The duality functor $F=\text{Hom}(-,k) \in Fun(\text{Vect}^{op},\text{Vect})$ (where $k$ is the ground field) gives the equivalence. In fact this maps every finite dimensional vector space $V$ to its dual $V'$. This maps every morphism $T:V\to W$ in $\text{Vect}$ to $T':W'\to V'$ by precomposing morphisms in $W'=\text{Hom}(W,k)$ with $T$.

Now in a bit more concrete term, each $T':W'\to V'$ is the dual map of $T$, and from elementary linear algebra there is bijective correspondence between $T'$ and $T$ because we are in finite dimensional case. So $F$ is fully faithful, and it is also essentially surjective (dense) because for every $V\in \text{Vect}$, $V'$ is isomorphic to $V$.

Hence this functor gives a equivalence.

Some questions I have about this:

It feels like I've done this in a very roundabout way, like I'm missing some obvious observations that would make problem a lot easier. For instance, some of the main advanages of having finite dimensional vector spaces is that $V$ is isomorphic to its bidual $V''$, and it is very tempting to say stuff like $F^2(V)=V''$ is isomorphic to $V$. Except that $F$ is $\text{Vect}^{op}\to \text{Vect}$, so we cannot talk about $F^2$. Is this just a good looking coincidence?

Also, what would be the other functor constituting the equivalence with $F$?

  • 3
    $\begingroup$ The other functor would surely be $F^{op}$ and it's $F^{op}\circ F$ rather than $F^2$ that maps $V$ to $V''$. $\endgroup$ Jun 7 '17 at 19:16
  • $\begingroup$ @LordSharktheUnknown: This is an answer. Not a comment. $\endgroup$
    – HeinrichD
    Jun 10 '17 at 9:11

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