Left adjoint of a strange functor I'm looking at the category (call it $\mathcal{C}$) of colimit preserving functors from $R$-mod to $Vect$, where $R$ is some commutative $k$-algebra. There is an obvious functor
$$
\mathcal{F} : \mathcal{C} \to Vect,
$$ 
which sends a functor $G$ to its value on $R$. And there is an obvious functor back $\mathcal{G}$ which sends a vector space to the colimit preserving functor $\mathcal{G}_V$ defined by $\mathcal{G}_V(M) = M \otimes_k V$. 
I want to show that $\mathcal{G}$ is left adjoint to $\mathcal{F}$, and I'm trying to use the unit counit approach, but I can't see an obvious counit $\epsilon : \mathcal{GF} \implies Id_\mathcal{C}$. Really, I need natural maps 
$$
M \otimes_k G(R) \to M
$$
for any $R$-module $M$ where $G(R)$ is just a vector space.
Can anyone help?
 A: First of all, the result you're trying to prove isn't true as stated.  It is only true if you restrict $\mathcal{C}$ to be the category of $k$-linear colimit-preserving functors.  Second, you don't need a natural map $M\otimes_k G(R)\to M$; you need a natural map $M\otimes_k G(R)\to G(M)$.
To construct such a map, first consider the case $M=R$.  In that case, note that $R$ acts on itself (as a module) by multiplication, and so since $G$ is a functor, $R$ acts on $G(R)$ as well.  Furthermore, since $G$ is $k$-linear, this action is compatible with the $k$-vector space structure on $G(R)$ and the $k$-algebra structure on $R$ (that is, elements of $R$ which are in $k$ act by the scalar multiplication on the vector space $G(R)$).  This gives us a $k$-bilinear map $R\times G(R)\to G(R)$, and hence a $k$-linear map $R\otimes_k G(R)\to G(R)$.
Now let $M$ be an arbitrary module.  Note that $M$ can canonically constructed from $R$ by colimits: the canonical presentation of $M$ as an $R$-module expresses $M$ as a cokernel of a map between coproducts of copies of $R$.  Note that both $M\mapsto M\otimes_k G(R)$ and $M\mapsto G(M)$ preserve colimits, and so applying them to these particular colimits which construct $M$ from $R$, we see that our map $R\otimes_k G(R)\to G(R)$ induces a map $M\otimes_k G(R)\to G(M)$.  This is functorial in $M$ because our expression of $M$ as a colimit is functorial in $M$ (if we choose the canonical presentation).
A bit more explicitly, there are functors $U,V:R\text{-mod}\to Set$ and an exact sequence $$\bigoplus_{V(M)} R\to \bigoplus_{U(M)} R\to M\to 0$$ which is natural in $M$ (here $U(M)$ is just the underlying set of $M$, and $V(M)$ is the underlying set of the kernel of the natural map $\bigoplus_{U(M)} R\to M$).  Since $M\mapsto M\otimes_k G(R)$ and $M\mapsto G(M)$ preserve colimits, there are exact sequences $$\bigoplus_{V(M)} R\otimes_k G(R)\to \bigoplus_{U(M)} R\otimes_k G(R)\to M\otimes_k G(R)\to 0$$ and $$\bigoplus_{V(M)} G(R)\to \bigoplus_{U(M)} G(R)\to G(M)\to 0,$$ natural in $M$.  Our map $R\otimes_k G(R)\to G(R)$ gives natural maps between the first two terms of these sequences which make the diagram 
$$\require{AMScd}
\begin{CD}
\bigoplus_{V(M)} R\otimes_k G(R) @>>> \bigoplus_{U(M)} R\otimes_k G(R) @>>> M\otimes_k G(R) @>>> 0\\
@V{}VV @V{}VV \\
\bigoplus_{V(M)} G(R) @>>> \bigoplus_{U(M)} G(R) @>>> G(M) @>>> 0
\end{CD}$$
commute (the fact that the diagram commutes comes from the fact that the map $R\otimes_k G(R)\to G(R)$ is not just $k$-linear but $R$-linear, where $R$ acts on the first coordinate of $R\otimes_k G(R)$ and on $G(R)$ as described above).  By exactness of the rows, there is then a unique map $M\otimes_k G(R)\to G(M)$ which makes the whole diagram commute.
