Left adjoint to the tensor functor Let $V$ be a vector space over a field $k$.
Define $V \otimes - :$ Vect $\to$ Vect the tensor endofunctor on the category of $k$ vector spaces.
Assume that $V \otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F \dashv V\otimes-$.
It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W \cong \oplus_{i \in I}k$ where a basis of $W$ is indexed by $I$.

Show that $F$ is given by tensoring with some vector space.

Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W \otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W \otimes Z$ for all $W$.
In order to prove something like that I'd need to find a canonical map $\phi: W \times V_W \to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.
I also have this information:
Hom$(F(W), Z) \cong$ Hom$(W, V\otimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.
Any guidance?
 A: $V \otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism
$$V \otimes (-) \cong \text{Hom}(V^{\ast}, -)$$
and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{\ast} \otimes (-)$. 
In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M \otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $\text{Hom}_S(M, S)$ over $R$; see this blog post.
A: To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $\kappa$, then $F(W)$ is the sum of $\kappa$ copies of $F(k)$. This is the same as $W\otimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)\cong \oplus_\kappa F(k)\cong F(k)\otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.
