Let $(C,\otimes , 1^C , \alpha , \lambda , \rho)$ be a monoidal category and let $\mathsf{Mon}_C$ be the category of monoids in $C$. And let $A \in \mathsf{Mon}_C$. Denote the category of $A$-modules by $A-\mathsf{Mod}_C$. Under what conditions does the forgetful functor $U : A-\mathsf{Mod}_C \to C$ have a left adjoint?
I am used to the free $R$-module on a set $S$. But this is different than the above situation because the explicit construction in this case makes reference to properties I do not know the categorical analogue. I have seen the definition of the free $R$-module as $R^{(S)}$, which is all finitely supported set function from $S$ to the underlying set of $R$. I am able to show that this is indeed the free $R$-module on $S$. But this does not generalize since it requires that $R$ have underlying set. It also requires a notion of finitely supported.
My idea to generalize this to a arbitrary monoidal category with monoid $A$ in $C$ is the following: for an object $c \in C$, we want the to speak of some sort of "coproduct of $A$ indexed by c". But I'm not sure how to go about making this precise. Any ideas I have on this require importing lots of extra structure into $C$.
My main question is under what conditions doeu $U$ admit a left adjoint? I am also interested in a construction of this adjoint, in cases where it exists. References are also appreciated.
The forgetful functor $U$ always has a left adjoint, which is given by just $F(X)=A\otimes X$. For any object $X$ of $C$, the object $A\otimes X$ has a canonical $A$-module structure, given by the composition $A\otimes (A\otimes X)\to (A\otimes A)\otimes X\to A\otimes X$ where the first map is the associator and the second is the multiplication of $A$. There is also a canonical $C$-morphism $X\to A\otimes X$ given by the composition $X\to 1\otimes X\to A\otimes X$ where the first map is the unitor and the second is the unit of $A$. This in fact makes $A\otimes X$ the free $A$-module on $X$: if $M$ is any other $A$-module with a $C$-morphism $f:X\to M$, this induces a map $A\otimes X\to M$ given by composing $A\otimes f:A\otimes X\to A\otimes M$ with the module structure map $A\otimes M\to M$. This composition is an $A$-module morphism (by associativity of the module structure on $M$). Moreover, it is the unique $A$-module morphism which factors $f$ through the canonical map $X\to A\otimes X$ (if you apply the functor $A\otimes -$ to that commutative diagram it forces the map $A\otimes X\to M$ to be defined as above by unitality of the $A$-module structure of $M$).