (in bijection with) the terminal object $\{*\}$ (any singleton). So the two categories are not equivalent.
If you want more details: suppose that $F : \mathsf{Set} \to \mathsf{Vect}$ were an equivalence. Then since $F$ is fully faithful, $\hom_{\mathsf{Vect}}(F(\varnothing), F(X)) \cong \hom_{\mathsf{Set}}(\varnothing, X) = \{ \varnothing \}$ would be a singleton for all sets $X$, and since $F$ is essentially surjective, it follows that $F(\varnothing)$ would be an initial object. Similarly you can prove that $F(\{*\})$ would be a terminal object. Thus $F(\varnothing)$ and $F(\{*\})$ would need to be isomorphic, in particular there would be an isomorphism $F(\{*\}) \to F(\varnothing)$. But in $\mathsf{Set}$ there isn't even a morphism (map) $\{*\} \to \varnothing$, contradicting the fullness of $F$.
If you want to use the characteristic zero assumption: since $k$ has characteristic zero, any nonzero vector space has an infinite number of endomorphisms (multiplication by $\lambda \in k$ for example). And the zero vector space only has one endomorphism. This means that the two-element set $\{0,1\}$ (which has $4$ self-maps) cannot be mapped to any vector space without contradicting the fully faithful property of $F$.
